{"id":3104,"date":"2018-07-30T14:30:57","date_gmt":"2018-07-30T14:30:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/?post_type=chapter&#038;p=3104"},"modified":"2018-07-30T14:30:57","modified_gmt":"2018-07-30T14:30:57","slug":"non-right-triangles-law-of-cosines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/chapter\/non-right-triangles-law-of-cosines\/","title":{"raw":"Non-right Triangles: Law of Cosines","rendered":"Non-right Triangles: Law of Cosines"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Use the Law of Cosines to solve oblique triangles.<\/li>\r\n \t<li>Solve applied problems using the Law of Cosines.<\/li>\r\n \t<li>Use Heron\u2019s formula to \ufb01nd the area of a triangle.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137693532\">Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_001\">(Figure)<\/a>. How far from port is the boat?<\/p>\r\n\r\n<div id=\"Figure_08_02_001\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145527\/CNX_Precalc_Figure_08_02_001.jpg\" alt=\"A triangle whose vertices are the boat, the port, and the turning point of the boat. The side between the port and the turning point is 10 mi, and the side between the turning point and the boat is 8 miles. The side between the port and the turning point is extended in a straight dotted line. The angle between the dotted line and the 8 mile side is 20 degrees.\" width=\"487\" height=\"517\" \/> <strong>Figure 1.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165135553590\">Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a <span class=\"no-emphasis\">SAS (side-angle-side) triangle<\/span>, or when all three sides are known, but no angles are known, a <span class=\"no-emphasis\">SSS (side-side-side) triangle<\/span>. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.<\/p>\r\n\r\n<div id=\"fs-id1165135190782\" class=\"bc-section section\">\r\n<h3>Using the Law of Cosines to Solve Oblique Triangles<\/h3>\r\n<p id=\"fs-id1165137437275\">The tool we need to solve the problem of the boat\u2019s distance from the port is the <strong>Law of Cosines<\/strong>, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level.<\/p>\r\n<p id=\"fs-id1165134079626\">Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the <span class=\"no-emphasis\">Pythagorean Theorem<\/span> to non-right triangles. Here is how it works: An arbitrary non-right triangle[latex]\\,ABC\\,[\/latex]is placed in the coordinate plane with vertex[latex]\\,A\\,[\/latex]at the origin, side[latex]\\,c\\,[\/latex]drawn along the <em>x<\/em>-axis, and vertex[latex]\\,C\\,[\/latex]located at some point[latex]\\,\\left(x,y\\right)\\,[\/latex]in the plane, as illustrated in <a class=\"autogenerated-content\" href=\"#Figure_08_02_002\">(Figure)<\/a>. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.<\/p>\r\n\r\n<div id=\"Figure_08_02_002\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145537\/CNX_Precalc_Figure_08_02_002.jpg\" alt=\"A triangle A B C plotted in quadrant 1 of the x,y plane. Angle A is theta degrees with opposite side a, angles B and C, with opposite sides b and c respectively, are unknown. Vertex A is located at the origin (0,0), vertex B is located at some point (x-c, 0) along the x-axis, and point C is located at some point in quadrant 1 at the point (b times the cos of theta, b times the sin of theta).\" width=\"487\" height=\"266\" \/> <strong>Figure 2.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165134363434\">We can drop a perpendicular from[latex]\\,C\\,[\/latex]to the <em>x-<\/em>axis (this is the altitude or height). Recalling the basic <span class=\"no-emphasis\">trigonometric identities<\/span>, we know that<\/p>\r\n\r\n<div class=\"unnumbered\">[latex]\\mathrm{cos}\\,\\theta =\\frac{x\\text{(adjacent)}}{b\\text{(hypotenuse)}}\\text{ and }\\mathrm{sin}\\,\\theta =\\frac{y\\text{(opposite)}}{b\\text{(hypotenuse)}}[\/latex]<\/div>\r\n<p id=\"fs-id1165134243120\">In terms of[latex]\\,\\theta ,\\text{ }x=b\\mathrm{cos}\\,\\theta \\,[\/latex]and[latex]y=b\\mathrm{sin}\\,\\theta .\\text{ }[\/latex]The[latex]\\,\\left(x,y\\right)\\,[\/latex]point located at[latex]\\,C\\,[\/latex]has coordinates[latex]\\,\\left(b\\mathrm{cos}\\,\\theta ,\\,\\,b\\mathrm{sin}\\,\\theta \\right).\\,[\/latex]Using the side[latex]\\,\\left(x-c\\right)\\,[\/latex]as one leg of a right triangle and[latex]\\,y\\,[\/latex]as the second leg, we can find the length of hypotenuse[latex]\\,a\\,[\/latex]using the Pythagorean Theorem. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165135471054\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{llllll} {a}^{2}={\\left(x-c\\right)}^{2}+{y}^{2}\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }={\\left(b\\mathrm{cos}\\,\\theta -c\\right)}^{2}+{\\left(b\\mathrm{sin}\\,\\theta \\right)}^{2}\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Substitute }\\left(b\\mathrm{cos}\\,\\theta \\right)\\text{ for}\\,x\\,\\,\\text{and }\\left(b\\mathrm{sin}\\,\\theta \\right)\\,\\text{for }y.\\hfill \\\\ \\text{ }=\\left({b}^{2}{\\mathrm{cos}}^{2}\\theta -2bc\\mathrm{cos}\\,\\theta +{c}^{2}\\right)+{b}^{2}{\\mathrm{sin}}^{2}\\theta \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Expand the perfect square}.\\hfill \\\\ \\text{ }={b}^{2}{\\mathrm{cos}}^{2}\\theta +{b}^{2}{\\mathrm{sin}}^{2}\\theta +{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Group terms noting that }{\\mathrm{cos}}^{2}\\theta +{\\mathrm{sin}}^{2}\\theta =1.\\hfill \\\\ \\text{ }={b}^{2}\\left({\\mathrm{cos}}^{2}\\theta +{\\mathrm{sin}}^{2}\\theta \\right)+{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Factor out }{b}^{2}.\\hfill \\\\ {a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165135481279\">The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.<\/p>\r\n<p id=\"fs-id1165137539618\">Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Law of Cosines<\/h3>\r\n<p id=\"fs-id1165137428181\">The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in <a class=\"autogenerated-content\" href=\"#Figure_08_02_003\">(Figure)<\/a>, with angles[latex]\\,\\alpha ,\\beta ,[\/latex] and[latex]\\,\\gamma ,[\/latex] and opposite corresponding sides[latex]\\,a,b,[\/latex] and[latex]\\,c,\\,[\/latex]respectively, the Law of Cosines is given as three equations.<\/p>\r\n\r\n<div id=\"fs-id1165137436222\">[latex]\\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\\,\\,\\mathrm{cos}\\,\\alpha \\\\ {b}^{2}={a}^{2}+{c}^{2}-2ac\\,\\,\\mathrm{cos}\\,\\beta \\\\ {c}^{2}={a}^{2}+{b}^{2}-2ab\\,\\,\\mathrm{cos}\\,\\gamma \\end{array}[\/latex]<\/div>\r\n<div id=\"Figure_08_02_003\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145545\/CNX_Precalc_Figure_08_02_003n.jpg\" alt=\"A triangle with standard labels: angles alpha, beta, and gamma with opposite sides a, b, and c respectively.\" width=\"487\" height=\"239\" \/> <strong>Figure 3.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137900234\">To solve for a missing side measurement, the corresponding opposite angle measure is needed.<\/p>\r\n<p id=\"fs-id1165134272589\">When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle.<\/p>\r\n\r\n<div id=\"fs-id1165137715303\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\mathrm{cos}\\text{ }\\alpha =\\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\\hfill \\end{array}\\hfill \\\\ \\mathrm{cos}\\text{ }\\beta =\\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\\hfill \\\\ \\mathrm{cos}\\text{ }\\gamma =\\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"precalculus howto\">\r\n<p id=\"fs-id1165137444540\"><strong>Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165132949849\" type=\"1\">\r\n \t<li>Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.<\/li>\r\n \t<li>Apply the Law of Cosines to find the length of the unknown side or angle.<\/li>\r\n \t<li>Apply the <span class=\"no-emphasis\">Law of Sines<\/span> or Cosines to find the measure of a second angle.<\/li>\r\n \t<li>Compute the measure of the remaining angle.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_08_02_01\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137442107\">\r\n<div id=\"fs-id1165134261612\">\r\n<h3>Finding the Unknown Side and Angles of a SAS Triangle<\/h3>\r\n<p id=\"fs-id1165137723520\">Find the unknown side and angles of the triangle in <a class=\"autogenerated-content\" href=\"#Figure_08_02_004\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"Figure_08_02_004\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145552\/CNX_Precalc_Figure_08_02_004.jpg\" alt=\"A triangle with standard labels. Side a = 10, side c = 12, and angle beta = 30 degrees.\" width=\"487\" height=\"189\" \/> <strong>Figure 4.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137810707\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137810707\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137810707\"]\r\n<p id=\"fs-id1165133192903\">First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.<\/p>\r\n<p id=\"fs-id1165137539491\">Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side[latex]\\,b,\\,[\/latex]as we know the measurement of the opposite angle[latex]\\,\\beta .[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165137768253\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ll}{b}^{2}={a}^{2}+{c}^{2}-2ac\\mathrm{cos}\\,\\beta \\hfill &amp; \\hfill \\\\ {b}^{2}={10}^{2}+{12}^{2}-2\\left(10\\right)\\left(12\\right)\\mathrm{cos}\\left({30}^{\\circ }\\right)\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Substitute the measurements for the known quantities}.\\hfill \\\\ {b}^{2}=100+144-240\\left(\\frac{\\sqrt{3}}{2}\\right)\\hfill &amp; \\text{Evaluate the cosine and begin to simplify}.\\hfill \\\\ {b}^{2}=244-120\\sqrt{3}\\hfill &amp; \\hfill \\\\ \\,\\,\\,b=\\sqrt{244-120\\sqrt{3}}\\hfill &amp; \\,\\text{Use the square root property}.\\hfill \\\\ \\,\\,\\,b\\approx 6.013\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165135542697\">Because we are solving for a length, we use only the positive square root. Now that we know the length[latex]\\,b,\\,[\/latex]we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle[latex]\\,\\alpha ,\\,[\/latex]we have<\/p>\r\n\r\n<div id=\"fs-id1165137574905\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ll}\\frac{\\mathrm{sin}\\,\\alpha }{a}=\\frac{\\mathrm{sin}\\,\\beta }{b}\\hfill &amp; \\hfill \\\\ \\frac{\\mathrm{sin}\\,\\alpha }{10}=\\frac{\\mathrm{sin}\\left(30\u00b0\\right)}{6.013}\\hfill &amp; \\hfill \\\\ \\,\\mathrm{sin}\\,\\alpha =\\frac{10\\mathrm{sin}\\left(30\u00b0\\right)}{6.013}\\hfill &amp; \\text{Multiply both sides of the equation by 10}.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\alpha ={\\mathrm{sin}}^{-1}\\left(\\frac{10\\mathrm{sin}\\left(30\u00b0\\right)}{6.013}\\right)\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Find the inverse sine of }\\frac{10\\mathrm{sin}\\left(30\u00b0\\right)}{6.013}.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\alpha \\approx 56.3\u00b0\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165133202446\">The other possibility for[latex]\\,\\alpha \\,[\/latex]would be[latex]\\,\\alpha =180\u00b0\u201356.3\u00b0\\approx 123.7\u00b0.\\,[\/latex]In the original diagram,[latex]\\,\\alpha \\,[\/latex]is adjacent to the longest side, so[latex]\\,\\alpha \\,[\/latex]is an acute angle and, therefore,[latex]\\,123.7\u00b0\\,[\/latex]does not make sense. Notice that if we choose to apply the <span class=\"no-emphasis\">Law of Cosines<\/span>, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between[latex]\\,0\u00b0\\,[\/latex]and[latex]\\,180\u00b0.\\,[\/latex]Proceeding with[latex]\\,\\alpha \\approx 56.3\u00b0,\\,[\/latex]we can then find the third angle of the triangle.<\/p>\r\n\r\n<div class=\"unnumbered\">[latex]\\gamma =180\u00b0-30\u00b0-56.3\u00b0\\approx 93.7\u00b0[\/latex]<\/div>\r\n<p id=\"fs-id1165137939414\">The complete set of angles and sides is<\/p>\r\n\r\n<div id=\"fs-id1165135263593\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ll}\\alpha \\approx 56.3\u00b0\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; a=10\\hfill \\\\ \\beta =30\u00b0\\hfill &amp; b\\approx 6.013\\hfill \\\\ \\,\\gamma \\approx 93.7\u00b0\\hfill &amp; c=12\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137734372\" class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n<div id=\"ti_08_02_01\">\r\n<div id=\"fs-id1165135553574\">\r\n<p id=\"fs-id1165135553575\">Find the missing side and angles of the given triangle:[latex]\\,\\alpha =30\u00b0,\\,\\,b=12,\\,\\,c=24.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135360387\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135360387\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135360387\"]\r\n<p id=\"fs-id1165135360388\">[latex]a\\approx 14.9,\\,\\,\\beta \\approx 23.8\u00b0,\\,\\,\\gamma \\approx 126.2\u00b0.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_08_02_02\" class=\"textbox examples\">\r\n<div id=\"fs-id1165133134878\">\r\n<div id=\"fs-id1165133134880\">\r\n<h3>Solving for an Angle of a SSS Triangle<\/h3>\r\n<p id=\"fs-id1165137758182\">Find the angle[latex]\\,\\alpha \\,[\/latex]for the given triangle if side[latex]\\,a=20,\\,[\/latex]side[latex]\\,b=25,\\,[\/latex]and side[latex]\\,c=18.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135361166\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135361166\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135361166\"]\r\n<p id=\"fs-id1165135361168\">For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle[latex]\\,\\alpha ,\\,[\/latex]we have<\/p>\r\n\r\n<div id=\"fs-id1165134116894\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{llll}\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\,\\,\\text{ }{a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\alpha \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }{20}^{2}={25}^{2}+{18}^{2}-2\\left(25\\right)\\left(18\\right)\\mathrm{cos}\\,\\alpha \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Substitute the appropriate measurements}.\\hfill \\\\ \\text{ }400=625+324-900\\mathrm{cos}\\,\\alpha \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Simplify in each step}.\\hfill \\\\ \\text{ }400=949-900\\mathrm{cos}\\,\\alpha \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\,\\text{ }-549=-900\\mathrm{cos}\\,\\alpha \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Isolate cos }\\alpha .\\hfill \\\\ \\text{ }\\frac{-549}{-900}=\\mathrm{cos}\\,\\alpha \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\,\\text{ }0.61\\approx \\mathrm{cos}\\,\\alpha \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ {\\mathrm{cos}}^{-1}\\left(0.61\\right)\\approx \\alpha \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Find the inverse cosine}.\\hfill \\\\ \\text{ }\\alpha \\approx 52.4\u00b0\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165133349386\">See <a class=\"autogenerated-content\" href=\"#Figure_08_02_005\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"Figure_08_02_005\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145558\/CNX_Precalc_Figure_08_02_005.jpg\" alt=\"A triangle with standard labels. Side b =25, side a = 20, side c = 18, and angle alpha = 52.4 degrees.\" width=\"487\" height=\"266\" \/> <strong>Figure 5.<\/strong>[\/caption]\r\n\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137807087\">\r\n<h4>Analysis<\/h4>\r\n<p id=\"fs-id1165137605771\">Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137605779\" class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n<div id=\"ti_08_02_02\">\r\n<div id=\"fs-id1165135648695\">\r\n<p id=\"fs-id1165135648696\">Given[latex]\\,a=5,b=7,\\,[\/latex]and[latex]\\,c=10,\\,[\/latex]find the missing angles.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165133210042\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133210042\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165133210042\"]\r\n<p id=\"fs-id1165133210043\">[latex]\\alpha \\approx 27.7\u00b0,\\,\\,\\beta \\approx 40.5\u00b0,\\,\\,\\gamma \\approx 111.8\u00b0[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165133215477\" class=\"bc-section section\">\r\n<h3>Solving Applied Problems Using the Law of Cosines<\/h3>\r\n<p id=\"fs-id1165133215482\">Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.<\/p>\r\n\r\n<div id=\"Example_08_02_03\" class=\"textbox examples\">\r\n<div id=\"fs-id1165133035957\">\r\n<div id=\"fs-id1165133035960\">\r\n<h3>Using the Law of Cosines to Solve a Communication Problem<\/h3>\r\n<p id=\"fs-id1165133035965\">On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137419121\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137419121\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137419121\"]\r\n<p id=\"fs-id1165137419123\">For simplicity, we start by drawing a diagram similar to <a class=\"autogenerated-content\" href=\"#Figure_08_02_006\">(Figure)<\/a> and labeling our given information.<\/p>\r\n\r\n<div id=\"Figure_08_02_006\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145605\/CNX_Precalc_Figure_08_02_006.jpg\" alt=\"A triangle formed between the two cell phone towers located on am east to west highway and the cellphone between and north of them. The side between the two towers is 6000 feet, the side between the left tower and the phone is 5050 feet, and the side between the right tower and the phone is 2420 feet. The angle between the 5050 and 6000 feet sides is labeled theta.\" width=\"487\" height=\"199\" \/> <strong>Figure 6.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165135428314\">Using the Law of Cosines, we can solve for the angle[latex]\\,\\theta .\\,[\/latex]Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let[latex]\\,a=2420,b=5050,\\,[\/latex]and[latex]\\,c=6000.\\,[\/latex]Thus,[latex]\\,\\theta \\,[\/latex]corresponds to the opposite side[latex]\\,a=2420.\\,[\/latex]<\/p>\r\n\r\n<div class=\"unnumbered\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\,\\text{ }{a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill \\end{array}\\hfill \\\\ \\text{ }{\\left(2420\\right)}^{2}={\\left(5050\\right)}^{2}+{\\left(6000\\right)}^{2}-2\\left(5050\\right)\\left(6000\\right)\\mathrm{cos}\\,\\theta \\hfill \\\\ \\,\\,\\,\\,\\,\\,{\\left(2420\\right)}^{2}-{\\left(5050\\right)}^{2}-{\\left(6000\\right)}^{2}=-2\\left(5050\\right)\\left(6000\\right)\\mathrm{cos}\\,\\theta \\hfill \\\\ \\text{ }\\frac{{\\left(2420\\right)}^{2}-{\\left(5050\\right)}^{2}-{\\left(6000\\right)}^{2}}{-2\\left(5050\\right)\\left(6000\\right)}=\\mathrm{cos}\\,\\theta \\hfill \\\\ \\text{ }\\mathrm{cos}\\,\\theta \\approx 0.9183\\hfill \\\\ \\text{ }\\theta \\approx {\\mathrm{cos}}^{-1}\\left(0.9183\\right)\\hfill \\\\ \\text{ }\\theta \\approx 23.3\u00b0\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165133178309\">To answer the questions about the phone\u2019s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in <a class=\"autogenerated-content\" href=\"#Figure_08_02_007\">(Figure)<\/a>. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.<\/p>\r\n\r\n<div id=\"Figure_08_02_007\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145616\/CNX_Precalc_Figure_08_02_007.jpg\" alt=\"The triangle between the phone, the left tower, and a point between the phone and the highway between the towers. The side between the phone and the highway is perpendicular to the highway and is y feet. The highway side is x feet. The angle at the tower, previously labeled theta, is 23.3 degrees.\" width=\"487\" height=\"177\" \/> <strong>Figure 7.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165135317539\">Using the angle[latex]\\,\\theta =23.3\u00b0\\,[\/latex]and the basic trigonometric identities, we can find the solutions. Thus<\/p>\r\n\r\n<div id=\"fs-id1165135407362\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\mathrm{cos}\\left(23.3\u00b0\\right)=\\frac{x}{5050}\\hfill \\end{array}\\hfill \\\\ \\text{ }x=5050\\mathrm{cos}\\left(23.3\u00b0\\right)\\hfill \\\\ \\text{ }x\\approx 4638.15\\,\\text{feet}\\hfill \\\\ \\text{ }\\mathrm{sin}\\left(23.3\u00b0\\right)=\\frac{y}{5050}\\hfill \\\\ \\text{ }y=5050\\mathrm{sin}\\left(23.3\u00b0\\right)\\hfill \\\\ \\text{ }y\\approx 1997.5\\,\\text{feet}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165135546940\">The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_08_02_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1165134113730\">\r\n<div id=\"fs-id1165134113733\">\r\n<h3>Calculating Distance Traveled Using a SAS Triangle<\/h3>\r\n<p id=\"fs-id1165134113738\">Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in <a class=\"autogenerated-content\" href=\"#Figure_08_02_009\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"Figure_08_02_009\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145627\/CNX_Precalc_Figure_08_02_009.jpg\" alt=\"A triangle whose vertices are the boat, the port, and the turning point of the boat. The side between the port and the turning point is 10 mi, and the side between the turning point and the boat is 8 miles. The side between the port and the turning point is extended in a straight dotted line. The angle between the dotted line and the 8 mile side is 20 degrees.\" width=\"487\" height=\"517\" \/> <strong>Figure 8.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134282021\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134282021\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134282021\"]\r\n<p id=\"fs-id1165134282023\">The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle,[latex]180\u00b0-20\u00b0=160\u00b0.\\,[\/latex]With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle\u2014the distance of the boat to the port.<\/p>\r\n\r\n<div id=\"fs-id1165135503577\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\,{x}^{2}={8}^{2}+{10}^{2}-2\\left(8\\right)\\left(10\\right)\\mathrm{cos}\\left(160\u00b0\\right)\\hfill \\\\ \\,{x}^{2}=314.35\\hfill \\\\ \\,\\,\\,\\,x=\\sqrt{314.35}\\hfill \\\\ \\,\\,\\,\\,x\\approx 17.7\\,\\text{miles}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165134393824\">The boat is about 17.7 miles from port.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bc-section section\">\r\n<h3>Using Heron\u2019s Formula to Find the Area of a Triangle<\/h3>\r\n<p id=\"fs-id1165135650692\">We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use <span class=\"no-emphasis\">Heron\u2019s formula<\/span> instead of finding the height. <span class=\"no-emphasis\">Heron of Alexandria<\/span> was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.<\/p>\r\n\r\n<div id=\"fs-id1165133306825\" class=\"textbox key-takeaways\">\r\n<h3>Heron\u2019s Formula<\/h3>\r\n<p id=\"fs-id1165137502458\">Heron\u2019s formula finds the area of oblique triangles in which sides[latex]\\,a,b\\text{,}[\/latex]and[latex]\\,c\\,[\/latex]are known.<\/p>\r\n\r\n<div>[latex]\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex]<\/div>\r\n<p id=\"fs-id1165134258499\">where[latex]\\,s=\\frac{\\left(a+b+c\\right)}{2}\\,[\/latex] is one half of the perimeter of the triangle, sometimes called the semi-perimeter.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_08_02_05\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135704848\">\r\n<div id=\"fs-id1165137723217\">\r\n<h3>Using Heron\u2019s Formula to Find the Area of a Given Triangle<\/h3>\r\n<p id=\"fs-id1165137723224\">Find the area of the triangle in <a class=\"autogenerated-content\" href=\"#Figure_08_02_010\">(Figure)<\/a> using Heron\u2019s formula.<\/p>\r\n\r\n<div id=\"Figure_08_02_010\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145634\/CNX_Precalc_Figure_08_02_010.jpg\" alt=\"A triangle with angles A, B, and C and opposite sides a, b, and c, respectively. Side a = 10, side b - 15, and side c = 7.\" width=\"487\" height=\"134\" \/> <strong>Figure 9.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"611728\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"611728\"]\r\n<p id=\"fs-id1165134314759\">First, we calculate[latex]\\,s.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165135408511\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ s=\\frac{\\left(a+b+c\\right)}{2}\\end{array}\\hfill \\\\ s=\\frac{\\left(10+15+7\\right)}{2}=16\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165134170047\">Then we apply the formula.<\/p>\r\n\r\n<div class=\"unnumbered\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ \\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}\\end{array}\\hfill \\\\ \\text{Area}=\\sqrt{16\\left(16-10\\right)\\left(16-15\\right)\\left(16-7\\right)}\\hfill \\\\ \\text{Area}\\approx 29.4\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165134373529\">The area is approximately 29.4 square units.<\/p>\r\n\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134547309\" class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n<div id=\"ti_08_02_04\">\r\n<div id=\"fs-id1165135300762\">\r\n<p id=\"fs-id1165135300763\">Use Heron\u2019s formula to find the area of a triangle with sides of lengths[latex]\\,a=29.7\\,\\text{ft},b=42.3\\,\\text{ft},\\,[\/latex]and[latex]\\,c=38.4\\,\\text{ft}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134102074\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134102074\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134102074\"]\r\n<p id=\"fs-id1165134102075\">Area = 552 square feet<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_08_02_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135347549\">\r\n<div id=\"fs-id1165137769676\">\r\n<h3>Applying Heron\u2019s Formula to a Real-World Problem<\/h3>\r\n<p id=\"fs-id1165137769681\">A Chicago city developer wants to construct a building consisting of artist\u2019s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See <a class=\"autogenerated-content\" href=\"#Figure_08_02_011\">(Figure)<\/a> for a view of the city property.<\/p>\r\n\r\n<div id=\"Figure_08_02_011\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145640\/CNX_Precalc_Figure_08_02_011.jpg\" alt=\"A triangle formed by sides Rush Street, N. Wabash Ave, and E. Pearson Street with lengths 62.4, 43.5, and 34.1, respectively.\" width=\"487\" height=\"520\" \/> <strong>Figure 10.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135518137\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135518137\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135518137\"]\r\n<p id=\"fs-id1165135518140\">Find the measurement for[latex]\\,s,\\,[\/latex]which is one-half of the perimeter.<\/p>\r\n\r\n<div class=\"unnumbered\">[latex]\\begin{array}{l}s=\\frac{\\left(62.4+43.5+34.1\\right)}{2}\\hfill \\\\ s=70\\,\\text{m}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137619098\">Apply Heron\u2019s formula.<\/p>\r\n\r\n<div id=\"fs-id1165137619101\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\text{Area}=\\sqrt{70\\left(70-62.4\\right)\\left(70-43.5\\right)\\left(70-34.1\\right)}\\hfill \\\\ \\text{Area}=\\sqrt{506,118.2}\\hfill \\\\ \\text{Area}\\approx 711.4\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165135497734\">The developer has about 711.4 square meters.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n<div id=\"ti_08_02_05\">\r\n<div id=\"fs-id1165135689456\">\r\n<p id=\"fs-id1165135689457\">Find the area of a triangle given[latex]\\,a=4.38\\,\\text{ft}\\,,b=3.79\\,\\text{ft,}\\,[\/latex]and[latex]\\,c=5.22\\,\\text{ft}\\text{.}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137394207\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137394207\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137394207\"]\r\n<p id=\"fs-id1165137394208\">about 8.15 square feet<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135532502\" class=\"precalculus media\">\r\n<p id=\"fs-id1165135532508\">Access these online resources for additional instruction and practice with the Law of Cosines.<\/p>\r\n\r\n<ul id=\"eip-id1165135445621\">\r\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/lawcosines\">Law of Cosines<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/cosineapp\">Law of Cosines: Applications<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/cosineapp2\">Law of Cosines: Applications 2<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135349445\" class=\"key-equations\">\r\n<h3>Key Equations<\/h3>\r\n<table id=\"eip-id1956425\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Law of Cosines<\/td>\r\n<td>[latex]\\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\alpha \\hfill \\\\ {b}^{2}={a}^{2}+{c}^{2}-2ac\\mathrm{cos}\\,\\beta \\hfill \\\\ {c}^{2}={a}^{2}+{b}^{2}-2abcos\\,\\gamma \\hfill \\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Heron\u2019s formula<\/td>\r\n<td>[latex]\\begin{array}{l}\\text{ Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}\\hfill \\\\ \\text{where }s=\\frac{\\left(a+b+c\\right)}{2}\\hfill \\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1165135442379\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1165135442385\">\r\n \t<li>The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.<\/li>\r\n \t<li>The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See <a class=\"autogenerated-content\" href=\"#Example_08_02_01\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_08_02_02\">(Figure)<\/a>.<\/li>\r\n \t<li>The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See <a class=\"autogenerated-content\" href=\"#Example_08_02_03\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_08_02_04\">(Figure)<\/a>.<\/li>\r\n \t<li>Heron\u2019s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron\u2019s formula. See <a class=\"autogenerated-content\" href=\"#Example_08_02_05\">(Figure)<\/a> and See <a class=\"autogenerated-content\" href=\"#Example_08_02_06\">(Figure)<\/a>.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165134212020\" class=\"textbox exercises\">\r\n<h3>Section Exercises<\/h3>\r\n<div class=\"bc-section section\">\r\n<h4>Verbal<\/h4>\r\n<div id=\"fs-id1165134212028\">\r\n<div id=\"fs-id1165137887389\">\r\n<p id=\"fs-id1165137887391\">If you are looking for a missing side of a triangle, what do you need to know when using the Law of Cosines?<\/p>\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1165137887398\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137887398\"]\r\n<p id=\"fs-id1165137887398\">two sides and the angle opposite the missing side.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135430914\">\r\n<div id=\"fs-id1165135430916\">\r\n<p id=\"fs-id1165135430919\">If you are looking for a missing angle of a triangle, what do you need to know when using the Law of Cosines?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134073090\">\r\n<div id=\"fs-id1165134073092\">\r\n<p id=\"fs-id1165134073095\">Explain what[latex]\\,s\\,[\/latex]represents in Heron\u2019s formula.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165133143109\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133143109\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165133143109\"]\r\n<p id=\"fs-id1165133143111\">[latex]\\,s\\,[\/latex]is the semi-perimeter, which is half the perimeter of the triangle.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134383705\">\r\n<div id=\"fs-id1165134383707\">\r\n<p id=\"fs-id1165134383710\">Explain the relationship between the Pythagorean Theorem and the Law of Cosines.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134383715\">\r\n<div id=\"fs-id1165134383717\">\r\n\r\nWhen must you use the Law of Cosines instead of the Pythagorean Theorem?\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135518102\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135518102\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135518102\"]\r\n<p id=\"fs-id1165135518104\">The Law of Cosines must be used for any oblique (non-right) triangle.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135518110\" class=\"bc-section section\">\r\n<h4>Algebraic<\/h4>\r\n<p id=\"fs-id1165137828289\">For the following exercises, assume[latex]\\,\\alpha \\,[\/latex]is opposite side[latex]\\,a,\\beta \\,[\/latex] is opposite side[latex]\\,b,\\,[\/latex]and[latex]\\,\\gamma \\,[\/latex] is opposite side[latex]\\,c.\\,[\/latex]If possible, solve each triangle for the unknown side. Round to the nearest tenth.<\/p>\r\n\r\n<div id=\"fs-id1165135426388\">\r\n<div id=\"fs-id1165135426391\">\r\n<p id=\"fs-id1165135452292\">[latex]\\gamma =41.2\u00b0,a=2.49,b=3.13[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134278648\">\r\n<div id=\"fs-id1165134278650\">\r\n<p id=\"fs-id1165134278652\">[latex]\\alpha =120\u00b0,b=6,c=7[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165133359517\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133359517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165133359517\"]11.3\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165133359525\">\r\n<div id=\"fs-id1165133359527\">[latex]\\beta =58.7\u00b0,a=10.6,c=15.7[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134272650\">\r\n<div id=\"fs-id1165133454475\">\r\n<p id=\"fs-id1165133454477\">[latex]\\gamma =115\u00b0,a=18,b=23[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134574953\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134574953\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134574953\"]\r\n<p id=\"fs-id1165132951595\">34.7<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165132951600\">\r\n<div id=\"fs-id1165132951602\">\r\n<p id=\"fs-id1165132951604\">[latex]\\alpha =119\u00b0,a=26,b=14[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134071612\">\r\n<div id=\"fs-id1165134071614\">\r\n<p id=\"fs-id1165132914223\">[latex]\\gamma =113\u00b0,b=10,c=32[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135354970\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135354970\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135354970\"]\r\n<p id=\"fs-id1165135329809\">26.7<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135329815\">\r\n<div id=\"fs-id1165135329817\">\r\n<p id=\"fs-id1165135329819\">[latex]\\beta =67\u00b0,a=49,b=38[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165131940267\">\r\n<div id=\"fs-id1165131940269\">\r\n<p id=\"fs-id1165131940271\">[latex]\\alpha =43.1\u00b0,a=184.2,b=242.8[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137938467\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137938467\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137938467\"]\r\n<p id=\"fs-id1165137938469\">257.4<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135538836\">\r\n<div id=\"fs-id1165135538838\">\r\n<p id=\"fs-id1165135538840\">[latex]\\alpha =36.6\u00b0,a=186.2,b=242.2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165132945436\">\r\n<div id=\"fs-id1165132945439\">\r\n<p id=\"fs-id1165132945441\">[latex]\\beta =50\u00b0,a=105,b=45{}_{}{}^{}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135515881\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135515881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135515881\"]\r\n<p id=\"fs-id1165135354976\">not possible<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165135354982\">For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth.<\/p>\r\n\r\n<div id=\"fs-id1165135354986\">\r\n<div id=\"fs-id1165135354988\">\r\n<p id=\"fs-id1165133306919\">[latex]\\,a=42,b=19,c=30;\\,[\/latex]find angle[latex]\\,A.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div id=\"fs-id1165135550002\">\r\n<p id=\"fs-id1165135550004\">[latex]\\,a=14,\\text{ }b=13,\\text{ }c=20;\\,[\/latex]find angle[latex]\\,C.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135208492\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135208492\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135208492\"]\r\n<p id=\"fs-id1165135208494\">95.5\u00b0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135208499\">\r\n<div id=\"fs-id1165135208501\">\r\n<p id=\"fs-id1165135208504\">[latex]\\,a=16,b=31,c=20;\\,[\/latex]find angle[latex]\\,B.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135381275\">\r\n<div id=\"fs-id1165135381277\">\r\n<p id=\"fs-id1165135154582\">[latex]\\,a=13,\\,b=22,\\,c=28;\\,[\/latex]find angle[latex]\\,A.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137899963\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137899963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137899963\"]\r\n<p id=\"fs-id1165135697813\">26.9\u00b0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134378611\">\r\n<div id=\"fs-id1165134378613\">\r\n\r\n[latex]a=108,\\,b=132,\\,c=160;\\,[\/latex]find angle[latex]\\,C.\\,[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165135707284\">For the following exercises, solve the triangle. Round to the nearest tenth.<\/p>\r\n\r\n<div id=\"fs-id1165135707288\">\r\n<div id=\"fs-id1165135707290\">\r\n<p id=\"fs-id1165135707292\">[latex]A=35\u00b0,b=8,c=11[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165133199358\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133199358\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165133199358\"]\r\n<p id=\"fs-id1165133199360\">[latex]B\\approx 45.9\u00b0,C\\approx 99.1\u00b0,a\\approx 6.4[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134257565\">\r\n<div id=\"fs-id1165134257567\">\r\n<p id=\"fs-id1165135341174\">[latex]B=88\u00b0,a=4.4,c=5.2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134156056\">\r\n<div id=\"fs-id1165134156059\">\r\n<p id=\"fs-id1165134156061\">[latex]C=121\u00b0,a=21,b=37[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137765785\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137765785\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137765785\"]\r\n<p id=\"fs-id1165137765788\">[latex]A\\approx 20.6\u00b0,B\\approx 38.4\u00b0,c\\approx 51.1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135246610\">\r\n<div id=\"fs-id1165135246612\">[latex]a=13,b=11,c=15[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134149779\">\r\n<div>\r\n<p id=\"fs-id1165134149784\">[latex]a=3.1,b=3.5,c=5[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137643163\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137643163\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137643163\"]\r\n<p id=\"fs-id1165133307603\">[latex]A\\approx 37.8\u00b0,B\\approx 43.8,C\\approx 98.4\u00b0[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137931166\">\r\n<div id=\"fs-id1165137931168\">\r\n<p id=\"fs-id1165137931170\">[latex]a=51,b=25,c=29[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165135442502\">For the following exercises, use Heron\u2019s formula to find the area of the triangle. Round to the nearest hundredth.<\/p>\r\n\r\n<div id=\"fs-id1165135442507\">\r\n<div id=\"fs-id1165135442509\">\r\n<p id=\"fs-id1165135442511\">Find the area of a triangle with sides of length 18 in, 21 in, and 32 in. Round to the nearest tenth.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134201647\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134201647\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134201647\"]\r\n<p id=\"fs-id1165134201649\">177.56 in2<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134201654\">\r\n<div id=\"fs-id1165134201656\">\r\n<p id=\"fs-id1165134201658\">Find the area of a triangle with sides of length 20 cm, 26 cm, and 37 cm. Round to the nearest tenth.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137770136\">\r\n<div id=\"fs-id1165137770138\">\r\n<p id=\"fs-id1165137770140\">[latex]a=\\frac{1}{2}\\,\\text{m},b=\\frac{1}{3}\\,\\text{m},c=\\frac{1}{4}\\,\\text{m}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134328321\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134328321\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134328321\"]\r\n<p id=\"fs-id1165134328323\">0.04 m2<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165131894561\">\r\n<div id=\"fs-id1165131894564\">\r\n<p id=\"fs-id1165131894566\">[latex]a=12.4\\text{ ft},\\text{ }b=13.7\\text{ ft},\\text{ }c=20.2\\text{ ft}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165133023583\">\r\n<div id=\"fs-id1165133023585\">\r\n<p id=\"fs-id1165133023587\">[latex]a=1.6\\text{ yd},\\text{ }b=2.6\\text{ yd},\\text{ }c=4.1\\text{ yd}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165133365504\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133365504\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165133365504\"]\r\n<p id=\"fs-id1165133365506\">0.91 yd2<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165131880282\" class=\"bc-section section\">\r\n<h4>Graphical<\/h4>\r\n<p id=\"fs-id1165131880287\">For the following exercises, find the length of side [latex]x.[\/latex] Round to the nearest tenth.<\/p>\r\n\r\n<div id=\"fs-id1165135600236\">\r\n<div id=\"fs-id1165135600238\"><span id=\"fs-id1165135600244\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145646\/CNX_Precalc_Figure_08_02_201.jpg\" alt=\"A triangle. One angle is 72 degrees, with opposite side = x. The other two sides are 5 and 6.5.\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135442441\">\r\n<div id=\"fs-id1165135442443\"><span id=\"fs-id1165135505452\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145648\/CNX_Precalc_Figure_08_02_202.jpg\" alt=\"A triangle. One angle is 42 degrees with opposite side = x. The other two sides are 4.5 and 3.4.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165135505463\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135505463\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135505463\"]\r\n<p id=\"fs-id1165135505465\">3.0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134272137\">\r\n<div id=\"fs-id1165134272139\"><span id=\"fs-id1165134272145\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145654\/CNX_Precalc_Figure_08_02_203.jpg\" alt=\"A triangle. One angle is 40 degrees with opposite side = 15. The other two sides are 12 and x.\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137450863\">\r\n<div id=\"fs-id1165137450865\"><span id=\"fs-id1165133249121\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145656\/CNX_Precalc_Figure_08_02_204.jpg\" alt=\"A triangle. One angle is 65 degrees with opposite side = x. The other two sides are 30 and 23.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165133249132\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133249132\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165133249132\"]\r\n<p id=\"fs-id1165133249134\">29.1<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135483114\">\r\n<div id=\"fs-id1165135483116\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145706\/CNX_Precalc_Figure_08_02_205.jpg\" alt=\"A triangle. One angle is 50 degrees with opposite side = x. The other two sides are 225 and 305.\" \/><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135556651\">\r\n<div id=\"fs-id1165135556653\"><span id=\"fs-id1165134370014\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145716\/CNX_Precalc_Figure_08_02_206.jpg\" alt=\"A triangle. One angle is 123 degrees with opposite side = x. The other two sides are 1\/5 and 1\/3.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165134370025\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134370025\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134370025\"]\r\n<p id=\"fs-id1165133306900\">0.5<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165133306905\">For the following exercises, find the measurement of angle[latex]\\,A.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165135502926\">\r\n<div id=\"fs-id1165135502928\"><span id=\"fs-id1165135502934\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145723\/CNX_Precalc_Figure_08_02_207.jpg\" alt=\"A triangle. Angle A is opposite a side of length 2.3. The other two sides are 1.5 and 2.5.\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137696000\">\r\n<div id=\"fs-id1165137696003\"><span id=\"fs-id1165134323614\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145725\/CNX_Precalc_Figure_08_02_208.jpg\" alt=\"A triangle. Angle A is opposite a side of length 125. The other two sides are 115 and 100.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165135545523\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135545523\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135545523\"]\r\n<p id=\"fs-id1165135545525\">70.7\u00b0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135545530\">\r\n<div id=\"fs-id1165135545532\"><span id=\"fs-id1165134384434\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145728\/CNX_Precalc_Figure_08_02_209.jpg\" alt=\"A triangle. Angle A is opposite a side of length 6.8. The other two sides are 4.3 and 8.2.\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134384446\">\r\n<div id=\"fs-id1165134068908\"><span id=\"fs-id1165134068914\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145735\/CNX_Precalc_Figure_08_02_210.jpg\" alt=\"A triangle. Angle A is opposite a side of length 40.6. The other two sides are 38.7 and 23.3.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165134401621\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134401621\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134401621\"]\r\n<p id=\"fs-id1165134401623\">77.4\u00b0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134401629\">\r\n<div id=\"fs-id1165134401631\">\r\n\r\nFind the measure of each angle in the triangle shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_211\">(Figure)<\/a>. Round to the nearest tenth.\r\n<div id=\"Figure_08_02_211\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145737\/CNX_Precalc_Figure_08_02_211.jpg\" alt=\"A triangle A B C. Angle A is opposite a side of length 10, angle B is opposite a side of length 12, and angle C is opposite a side of length 7.\" width=\"487\" height=\"171\" \/> <strong>Figure 11.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"eip-483\">For the following exercises, solve for the unknown side. Round to the nearest tenth.<\/p>\r\n\r\n<div id=\"fs-id1165137731858\">\r\n<div id=\"fs-id1165137731860\"><span id=\"fs-id1165133236441\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145743\/CNX_Precalc_Figure_08_02_212.jpg\" alt=\"A triangle. One angle is 60 degrees with opposite side unknown. The other two sides are 20 and 28.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165133236451\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133236451\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165133236451\"]\r\n<p id=\"fs-id1165135400873\">25.0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135400878\">\r\n<div id=\"fs-id1165135400880\"><span id=\"fs-id1165135400887\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145751\/CNX_Precalc_Figure_08_02_213.jpg\" alt=\"A triangle. One angle is 30 degrees with opposite side unknown. The other two sides are 16 and 10.\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135186079\">\r\n<div id=\"fs-id1165135186081\"><span id=\"fs-id1165134087532\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145816\/CNX_Precalc_Figure_08_02_214.jpg\" alt=\"A triangle. One angle is 22 degrees with opposite side unknown. The other two sides are 20 and 13.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165135698477\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135698477\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135698477\"]\r\n<p id=\"fs-id1165135698480\">9.3<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135698484\">\r\n<div id=\"fs-id1165135698486\"><span id=\"fs-id1165135385603\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145825\/CNX_Precalc_Figure_08_02_215.jpg\" alt=\"A triangle. One angle is 88 degrees with opposite side = 9. Another side is 5.\" \/><\/span><\/div>\r\n<\/div>\r\nFor the following exercises, find the area of the triangle. Round to the nearest hundredth.\r\n<div id=\"fs-id1165135385615\">\r\n<div id=\"fs-id1165135385617\"><span id=\"fs-id1165133088150\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145827\/CNX_Precalc_Figure_08_02_216.jpg\" alt=\"A triangle with sides 8, 12, and 17. Angles unknown.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165135525901\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135525901\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135525901\"]\r\n<p id=\"fs-id1165135525903\">43.52<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div id=\"fs-id1165135525911\"><span id=\"fs-id1165135512604\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145833\/CNX_Precalc_Figure_08_02_217.jpg\" alt=\"A triangle with sides 50, 22, and 36. Angles unknown.\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137507971\">\r\n<div id=\"fs-id1165137507973\"><span id=\"fs-id1165137507980\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145840\/CNX_Precalc_Figure_08_02_218.jpg\" alt=\"A triangle with sides 1.9, 2.6, and 4.3. Angles unknown.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165134239622\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134239622\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134239622\"]\r\n<p id=\"fs-id1165134239625\">1.41<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134239629\">\r\n<div><span id=\"fs-id1165135149921\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145851\/CNX_Precalc_Figure_08_02_219.jpg\" alt=\"A triangle with sides 8.9, 12.5, and 16.2. Angles unknown.\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135450305\">\r\n<div id=\"fs-id1165135450307\"><span id=\"fs-id1165135450313\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145859\/CNX_Precalc_Figure_08_02_220.jpg\" alt=\"A triangle with sides 1\/2, 2\/3, and 3\/5. Angles unknown.\" \/><\/span><\/div>\r\n<div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1165134263937\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134263937\"]\r\n<p id=\"fs-id1165134263937\">0.14<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134391071\" class=\"bc-section section\">\r\n<h4>Extensions<\/h4>\r\n<div id=\"fs-id1165134391076\">\r\n<div>\r\n\r\nA parallelogram has sides of length 16 units and 10 units. The shorter diagonal is 12 units. Find the measure of the longer diagonal.\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165133091581\">\r\n<div id=\"fs-id1165133091584\">\r\n<p id=\"fs-id1165133091586\">The sides of a parallelogram are 11 feet and 17 feet. The longer diagonal is 22 feet. Find the length of the shorter diagonal.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165133091591\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165133091591\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165133091591\"]\r\n<p id=\"fs-id1165133091593\">18.3<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135666704\">\r\n<div id=\"fs-id1165135666707\">\r\n<p id=\"fs-id1165135666709\">The sides of a parallelogram are 28 centimeters and 40 centimeters. The measure of the larger angle is 100\u00b0. Find the length of the shorter diagonal.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134383775\">\r\n<div id=\"fs-id1165134383777\">\r\n<p id=\"fs-id1165134383779\">A regular octagon is inscribed in a circle with a radius of 8 inches. (See <a class=\"autogenerated-content\" href=\"#Figure_08_02_221\">(Figure)<\/a>.) Find the perimeter of the octagon.<\/p>\r\n\r\n<div id=\"Figure_08_02_221\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145905\/CNX_Precalc_Figure_08_02_221.jpg\" alt=\"An octagon inscribed in a circle.\" width=\"487\" height=\"197\" \/> <strong>Figure 12.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135697763\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135697763\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135697763\"]\r\n<p id=\"fs-id1165135697765\">48.98<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134331096\">\r\n<div id=\"fs-id1165134331098\">\r\n\r\nA regular pentagon is inscribed in a circle of radius 12 cm. (See <a class=\"autogenerated-content\" href=\"#Figure_08_02_222\">(Figure)<\/a>.) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter.\r\n<div id=\"Figure_08_02_222\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145914\/CNX_Precalc_Figure_08_02_222.jpg\" alt=\"A pentagon inscribed in a circle.\" width=\"487\" height=\"211\" \/> <strong>Figure 13.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165133017581\">For the following exercises, suppose that[latex]\\,{x}^{2}=25+36-60\\mathrm{cos}\\left(52\\right)\\,[\/latex]represents the relationship of three sides of a triangle and the cosine of an angle.<\/p>\r\n\r\n<div id=\"fs-id1165131837012\">\r\n<div id=\"fs-id1165131837014\">\r\n<p id=\"fs-id1165131837016\">Draw the triangle.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165131837019\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165131837019\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165131837019\"]<span id=\"fs-id1165135339571\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145916\/CNX_Precalc_Figure_08_02_223.jpg\" alt=\"A triangle. One angle is 52 degrees with opposite side = x. The other two sides are 5 and 6.\" \/><\/span>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135339583\">\r\n<div id=\"fs-id1165135388758\">\r\n<p id=\"fs-id1165135388760\">Find the length of the third side.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nFor the following exercises, find the area of the triangle.\r\n<div id=\"fs-id1165135388769\">\r\n<div id=\"fs-id1165135388771\"><span id=\"fs-id1165135390787\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145918\/CNX_Precalc_Figure_08_02_224.jpg\" alt=\"A triangle. One angle is 22 degrees with opposite side = 3.4. Another side is 5.3.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165134130014\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134130014\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134130014\"]\r\n<p id=\"fs-id1165134130016\">7.62<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134130020\">\r\n<div id=\"fs-id1165134130023\"><span id=\"fs-id1165134260427\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145920\/CNX_Precalc_Figure_08_02_225.jpg\" alt=\"A triangle. One angle is 80 degrees with opposite side unknown. The other two sides are 8 and 6.\" \/><\/span><\/div>\r\n<\/div>\r\n<div>\r\n<div><span id=\"fs-id1165135186219\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145933\/CNX_Precalc_Figure_08_02_226.jpg\" alt=\"A triangle. One angle is 18 degrees with opposite side = 12.8. Another side is 18.8.\" \/><\/span><\/div>\r\n<div id=\"fs-id1165135633910\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135633910\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135633910\"]85.1\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135633919\" class=\"bc-section section\">\r\n<h4>Real-World Applications<\/h4>\r\n<div id=\"fs-id1165132957157\">\r\n<div id=\"fs-id1165132957158\">\r\n<p id=\"fs-id1165132957160\">A surveyor has taken the measurements shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_227\">(Figure)<\/a>. Find the distance across the lake. Round answers to the nearest tenth.<\/p>\r\n\r\n<div id=\"Figure_08_02_227\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145952\/CNX_Precalc_Figure_08_02_227.jpg\" alt=\"A triangle. One angle is 70 degrees with opposite side unknown, which is the length of the lake. The other two sides are 800 and 900 feet.\" width=\"487\" height=\"262\" \/> <strong>Figure 14.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134540068\">\r\n<div id=\"fs-id1165134540070\">\r\n<p id=\"fs-id1165134540073\">A satellite calculates the distances and angle shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_228\">(Figure)<\/a> (not to scale). Find the distance between the two cities. Round answers to the nearest tenth.<\/p>\r\n\r\n<div id=\"Figure_08_02_228\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145954\/CNX_Precalc_Figure_08_02_228.jpg\" alt=\"Insert figure(table) alt text: A triangle formed by two cities on the ground and a satellite above them. The angle by the satellite is 2.1 degrees with opposite side unknown, which is the distance between the two cities. The lengths of the other sides are 370 and 350 km.\" width=\"488\" height=\"307\" \/> <strong>Figure 15.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137827744\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137827744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137827744\"]\r\n<p id=\"fs-id1165137827746\">24.0 km<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137827752\">\r\n<div id=\"fs-id1165137827754\">\r\n<p id=\"fs-id1165137827756\">An airplane flies 220 miles with a heading of 40\u00b0, and then flies 180 miles with a heading of 170\u00b0. How far is the plane from its starting point, and at what heading? Round answers to the nearest tenth.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135181803\">\r\n<div id=\"fs-id1165135181805\">\r\n\r\nA 113-foot tower is located on a hill that is inclined 34\u00b0 to the horizontal, as shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_230\">(Figure)<\/a>. A guy-wire is to be attached to the top of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed.\r\n<div id=\"Figure_08_02_230\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19150005\/CNX_Precalc_Figure_08_02_230.jpg\" alt=\"Insert figure(table) alt text: Two triangles, one on top of the other. The bottom triangle is the hill inclined 34 degrees to the horizontal. The second is formed by the base of the tower on the incline of the hill, the top of the tower, and the wire anchor point uphill from the tower on the incline. The sides are the tower, the incline of the hill, and the wire. The tower side is 113 feet and the incline side is 98 feet.\" width=\"487\" height=\"267\" \/> <strong>Figure 16.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135500669\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135500669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135500669\"]\r\n<p id=\"fs-id1165135500671\">99.9 ft<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135500676\">\r\n<div id=\"fs-id1165135500678\">\r\n<p id=\"fs-id1165137921478\">Two ships left a port at the same time. One ship traveled at a speed of 18 miles per hour at a heading of 320\u00b0. The other ship traveled at a speed of 22 miles per hour at a heading of 194\u00b0. Find the distance between the two ships after 10 hours of travel.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137921488\">\r\n<div id=\"fs-id1165137921490\">\r\n<p id=\"fs-id1165135510887\">The graph in <a class=\"autogenerated-content\" href=\"#Figure_08_02_231\">(Figure)<\/a> represents two boats departing at the same time from the same dock. The first boat is traveling at 18 miles per hour at a heading of 327\u00b0 and the second boat is traveling at 4 miles per hour at a heading of 60\u00b0. Find the distance between the two boats after 2 hours.<\/p>\r\n\r\n<div id=\"Figure_08_02_231\" class=\"small wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19150007\/CNX_Precalc_Figure_08_02_231.jpg\" alt=\"Insert figure(table) alt text: A graph of two rays, which represent the paths of the two boats. Both rays start at the origin. The first goes into the first quadrant at a 60 degree angle at 4 mph. The second goes into the fourth quadrant at a 327 degree angle from the origin. The second travels at 18 mph.\" width=\"487\" height=\"404\" \/> <strong>Figure 17.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135695099\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135695099\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135695099\"]\r\n<p id=\"fs-id1165135695102\">37.3 miles<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165133340980\">\r\n<div id=\"fs-id1165133340982\">\r\n<p id=\"fs-id1165133340985\">A triangular swimming pool measures 40 feet on one side and 65 feet on another side. These sides form an angle that measures 50\u00b0. How long is the third side (to the nearest tenth)?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135453145\">\r\n<div id=\"fs-id1165135453147\">\r\n<p id=\"fs-id1165135453149\">A pilot flies in a straight path for 1 hour 30 min. She then makes a course correction, heading 10\u00b0 to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position?<\/p>\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1165135453158\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135453158\"]\r\n<p id=\"fs-id1165135453158\">2371 miles<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134331114\">\r\n<div id=\"fs-id1165134331116\">\r\n<p id=\"fs-id1165134331118\">Los Angeles is 1,744 miles from Chicago, Chicago is 714 miles from New York, and New York is 2,451 miles from Los Angeles. Draw a triangle connecting these three cities, and find the angles in the triangle.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135200372\">\r\n<div id=\"fs-id1165135200373\">\r\n<p id=\"fs-id1165135200374\">Philadelphia is 140 miles from Washington, D.C., Washington, D.C. is 442 miles from Boston, and Boston is 315 miles from Philadelphia. Draw a triangle connecting these three cities and find the angles in the triangle.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135200381\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135200381\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135200381\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19150009\/CNX_Precalc_Figure_08_02_233.jpg\" alt=\"Angle BO is 9.1 degrees, angle PH is 150.2 degrees, and angle DC is 20.7 degrees.\" \/>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135659166\">\r\n<div id=\"fs-id1165135659168\">\r\n<p id=\"fs-id1165134033219\">Two planes leave the same airport at the same time. One flies at 20\u00b0 east of north at 500 miles per hour. The second flies at 30\u00b0 east of south at 600 miles per hour. How far apart are the planes after 2 hours?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134033227\">\r\n<div id=\"fs-id1165134033229\">\r\n<p id=\"fs-id1165134033231\">Two airplanes take off in different directions. One travels 300 mph due west and the other travels 25\u00b0 north of west at 420 mph. After 90 minutes, how far apart are they, assuming they are flying at the same altitude?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135694395\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135694395\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135694395\"]599.8 miles\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135694402\">\r\n<div id=\"fs-id1165134408478\">\r\n<p id=\"fs-id1165134408480\">A parallelogram has sides of length 15.4 units and 9.8 units. Its area is 72.9 square units. Find the measure of the longer diagonal.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134408487\">\r\n<div id=\"fs-id1165134408488\">\r\n<p id=\"fs-id1165134408489\">The four sequential sides of a quadrilateral have lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117\u00b0. What is the area of this quadrilateral?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134408492\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134408492\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134408492\"]\r\n<p id=\"fs-id1165137810097\">65.4 cm2<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137810105\">\r\n<div id=\"fs-id1165137810107\">\r\n<p id=\"fs-id1165133359474\">The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106\u00b0. What is the area of this quadrilateral?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165133359483\">\r\n<div id=\"fs-id1165133359485\">\r\n<p id=\"fs-id1165133359487\">Find the area of a triangular piece of land that measures 30 feet on one side and 42 feet on another; the included angle measures 132\u00b0. Round to the nearest whole square foot.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135506390\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135506390\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135506390\"]\r\n<p id=\"fs-id1165135506393\">468 ft2<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165131991396\">\r\n<div id=\"fs-id1165131991398\">\r\n<p id=\"fs-id1165131991400\">Find the area of a triangular piece of land that measures 110 feet on one side and 250 feet on another; the included angle measures 85\u00b0. Round to the nearest whole square foot.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1165135690077\">\r\n \t<dt>Law of Cosines<\/dt>\r\n \t<dd id=\"fs-id1165135690083\">states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133184256\">\r\n \t<dt>Generalized Pythagorean Theorem<\/dt>\r\n \t<dd id=\"fs-id1165133184261\">an extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and SSS triangles<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Use the Law of Cosines to solve oblique triangles.<\/li>\n<li>Solve applied problems using the Law of Cosines.<\/li>\n<li>Use Heron\u2019s formula to \ufb01nd the area of a triangle.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137693532\">Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_001\">(Figure)<\/a>. How far from port is the boat?<\/p>\n<div id=\"Figure_08_02_001\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145527\/CNX_Precalc_Figure_08_02_001.jpg\" alt=\"A triangle whose vertices are the boat, the port, and the turning point of the boat. The side between the port and the turning point is 10 mi, and the side between the turning point and the boat is 8 miles. The side between the port and the turning point is extended in a straight dotted line. The angle between the dotted line and the 8 mile side is 20 degrees.\" width=\"487\" height=\"517\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.<\/strong><\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135553590\">Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a <span class=\"no-emphasis\">SAS (side-angle-side) triangle<\/span>, or when all three sides are known, but no angles are known, a <span class=\"no-emphasis\">SSS (side-side-side) triangle<\/span>. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.<\/p>\n<div id=\"fs-id1165135190782\" class=\"bc-section section\">\n<h3>Using the Law of Cosines to Solve Oblique Triangles<\/h3>\n<p id=\"fs-id1165137437275\">The tool we need to solve the problem of the boat\u2019s distance from the port is the <strong>Law of Cosines<\/strong>, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level.<\/p>\n<p id=\"fs-id1165134079626\">Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the <span class=\"no-emphasis\">Pythagorean Theorem<\/span> to non-right triangles. Here is how it works: An arbitrary non-right triangle[latex]\\,ABC\\,[\/latex]is placed in the coordinate plane with vertex[latex]\\,A\\,[\/latex]at the origin, side[latex]\\,c\\,[\/latex]drawn along the <em>x<\/em>-axis, and vertex[latex]\\,C\\,[\/latex]located at some point[latex]\\,\\left(x,y\\right)\\,[\/latex]in the plane, as illustrated in <a class=\"autogenerated-content\" href=\"#Figure_08_02_002\">(Figure)<\/a>. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.<\/p>\n<div id=\"Figure_08_02_002\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145537\/CNX_Precalc_Figure_08_02_002.jpg\" alt=\"A triangle A B C plotted in quadrant 1 of the x,y plane. Angle A is theta degrees with opposite side a, angles B and C, with opposite sides b and c respectively, are unknown. Vertex A is located at the origin (0,0), vertex B is located at some point (x-c, 0) along the x-axis, and point C is located at some point in quadrant 1 at the point (b times the cos of theta, b times the sin of theta).\" width=\"487\" height=\"266\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.<\/strong><\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165134363434\">We can drop a perpendicular from[latex]\\,C\\,[\/latex]to the <em>x-<\/em>axis (this is the altitude or height). Recalling the basic <span class=\"no-emphasis\">trigonometric identities<\/span>, we know that<\/p>\n<div class=\"unnumbered\">[latex]\\mathrm{cos}\\,\\theta =\\frac{x\\text{(adjacent)}}{b\\text{(hypotenuse)}}\\text{ and }\\mathrm{sin}\\,\\theta =\\frac{y\\text{(opposite)}}{b\\text{(hypotenuse)}}[\/latex]<\/div>\n<p id=\"fs-id1165134243120\">In terms of[latex]\\,\\theta ,\\text{ }x=b\\mathrm{cos}\\,\\theta \\,[\/latex]and[latex]y=b\\mathrm{sin}\\,\\theta .\\text{ }[\/latex]The[latex]\\,\\left(x,y\\right)\\,[\/latex]point located at[latex]\\,C\\,[\/latex]has coordinates[latex]\\,\\left(b\\mathrm{cos}\\,\\theta ,\\,\\,b\\mathrm{sin}\\,\\theta \\right).\\,[\/latex]Using the side[latex]\\,\\left(x-c\\right)\\,[\/latex]as one leg of a right triangle and[latex]\\,y\\,[\/latex]as the second leg, we can find the length of hypotenuse[latex]\\,a\\,[\/latex]using the Pythagorean Theorem. Thus,<\/p>\n<div id=\"fs-id1165135471054\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{llllll} {a}^{2}={\\left(x-c\\right)}^{2}+{y}^{2}\\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }={\\left(b\\mathrm{cos}\\,\\theta -c\\right)}^{2}+{\\left(b\\mathrm{sin}\\,\\theta \\right)}^{2}\\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Substitute }\\left(b\\mathrm{cos}\\,\\theta \\right)\\text{ for}\\,x\\,\\,\\text{and }\\left(b\\mathrm{sin}\\,\\theta \\right)\\,\\text{for }y.\\hfill \\\\ \\text{ }=\\left({b}^{2}{\\mathrm{cos}}^{2}\\theta -2bc\\mathrm{cos}\\,\\theta +{c}^{2}\\right)+{b}^{2}{\\mathrm{sin}}^{2}\\theta \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Expand the perfect square}.\\hfill \\\\ \\text{ }={b}^{2}{\\mathrm{cos}}^{2}\\theta +{b}^{2}{\\mathrm{sin}}^{2}\\theta +{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Group terms noting that }{\\mathrm{cos}}^{2}\\theta +{\\mathrm{sin}}^{2}\\theta =1.\\hfill \\\\ \\text{ }={b}^{2}\\left({\\mathrm{cos}}^{2}\\theta +{\\mathrm{sin}}^{2}\\theta \\right)+{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Factor out }{b}^{2}.\\hfill \\\\ {a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135481279\">The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.<\/p>\n<p id=\"fs-id1165137539618\">Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Law of Cosines<\/h3>\n<p id=\"fs-id1165137428181\">The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in <a class=\"autogenerated-content\" href=\"#Figure_08_02_003\">(Figure)<\/a>, with angles[latex]\\,\\alpha ,\\beta ,[\/latex] and[latex]\\,\\gamma ,[\/latex] and opposite corresponding sides[latex]\\,a,b,[\/latex] and[latex]\\,c,\\,[\/latex]respectively, the Law of Cosines is given as three equations.<\/p>\n<div id=\"fs-id1165137436222\">[latex]\\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\\,\\,\\mathrm{cos}\\,\\alpha \\\\ {b}^{2}={a}^{2}+{c}^{2}-2ac\\,\\,\\mathrm{cos}\\,\\beta \\\\ {c}^{2}={a}^{2}+{b}^{2}-2ab\\,\\,\\mathrm{cos}\\,\\gamma \\end{array}[\/latex]<\/div>\n<div id=\"Figure_08_02_003\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145545\/CNX_Precalc_Figure_08_02_003n.jpg\" alt=\"A triangle with standard labels: angles alpha, beta, and gamma with opposite sides a, b, and c respectively.\" width=\"487\" height=\"239\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.<\/strong><\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137900234\">To solve for a missing side measurement, the corresponding opposite angle measure is needed.<\/p>\n<p id=\"fs-id1165134272589\">When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle.<\/p>\n<div id=\"fs-id1165137715303\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\mathrm{cos}\\text{ }\\alpha =\\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\\hfill \\end{array}\\hfill \\\\ \\mathrm{cos}\\text{ }\\beta =\\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\\hfill \\\\ \\mathrm{cos}\\text{ }\\gamma =\\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div class=\"precalculus howto\">\n<p id=\"fs-id1165137444540\"><strong>Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.<\/strong><\/p>\n<ol id=\"fs-id1165132949849\" type=\"1\">\n<li>Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.<\/li>\n<li>Apply the Law of Cosines to find the length of the unknown side or angle.<\/li>\n<li>Apply the <span class=\"no-emphasis\">Law of Sines<\/span> or Cosines to find the measure of a second angle.<\/li>\n<li>Compute the measure of the remaining angle.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_08_02_01\" class=\"textbox examples\">\n<div id=\"fs-id1165137442107\">\n<div id=\"fs-id1165134261612\">\n<h3>Finding the Unknown Side and Angles of a SAS Triangle<\/h3>\n<p id=\"fs-id1165137723520\">Find the unknown side and angles of the triangle in <a class=\"autogenerated-content\" href=\"#Figure_08_02_004\">(Figure)<\/a>.<\/p>\n<div id=\"Figure_08_02_004\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145552\/CNX_Precalc_Figure_08_02_004.jpg\" alt=\"A triangle with standard labels. Side a = 10, side c = 12, and angle beta = 30 degrees.\" width=\"487\" height=\"189\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 4.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137810707\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137810707\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137810707\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133192903\">First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.<\/p>\n<p id=\"fs-id1165137539491\">Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side[latex]\\,b,\\,[\/latex]as we know the measurement of the opposite angle[latex]\\,\\beta .[\/latex]<\/p>\n<div id=\"fs-id1165137768253\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ll}{b}^{2}={a}^{2}+{c}^{2}-2ac\\mathrm{cos}\\,\\beta \\hfill & \\hfill \\\\ {b}^{2}={10}^{2}+{12}^{2}-2\\left(10\\right)\\left(12\\right)\\mathrm{cos}\\left({30}^{\\circ }\\right)\\begin{array}{cccc}& & & \\end{array}\\hfill & \\text{Substitute the measurements for the known quantities}.\\hfill \\\\ {b}^{2}=100+144-240\\left(\\frac{\\sqrt{3}}{2}\\right)\\hfill & \\text{Evaluate the cosine and begin to simplify}.\\hfill \\\\ {b}^{2}=244-120\\sqrt{3}\\hfill & \\hfill \\\\ \\,\\,\\,b=\\sqrt{244-120\\sqrt{3}}\\hfill & \\,\\text{Use the square root property}.\\hfill \\\\ \\,\\,\\,b\\approx 6.013\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135542697\">Because we are solving for a length, we use only the positive square root. Now that we know the length[latex]\\,b,\\,[\/latex]we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle[latex]\\,\\alpha ,\\,[\/latex]we have<\/p>\n<div id=\"fs-id1165137574905\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ll}\\frac{\\mathrm{sin}\\,\\alpha }{a}=\\frac{\\mathrm{sin}\\,\\beta }{b}\\hfill & \\hfill \\\\ \\frac{\\mathrm{sin}\\,\\alpha }{10}=\\frac{\\mathrm{sin}\\left(30\u00b0\\right)}{6.013}\\hfill & \\hfill \\\\ \\,\\mathrm{sin}\\,\\alpha =\\frac{10\\mathrm{sin}\\left(30\u00b0\\right)}{6.013}\\hfill & \\text{Multiply both sides of the equation by 10}.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\alpha ={\\mathrm{sin}}^{-1}\\left(\\frac{10\\mathrm{sin}\\left(30\u00b0\\right)}{6.013}\\right)\\begin{array}{cccc}& & & \\end{array}\\hfill & \\text{Find the inverse sine of }\\frac{10\\mathrm{sin}\\left(30\u00b0\\right)}{6.013}.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\alpha \\approx 56.3\u00b0\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165133202446\">The other possibility for[latex]\\,\\alpha \\,[\/latex]would be[latex]\\,\\alpha =180\u00b0\u201356.3\u00b0\\approx 123.7\u00b0.\\,[\/latex]In the original diagram,[latex]\\,\\alpha \\,[\/latex]is adjacent to the longest side, so[latex]\\,\\alpha \\,[\/latex]is an acute angle and, therefore,[latex]\\,123.7\u00b0\\,[\/latex]does not make sense. Notice that if we choose to apply the <span class=\"no-emphasis\">Law of Cosines<\/span>, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between[latex]\\,0\u00b0\\,[\/latex]and[latex]\\,180\u00b0.\\,[\/latex]Proceeding with[latex]\\,\\alpha \\approx 56.3\u00b0,\\,[\/latex]we can then find the third angle of the triangle.<\/p>\n<div class=\"unnumbered\">[latex]\\gamma =180\u00b0-30\u00b0-56.3\u00b0\\approx 93.7\u00b0[\/latex]<\/div>\n<p id=\"fs-id1165137939414\">The complete set of angles and sides is<\/p>\n<div id=\"fs-id1165135263593\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ll}\\alpha \\approx 56.3\u00b0\\begin{array}{cccc}& & & \\end{array}\\hfill & a=10\\hfill \\\\ \\beta =30\u00b0\\hfill & b\\approx 6.013\\hfill \\\\ \\,\\gamma \\approx 93.7\u00b0\\hfill & c=12\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137734372\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_08_02_01\">\n<div id=\"fs-id1165135553574\">\n<p id=\"fs-id1165135553575\">Find the missing side and angles of the given triangle:[latex]\\,\\alpha =30\u00b0,\\,\\,b=12,\\,\\,c=24.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135360387\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135360387\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135360387\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135360388\">[latex]a\\approx 14.9,\\,\\,\\beta \\approx 23.8\u00b0,\\,\\,\\gamma \\approx 126.2\u00b0.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_08_02_02\" class=\"textbox examples\">\n<div id=\"fs-id1165133134878\">\n<div id=\"fs-id1165133134880\">\n<h3>Solving for an Angle of a SSS Triangle<\/h3>\n<p id=\"fs-id1165137758182\">Find the angle[latex]\\,\\alpha \\,[\/latex]for the given triangle if side[latex]\\,a=20,\\,[\/latex]side[latex]\\,b=25,\\,[\/latex]and side[latex]\\,c=18.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135361166\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135361166\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135361166\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135361168\">For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle[latex]\\,\\alpha ,\\,[\/latex]we have<\/p>\n<div id=\"fs-id1165134116894\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{llll}\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\,\\,\\text{ }{a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\alpha \\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }{20}^{2}={25}^{2}+{18}^{2}-2\\left(25\\right)\\left(18\\right)\\mathrm{cos}\\,\\alpha \\hfill & \\hfill & \\hfill & \\text{Substitute the appropriate measurements}.\\hfill \\\\ \\text{ }400=625+324-900\\mathrm{cos}\\,\\alpha \\hfill & \\hfill & \\hfill & \\text{Simplify in each step}.\\hfill \\\\ \\text{ }400=949-900\\mathrm{cos}\\,\\alpha \\hfill & \\hfill & \\hfill & \\hfill \\\\ \\,\\text{ }-549=-900\\mathrm{cos}\\,\\alpha \\hfill & \\hfill & \\hfill & \\text{Isolate cos }\\alpha .\\hfill \\\\ \\text{ }\\frac{-549}{-900}=\\mathrm{cos}\\,\\alpha \\hfill & \\hfill & \\hfill & \\hfill \\\\ \\,\\text{ }0.61\\approx \\mathrm{cos}\\,\\alpha \\hfill & \\hfill & \\hfill & \\hfill \\\\ {\\mathrm{cos}}^{-1}\\left(0.61\\right)\\approx \\alpha \\hfill & \\hfill & \\hfill & \\text{Find the inverse cosine}.\\hfill \\\\ \\text{ }\\alpha \\approx 52.4\u00b0\\hfill & \\hfill & \\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165133349386\">See <a class=\"autogenerated-content\" href=\"#Figure_08_02_005\">(Figure)<\/a>.<\/p>\n<div id=\"Figure_08_02_005\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145558\/CNX_Precalc_Figure_08_02_005.jpg\" alt=\"A triangle with standard labels. Side b =25, side a = 20, side c = 18, and angle alpha = 52.4 degrees.\" width=\"487\" height=\"266\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 5.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137807087\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id1165137605771\">Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137605779\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_08_02_02\">\n<div id=\"fs-id1165135648695\">\n<p id=\"fs-id1165135648696\">Given[latex]\\,a=5,b=7,\\,[\/latex]and[latex]\\,c=10,\\,[\/latex]find the missing angles.<\/p>\n<\/div>\n<div id=\"fs-id1165133210042\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165133210042\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165133210042\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133210043\">[latex]\\alpha \\approx 27.7\u00b0,\\,\\,\\beta \\approx 40.5\u00b0,\\,\\,\\gamma \\approx 111.8\u00b0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133215477\" class=\"bc-section section\">\n<h3>Solving Applied Problems Using the Law of Cosines<\/h3>\n<p id=\"fs-id1165133215482\">Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.<\/p>\n<div id=\"Example_08_02_03\" class=\"textbox examples\">\n<div id=\"fs-id1165133035957\">\n<div id=\"fs-id1165133035960\">\n<h3>Using the Law of Cosines to Solve a Communication Problem<\/h3>\n<p id=\"fs-id1165133035965\">On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.<\/p>\n<\/div>\n<div id=\"fs-id1165137419121\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137419121\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137419121\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137419123\">For simplicity, we start by drawing a diagram similar to <a class=\"autogenerated-content\" href=\"#Figure_08_02_006\">(Figure)<\/a> and labeling our given information.<\/p>\n<div id=\"Figure_08_02_006\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145605\/CNX_Precalc_Figure_08_02_006.jpg\" alt=\"A triangle formed between the two cell phone towers located on am east to west highway and the cellphone between and north of them. The side between the two towers is 6000 feet, the side between the left tower and the phone is 5050 feet, and the side between the right tower and the phone is 2420 feet. The angle between the 5050 and 6000 feet sides is labeled theta.\" width=\"487\" height=\"199\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 6.<\/strong><\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135428314\">Using the Law of Cosines, we can solve for the angle[latex]\\,\\theta .\\,[\/latex]Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let[latex]\\,a=2420,b=5050,\\,[\/latex]and[latex]\\,c=6000.\\,[\/latex]Thus,[latex]\\,\\theta \\,[\/latex]corresponds to the opposite side[latex]\\,a=2420.\\,[\/latex]<\/p>\n<div class=\"unnumbered\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\,\\text{ }{a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill \\end{array}\\hfill \\\\ \\text{ }{\\left(2420\\right)}^{2}={\\left(5050\\right)}^{2}+{\\left(6000\\right)}^{2}-2\\left(5050\\right)\\left(6000\\right)\\mathrm{cos}\\,\\theta \\hfill \\\\ \\,\\,\\,\\,\\,\\,{\\left(2420\\right)}^{2}-{\\left(5050\\right)}^{2}-{\\left(6000\\right)}^{2}=-2\\left(5050\\right)\\left(6000\\right)\\mathrm{cos}\\,\\theta \\hfill \\\\ \\text{ }\\frac{{\\left(2420\\right)}^{2}-{\\left(5050\\right)}^{2}-{\\left(6000\\right)}^{2}}{-2\\left(5050\\right)\\left(6000\\right)}=\\mathrm{cos}\\,\\theta \\hfill \\\\ \\text{ }\\mathrm{cos}\\,\\theta \\approx 0.9183\\hfill \\\\ \\text{ }\\theta \\approx {\\mathrm{cos}}^{-1}\\left(0.9183\\right)\\hfill \\\\ \\text{ }\\theta \\approx 23.3\u00b0\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165133178309\">To answer the questions about the phone\u2019s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in <a class=\"autogenerated-content\" href=\"#Figure_08_02_007\">(Figure)<\/a>. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.<\/p>\n<div id=\"Figure_08_02_007\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145616\/CNX_Precalc_Figure_08_02_007.jpg\" alt=\"The triangle between the phone, the left tower, and a point between the phone and the highway between the towers. The side between the phone and the highway is perpendicular to the highway and is y feet. The highway side is x feet. The angle at the tower, previously labeled theta, is 23.3 degrees.\" width=\"487\" height=\"177\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 7.<\/strong><\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135317539\">Using the angle[latex]\\,\\theta =23.3\u00b0\\,[\/latex]and the basic trigonometric identities, we can find the solutions. Thus<\/p>\n<div id=\"fs-id1165135407362\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\mathrm{cos}\\left(23.3\u00b0\\right)=\\frac{x}{5050}\\hfill \\end{array}\\hfill \\\\ \\text{ }x=5050\\mathrm{cos}\\left(23.3\u00b0\\right)\\hfill \\\\ \\text{ }x\\approx 4638.15\\,\\text{feet}\\hfill \\\\ \\text{ }\\mathrm{sin}\\left(23.3\u00b0\\right)=\\frac{y}{5050}\\hfill \\\\ \\text{ }y=5050\\mathrm{sin}\\left(23.3\u00b0\\right)\\hfill \\\\ \\text{ }y\\approx 1997.5\\,\\text{feet}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135546940\">The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_08_02_04\" class=\"textbox examples\">\n<div id=\"fs-id1165134113730\">\n<div id=\"fs-id1165134113733\">\n<h3>Calculating Distance Traveled Using a SAS Triangle<\/h3>\n<p id=\"fs-id1165134113738\">Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in <a class=\"autogenerated-content\" href=\"#Figure_08_02_009\">(Figure)<\/a>.<\/p>\n<div id=\"Figure_08_02_009\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145627\/CNX_Precalc_Figure_08_02_009.jpg\" alt=\"A triangle whose vertices are the boat, the port, and the turning point of the boat. The side between the port and the turning point is 10 mi, and the side between the turning point and the boat is 8 miles. The side between the port and the turning point is extended in a straight dotted line. The angle between the dotted line and the 8 mile side is 20 degrees.\" width=\"487\" height=\"517\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 8.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134282021\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134282021\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134282021\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134282023\">The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle,[latex]180\u00b0-20\u00b0=160\u00b0.\\,[\/latex]With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle\u2014the distance of the boat to the port.<\/p>\n<div id=\"fs-id1165135503577\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\,{x}^{2}={8}^{2}+{10}^{2}-2\\left(8\\right)\\left(10\\right)\\mathrm{cos}\\left(160\u00b0\\right)\\hfill \\\\ \\,{x}^{2}=314.35\\hfill \\\\ \\,\\,\\,\\,x=\\sqrt{314.35}\\hfill \\\\ \\,\\,\\,\\,x\\approx 17.7\\,\\text{miles}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165134393824\">The boat is about 17.7 miles from port.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h3>Using Heron\u2019s Formula to Find the Area of a Triangle<\/h3>\n<p id=\"fs-id1165135650692\">We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use <span class=\"no-emphasis\">Heron\u2019s formula<\/span> instead of finding the height. <span class=\"no-emphasis\">Heron of Alexandria<\/span> was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.<\/p>\n<div id=\"fs-id1165133306825\" class=\"textbox key-takeaways\">\n<h3>Heron\u2019s Formula<\/h3>\n<p id=\"fs-id1165137502458\">Heron\u2019s formula finds the area of oblique triangles in which sides[latex]\\,a,b\\text{,}[\/latex]and[latex]\\,c\\,[\/latex]are known.<\/p>\n<div>[latex]\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex]<\/div>\n<p id=\"fs-id1165134258499\">where[latex]\\,s=\\frac{\\left(a+b+c\\right)}{2}\\,[\/latex] is one half of the perimeter of the triangle, sometimes called the semi-perimeter.<\/p>\n<\/div>\n<div id=\"Example_08_02_05\" class=\"textbox examples\">\n<div id=\"fs-id1165135704848\">\n<div id=\"fs-id1165137723217\">\n<h3>Using Heron\u2019s Formula to Find the Area of a Given Triangle<\/h3>\n<p id=\"fs-id1165137723224\">Find the area of the triangle in <a class=\"autogenerated-content\" href=\"#Figure_08_02_010\">(Figure)<\/a> using Heron\u2019s formula.<\/p>\n<div id=\"Figure_08_02_010\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145634\/CNX_Precalc_Figure_08_02_010.jpg\" alt=\"A triangle with angles A, B, and C and opposite sides a, b, and c, respectively. Side a = 10, side b - 15, and side c = 7.\" width=\"487\" height=\"134\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 9.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q611728\">Show Solution<\/span><\/p>\n<div id=\"q611728\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134314759\">First, we calculate[latex]\\,s.[\/latex]<\/p>\n<div id=\"fs-id1165135408511\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ s=\\frac{\\left(a+b+c\\right)}{2}\\end{array}\\hfill \\\\ s=\\frac{\\left(10+15+7\\right)}{2}=16\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165134170047\">Then we apply the formula.<\/p>\n<div class=\"unnumbered\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ \\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}\\end{array}\\hfill \\\\ \\text{Area}=\\sqrt{16\\left(16-10\\right)\\left(16-15\\right)\\left(16-7\\right)}\\hfill \\\\ \\text{Area}\\approx 29.4\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165134373529\">The area is approximately 29.4 square units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134547309\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_08_02_04\">\n<div id=\"fs-id1165135300762\">\n<p id=\"fs-id1165135300763\">Use Heron\u2019s formula to find the area of a triangle with sides of lengths[latex]\\,a=29.7\\,\\text{ft},b=42.3\\,\\text{ft},\\,[\/latex]and[latex]\\,c=38.4\\,\\text{ft}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134102074\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134102074\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134102074\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134102075\">Area = 552 square feet<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_08_02_06\" class=\"textbox examples\">\n<div id=\"fs-id1165135347549\">\n<div id=\"fs-id1165137769676\">\n<h3>Applying Heron\u2019s Formula to a Real-World Problem<\/h3>\n<p id=\"fs-id1165137769681\">A Chicago city developer wants to construct a building consisting of artist\u2019s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See <a class=\"autogenerated-content\" href=\"#Figure_08_02_011\">(Figure)<\/a> for a view of the city property.<\/p>\n<div id=\"Figure_08_02_011\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145640\/CNX_Precalc_Figure_08_02_011.jpg\" alt=\"A triangle formed by sides Rush Street, N. Wabash Ave, and E. Pearson Street with lengths 62.4, 43.5, and 34.1, respectively.\" width=\"487\" height=\"520\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 10.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135518137\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135518137\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135518137\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135518140\">Find the measurement for[latex]\\,s,\\,[\/latex]which is one-half of the perimeter.<\/p>\n<div class=\"unnumbered\">[latex]\\begin{array}{l}s=\\frac{\\left(62.4+43.5+34.1\\right)}{2}\\hfill \\\\ s=70\\,\\text{m}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137619098\">Apply Heron\u2019s formula.<\/p>\n<div id=\"fs-id1165137619101\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\text{Area}=\\sqrt{70\\left(70-62.4\\right)\\left(70-43.5\\right)\\left(70-34.1\\right)}\\hfill \\\\ \\text{Area}=\\sqrt{506,118.2}\\hfill \\\\ \\text{Area}\\approx 711.4\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135497734\">The developer has about 711.4 square meters.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_08_02_05\">\n<div id=\"fs-id1165135689456\">\n<p id=\"fs-id1165135689457\">Find the area of a triangle given[latex]\\,a=4.38\\,\\text{ft}\\,,b=3.79\\,\\text{ft,}\\,[\/latex]and[latex]\\,c=5.22\\,\\text{ft}\\text{.}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137394207\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137394207\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137394207\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137394208\">about 8.15 square feet<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135532502\" class=\"precalculus media\">\n<p id=\"fs-id1165135532508\">Access these online resources for additional instruction and practice with the Law of Cosines.<\/p>\n<ul id=\"eip-id1165135445621\">\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/lawcosines\">Law of Cosines<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/cosineapp\">Law of Cosines: Applications<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/cosineapp2\">Law of Cosines: Applications 2<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135349445\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"eip-id1956425\" summary=\"..\">\n<tbody>\n<tr>\n<td>Law of Cosines<\/td>\n<td>[latex]\\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\alpha \\hfill \\\\ {b}^{2}={a}^{2}+{c}^{2}-2ac\\mathrm{cos}\\,\\beta \\hfill \\\\ {c}^{2}={a}^{2}+{b}^{2}-2abcos\\,\\gamma \\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Heron\u2019s formula<\/td>\n<td>[latex]\\begin{array}{l}\\text{ Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}\\hfill \\\\ \\text{where }s=\\frac{\\left(a+b+c\\right)}{2}\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165135442379\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165135442385\">\n<li>The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.<\/li>\n<li>The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See <a class=\"autogenerated-content\" href=\"#Example_08_02_01\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_08_02_02\">(Figure)<\/a>.<\/li>\n<li>The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See <a class=\"autogenerated-content\" href=\"#Example_08_02_03\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_08_02_04\">(Figure)<\/a>.<\/li>\n<li>Heron\u2019s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron\u2019s formula. See <a class=\"autogenerated-content\" href=\"#Example_08_02_05\">(Figure)<\/a> and See <a class=\"autogenerated-content\" href=\"#Example_08_02_06\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165134212020\" class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id1165134212028\">\n<div id=\"fs-id1165137887389\">\n<p id=\"fs-id1165137887391\">If you are looking for a missing side of a triangle, what do you need to know when using the Law of Cosines?<\/p>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137887398\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137887398\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137887398\">two sides and the angle opposite the missing side.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135430914\">\n<div id=\"fs-id1165135430916\">\n<p id=\"fs-id1165135430919\">If you are looking for a missing angle of a triangle, what do you need to know when using the Law of Cosines?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134073090\">\n<div id=\"fs-id1165134073092\">\n<p id=\"fs-id1165134073095\">Explain what[latex]\\,s\\,[\/latex]represents in Heron\u2019s formula.<\/p>\n<\/div>\n<div id=\"fs-id1165133143109\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165133143109\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165133143109\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133143111\">[latex]\\,s\\,[\/latex]is the semi-perimeter, which is half the perimeter of the triangle.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134383705\">\n<div id=\"fs-id1165134383707\">\n<p id=\"fs-id1165134383710\">Explain the relationship between the Pythagorean Theorem and the Law of Cosines.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134383715\">\n<div id=\"fs-id1165134383717\">\n<p>When must you use the Law of Cosines instead of the Pythagorean Theorem?<\/p>\n<\/div>\n<div id=\"fs-id1165135518102\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135518102\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135518102\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135518104\">The Law of Cosines must be used for any oblique (non-right) triangle.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135518110\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\n<p id=\"fs-id1165137828289\">For the following exercises, assume[latex]\\,\\alpha \\,[\/latex]is opposite side[latex]\\,a,\\beta \\,[\/latex] is opposite side[latex]\\,b,\\,[\/latex]and[latex]\\,\\gamma \\,[\/latex] is opposite side[latex]\\,c.\\,[\/latex]If possible, solve each triangle for the unknown side. Round to the nearest tenth.<\/p>\n<div id=\"fs-id1165135426388\">\n<div id=\"fs-id1165135426391\">\n<p id=\"fs-id1165135452292\">[latex]\\gamma =41.2\u00b0,a=2.49,b=3.13[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134278648\">\n<div id=\"fs-id1165134278650\">\n<p id=\"fs-id1165134278652\">[latex]\\alpha =120\u00b0,b=6,c=7[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165133359517\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165133359517\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165133359517\" class=\"hidden-answer\" style=\"display: none\">11.3<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133359525\">\n<div id=\"fs-id1165133359527\">[latex]\\beta =58.7\u00b0,a=10.6,c=15.7[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134272650\">\n<div id=\"fs-id1165133454475\">\n<p id=\"fs-id1165133454477\">[latex]\\gamma =115\u00b0,a=18,b=23[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134574953\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134574953\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134574953\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165132951595\">34.7<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165132951600\">\n<div id=\"fs-id1165132951602\">\n<p id=\"fs-id1165132951604\">[latex]\\alpha =119\u00b0,a=26,b=14[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134071612\">\n<div id=\"fs-id1165134071614\">\n<p id=\"fs-id1165132914223\">[latex]\\gamma =113\u00b0,b=10,c=32[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135354970\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135354970\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135354970\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135329809\">26.7<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135329815\">\n<div id=\"fs-id1165135329817\">\n<p id=\"fs-id1165135329819\">[latex]\\beta =67\u00b0,a=49,b=38[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131940267\">\n<div id=\"fs-id1165131940269\">\n<p id=\"fs-id1165131940271\">[latex]\\alpha =43.1\u00b0,a=184.2,b=242.8[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137938467\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137938467\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137938467\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137938469\">257.4<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135538836\">\n<div id=\"fs-id1165135538838\">\n<p id=\"fs-id1165135538840\">[latex]\\alpha =36.6\u00b0,a=186.2,b=242.2[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165132945436\">\n<div id=\"fs-id1165132945439\">\n<p id=\"fs-id1165132945441\">[latex]\\beta =50\u00b0,a=105,b=45{}_{}{}^{}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135515881\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135515881\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135515881\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135354976\">not possible<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135354982\">For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth.<\/p>\n<div id=\"fs-id1165135354986\">\n<div id=\"fs-id1165135354988\">\n<p id=\"fs-id1165133306919\">[latex]\\,a=42,b=19,c=30;\\,[\/latex]find angle[latex]\\,A.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165135550002\">\n<p id=\"fs-id1165135550004\">[latex]\\,a=14,\\text{ }b=13,\\text{ }c=20;\\,[\/latex]find angle[latex]\\,C.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135208492\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135208492\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135208492\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135208494\">95.5\u00b0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135208499\">\n<div id=\"fs-id1165135208501\">\n<p id=\"fs-id1165135208504\">[latex]\\,a=16,b=31,c=20;\\,[\/latex]find angle[latex]\\,B.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135381275\">\n<div id=\"fs-id1165135381277\">\n<p id=\"fs-id1165135154582\">[latex]\\,a=13,\\,b=22,\\,c=28;\\,[\/latex]find angle[latex]\\,A.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137899963\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137899963\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137899963\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135697813\">26.9\u00b0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134378611\">\n<div id=\"fs-id1165134378613\">\n<p>[latex]a=108,\\,b=132,\\,c=160;\\,[\/latex]find angle[latex]\\,C.\\,[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135707284\">For the following exercises, solve the triangle. Round to the nearest tenth.<\/p>\n<div id=\"fs-id1165135707288\">\n<div id=\"fs-id1165135707290\">\n<p id=\"fs-id1165135707292\">[latex]A=35\u00b0,b=8,c=11[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165133199358\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165133199358\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165133199358\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133199360\">[latex]B\\approx 45.9\u00b0,C\\approx 99.1\u00b0,a\\approx 6.4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134257565\">\n<div id=\"fs-id1165134257567\">\n<p id=\"fs-id1165135341174\">[latex]B=88\u00b0,a=4.4,c=5.2[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134156056\">\n<div id=\"fs-id1165134156059\">\n<p id=\"fs-id1165134156061\">[latex]C=121\u00b0,a=21,b=37[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137765785\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137765785\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137765785\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137765788\">[latex]A\\approx 20.6\u00b0,B\\approx 38.4\u00b0,c\\approx 51.1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135246610\">\n<div id=\"fs-id1165135246612\">[latex]a=13,b=11,c=15[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165134149779\">\n<div>\n<p id=\"fs-id1165134149784\">[latex]a=3.1,b=3.5,c=5[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137643163\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137643163\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137643163\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133307603\">[latex]A\\approx 37.8\u00b0,B\\approx 43.8,C\\approx 98.4\u00b0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137931166\">\n<div id=\"fs-id1165137931168\">\n<p id=\"fs-id1165137931170\">[latex]a=51,b=25,c=29[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135442502\">For the following exercises, use Heron\u2019s formula to find the area of the triangle. Round to the nearest hundredth.<\/p>\n<div id=\"fs-id1165135442507\">\n<div id=\"fs-id1165135442509\">\n<p id=\"fs-id1165135442511\">Find the area of a triangle with sides of length 18 in, 21 in, and 32 in. Round to the nearest tenth.<\/p>\n<\/div>\n<div id=\"fs-id1165134201647\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134201647\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134201647\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134201649\">177.56 in2<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134201654\">\n<div id=\"fs-id1165134201656\">\n<p id=\"fs-id1165134201658\">Find the area of a triangle with sides of length 20 cm, 26 cm, and 37 cm. Round to the nearest tenth.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137770136\">\n<div id=\"fs-id1165137770138\">\n<p id=\"fs-id1165137770140\">[latex]a=\\frac{1}{2}\\,\\text{m},b=\\frac{1}{3}\\,\\text{m},c=\\frac{1}{4}\\,\\text{m}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134328321\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134328321\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134328321\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134328323\">0.04 m2<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131894561\">\n<div id=\"fs-id1165131894564\">\n<p id=\"fs-id1165131894566\">[latex]a=12.4\\text{ ft},\\text{ }b=13.7\\text{ ft},\\text{ }c=20.2\\text{ ft}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133023583\">\n<div id=\"fs-id1165133023585\">\n<p id=\"fs-id1165133023587\">[latex]a=1.6\\text{ yd},\\text{ }b=2.6\\text{ yd},\\text{ }c=4.1\\text{ yd}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165133365504\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165133365504\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165133365504\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133365506\">0.91 yd2<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131880282\" class=\"bc-section section\">\n<h4>Graphical<\/h4>\n<p id=\"fs-id1165131880287\">For the following exercises, find the length of side [latex]x.[\/latex] Round to the nearest tenth.<\/p>\n<div id=\"fs-id1165135600236\">\n<div id=\"fs-id1165135600238\"><span id=\"fs-id1165135600244\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145646\/CNX_Precalc_Figure_08_02_201.jpg\" alt=\"A triangle. One angle is 72 degrees, with opposite side = x. The other two sides are 5 and 6.5.\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-id1165135442441\">\n<div id=\"fs-id1165135442443\"><span id=\"fs-id1165135505452\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145648\/CNX_Precalc_Figure_08_02_202.jpg\" alt=\"A triangle. One angle is 42 degrees with opposite side = x. The other two sides are 4.5 and 3.4.\" \/><\/span><\/div>\n<div id=\"fs-id1165135505463\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135505463\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135505463\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135505465\">3.0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134272137\">\n<div id=\"fs-id1165134272139\"><span id=\"fs-id1165134272145\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145654\/CNX_Precalc_Figure_08_02_203.jpg\" alt=\"A triangle. One angle is 40 degrees with opposite side = 15. The other two sides are 12 and x.\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-id1165137450863\">\n<div id=\"fs-id1165137450865\"><span id=\"fs-id1165133249121\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145656\/CNX_Precalc_Figure_08_02_204.jpg\" alt=\"A triangle. One angle is 65 degrees with opposite side = x. The other two sides are 30 and 23.\" \/><\/span><\/div>\n<div id=\"fs-id1165133249132\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165133249132\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165133249132\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133249134\">29.1<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135483114\">\n<div id=\"fs-id1165135483116\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145706\/CNX_Precalc_Figure_08_02_205.jpg\" alt=\"A triangle. One angle is 50 degrees with opposite side = x. The other two sides are 225 and 305.\" \/><\/div>\n<\/div>\n<div id=\"fs-id1165135556651\">\n<div id=\"fs-id1165135556653\"><span id=\"fs-id1165134370014\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145716\/CNX_Precalc_Figure_08_02_206.jpg\" alt=\"A triangle. One angle is 123 degrees with opposite side = x. The other two sides are 1\/5 and 1\/3.\" \/><\/span><\/div>\n<div id=\"fs-id1165134370025\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134370025\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134370025\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133306900\">0.5<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165133306905\">For the following exercises, find the measurement of angle[latex]\\,A.[\/latex]<\/p>\n<div id=\"fs-id1165135502926\">\n<div id=\"fs-id1165135502928\"><span id=\"fs-id1165135502934\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145723\/CNX_Precalc_Figure_08_02_207.jpg\" alt=\"A triangle. Angle A is opposite a side of length 2.3. The other two sides are 1.5 and 2.5.\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-id1165137696000\">\n<div id=\"fs-id1165137696003\"><span id=\"fs-id1165134323614\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145725\/CNX_Precalc_Figure_08_02_208.jpg\" alt=\"A triangle. Angle A is opposite a side of length 125. The other two sides are 115 and 100.\" \/><\/span><\/div>\n<div id=\"fs-id1165135545523\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135545523\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135545523\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135545525\">70.7\u00b0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135545530\">\n<div id=\"fs-id1165135545532\"><span id=\"fs-id1165134384434\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145728\/CNX_Precalc_Figure_08_02_209.jpg\" alt=\"A triangle. Angle A is opposite a side of length 6.8. The other two sides are 4.3 and 8.2.\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-id1165134384446\">\n<div id=\"fs-id1165134068908\"><span id=\"fs-id1165134068914\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145735\/CNX_Precalc_Figure_08_02_210.jpg\" alt=\"A triangle. Angle A is opposite a side of length 40.6. The other two sides are 38.7 and 23.3.\" \/><\/span><\/div>\n<div id=\"fs-id1165134401621\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134401621\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134401621\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134401623\">77.4\u00b0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134401629\">\n<div id=\"fs-id1165134401631\">\n<p>Find the measure of each angle in the triangle shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_211\">(Figure)<\/a>. Round to the nearest tenth.<\/p>\n<div id=\"Figure_08_02_211\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145737\/CNX_Precalc_Figure_08_02_211.jpg\" alt=\"A triangle A B C. Angle A is opposite a side of length 10, angle B is opposite a side of length 12, and angle C is opposite a side of length 7.\" width=\"487\" height=\"171\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 11.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"eip-483\">For the following exercises, solve for the unknown side. Round to the nearest tenth.<\/p>\n<div id=\"fs-id1165137731858\">\n<div id=\"fs-id1165137731860\"><span id=\"fs-id1165133236441\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145743\/CNX_Precalc_Figure_08_02_212.jpg\" alt=\"A triangle. One angle is 60 degrees with opposite side unknown. The other two sides are 20 and 28.\" \/><\/span><\/div>\n<div id=\"fs-id1165133236451\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165133236451\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165133236451\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135400873\">25.0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135400878\">\n<div id=\"fs-id1165135400880\"><span id=\"fs-id1165135400887\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145751\/CNX_Precalc_Figure_08_02_213.jpg\" alt=\"A triangle. One angle is 30 degrees with opposite side unknown. The other two sides are 16 and 10.\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-id1165135186079\">\n<div id=\"fs-id1165135186081\"><span id=\"fs-id1165134087532\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145816\/CNX_Precalc_Figure_08_02_214.jpg\" alt=\"A triangle. One angle is 22 degrees with opposite side unknown. The other two sides are 20 and 13.\" \/><\/span><\/div>\n<div id=\"fs-id1165135698477\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135698477\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135698477\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135698480\">9.3<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135698484\">\n<div id=\"fs-id1165135698486\"><span id=\"fs-id1165135385603\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145825\/CNX_Precalc_Figure_08_02_215.jpg\" alt=\"A triangle. One angle is 88 degrees with opposite side = 9. Another side is 5.\" \/><\/span><\/div>\n<\/div>\n<p>For the following exercises, find the area of the triangle. Round to the nearest hundredth.<\/p>\n<div id=\"fs-id1165135385615\">\n<div id=\"fs-id1165135385617\"><span id=\"fs-id1165133088150\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145827\/CNX_Precalc_Figure_08_02_216.jpg\" alt=\"A triangle with sides 8, 12, and 17. Angles unknown.\" \/><\/span><\/div>\n<div id=\"fs-id1165135525901\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135525901\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135525901\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135525903\">43.52<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div id=\"fs-id1165135525911\"><span id=\"fs-id1165135512604\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145833\/CNX_Precalc_Figure_08_02_217.jpg\" alt=\"A triangle with sides 50, 22, and 36. Angles unknown.\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-id1165137507971\">\n<div id=\"fs-id1165137507973\"><span id=\"fs-id1165137507980\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145840\/CNX_Precalc_Figure_08_02_218.jpg\" alt=\"A triangle with sides 1.9, 2.6, and 4.3. Angles unknown.\" \/><\/span><\/div>\n<div id=\"fs-id1165134239622\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134239622\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134239622\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134239625\">1.41<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134239629\">\n<div><span id=\"fs-id1165135149921\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145851\/CNX_Precalc_Figure_08_02_219.jpg\" alt=\"A triangle with sides 8.9, 12.5, and 16.2. Angles unknown.\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-id1165135450305\">\n<div id=\"fs-id1165135450307\"><span id=\"fs-id1165135450313\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145859\/CNX_Precalc_Figure_08_02_220.jpg\" alt=\"A triangle with sides 1\/2, 2\/3, and 3\/5. Angles unknown.\" \/><\/span><\/div>\n<div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134263937\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134263937\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134263937\">0.14<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134391071\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<div id=\"fs-id1165134391076\">\n<div>\n<p>A parallelogram has sides of length 16 units and 10 units. The shorter diagonal is 12 units. Find the measure of the longer diagonal.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133091581\">\n<div id=\"fs-id1165133091584\">\n<p id=\"fs-id1165133091586\">The sides of a parallelogram are 11 feet and 17 feet. The longer diagonal is 22 feet. Find the length of the shorter diagonal.<\/p>\n<\/div>\n<div id=\"fs-id1165133091591\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165133091591\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165133091591\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165133091593\">18.3<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135666704\">\n<div id=\"fs-id1165135666707\">\n<p id=\"fs-id1165135666709\">The sides of a parallelogram are 28 centimeters and 40 centimeters. The measure of the larger angle is 100\u00b0. Find the length of the shorter diagonal.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134383775\">\n<div id=\"fs-id1165134383777\">\n<p id=\"fs-id1165134383779\">A regular octagon is inscribed in a circle with a radius of 8 inches. (See <a class=\"autogenerated-content\" href=\"#Figure_08_02_221\">(Figure)<\/a>.) Find the perimeter of the octagon.<\/p>\n<div id=\"Figure_08_02_221\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145905\/CNX_Precalc_Figure_08_02_221.jpg\" alt=\"An octagon inscribed in a circle.\" width=\"487\" height=\"197\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 12.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135697763\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135697763\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135697763\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135697765\">48.98<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134331096\">\n<div id=\"fs-id1165134331098\">\n<p>A regular pentagon is inscribed in a circle of radius 12 cm. (See <a class=\"autogenerated-content\" href=\"#Figure_08_02_222\">(Figure)<\/a>.) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter.<\/p>\n<div id=\"Figure_08_02_222\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145914\/CNX_Precalc_Figure_08_02_222.jpg\" alt=\"A pentagon inscribed in a circle.\" width=\"487\" height=\"211\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 13.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165133017581\">For the following exercises, suppose that[latex]\\,{x}^{2}=25+36-60\\mathrm{cos}\\left(52\\right)\\,[\/latex]represents the relationship of three sides of a triangle and the cosine of an angle.<\/p>\n<div id=\"fs-id1165131837012\">\n<div id=\"fs-id1165131837014\">\n<p id=\"fs-id1165131837016\">Draw the triangle.<\/p>\n<\/div>\n<div id=\"fs-id1165131837019\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165131837019\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165131837019\" class=\"hidden-answer\" style=\"display: none\"><span id=\"fs-id1165135339571\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145916\/CNX_Precalc_Figure_08_02_223.jpg\" alt=\"A triangle. One angle is 52 degrees with opposite side = x. The other two sides are 5 and 6.\" \/><\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135339583\">\n<div id=\"fs-id1165135388758\">\n<p id=\"fs-id1165135388760\">Find the length of the third side.<\/p>\n<\/div>\n<\/div>\n<p>For the following exercises, find the area of the triangle.<\/p>\n<div id=\"fs-id1165135388769\">\n<div id=\"fs-id1165135388771\"><span id=\"fs-id1165135390787\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145918\/CNX_Precalc_Figure_08_02_224.jpg\" alt=\"A triangle. One angle is 22 degrees with opposite side = 3.4. Another side is 5.3.\" \/><\/span><\/div>\n<div id=\"fs-id1165134130014\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134130014\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134130014\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134130016\">7.62<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134130020\">\n<div id=\"fs-id1165134130023\"><span id=\"fs-id1165134260427\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145920\/CNX_Precalc_Figure_08_02_225.jpg\" alt=\"A triangle. One angle is 80 degrees with opposite side unknown. The other two sides are 8 and 6.\" \/><\/span><\/div>\n<\/div>\n<div>\n<div><span id=\"fs-id1165135186219\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145933\/CNX_Precalc_Figure_08_02_226.jpg\" alt=\"A triangle. One angle is 18 degrees with opposite side = 12.8. Another side is 18.8.\" \/><\/span><\/div>\n<div id=\"fs-id1165135633910\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135633910\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135633910\" class=\"hidden-answer\" style=\"display: none\">85.1<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135633919\" class=\"bc-section section\">\n<h4>Real-World Applications<\/h4>\n<div id=\"fs-id1165132957157\">\n<div id=\"fs-id1165132957158\">\n<p id=\"fs-id1165132957160\">A surveyor has taken the measurements shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_227\">(Figure)<\/a>. Find the distance across the lake. Round answers to the nearest tenth.<\/p>\n<div id=\"Figure_08_02_227\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145952\/CNX_Precalc_Figure_08_02_227.jpg\" alt=\"A triangle. One angle is 70 degrees with opposite side unknown, which is the length of the lake. The other two sides are 800 and 900 feet.\" width=\"487\" height=\"262\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 14.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134540068\">\n<div id=\"fs-id1165134540070\">\n<p id=\"fs-id1165134540073\">A satellite calculates the distances and angle shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_228\">(Figure)<\/a> (not to scale). Find the distance between the two cities. Round answers to the nearest tenth.<\/p>\n<div id=\"Figure_08_02_228\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 498px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145954\/CNX_Precalc_Figure_08_02_228.jpg\" alt=\"Insert figure(table) alt text: A triangle formed by two cities on the ground and a satellite above them. The angle by the satellite is 2.1 degrees with opposite side unknown, which is the distance between the two cities. The lengths of the other sides are 370 and 350 km.\" width=\"488\" height=\"307\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 15.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137827744\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137827744\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137827744\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137827746\">24.0 km<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137827752\">\n<div id=\"fs-id1165137827754\">\n<p id=\"fs-id1165137827756\">An airplane flies 220 miles with a heading of 40\u00b0, and then flies 180 miles with a heading of 170\u00b0. How far is the plane from its starting point, and at what heading? Round answers to the nearest tenth.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135181803\">\n<div id=\"fs-id1165135181805\">\n<p>A 113-foot tower is located on a hill that is inclined 34\u00b0 to the horizontal, as shown in <a class=\"autogenerated-content\" href=\"#Figure_08_02_230\">(Figure)<\/a>. A guy-wire is to be attached to the top of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed.<\/p>\n<div id=\"Figure_08_02_230\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19150005\/CNX_Precalc_Figure_08_02_230.jpg\" alt=\"Insert figure(table) alt text: Two triangles, one on top of the other. The bottom triangle is the hill inclined 34 degrees to the horizontal. The second is formed by the base of the tower on the incline of the hill, the top of the tower, and the wire anchor point uphill from the tower on the incline. The sides are the tower, the incline of the hill, and the wire. The tower side is 113 feet and the incline side is 98 feet.\" width=\"487\" height=\"267\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 16.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135500669\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135500669\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135500669\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135500671\">99.9 ft<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135500676\">\n<div id=\"fs-id1165135500678\">\n<p id=\"fs-id1165137921478\">Two ships left a port at the same time. One ship traveled at a speed of 18 miles per hour at a heading of 320\u00b0. The other ship traveled at a speed of 22 miles per hour at a heading of 194\u00b0. Find the distance between the two ships after 10 hours of travel.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137921488\">\n<div id=\"fs-id1165137921490\">\n<p id=\"fs-id1165135510887\">The graph in <a class=\"autogenerated-content\" href=\"#Figure_08_02_231\">(Figure)<\/a> represents two boats departing at the same time from the same dock. The first boat is traveling at 18 miles per hour at a heading of 327\u00b0 and the second boat is traveling at 4 miles per hour at a heading of 60\u00b0. Find the distance between the two boats after 2 hours.<\/p>\n<div id=\"Figure_08_02_231\" class=\"small wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19150007\/CNX_Precalc_Figure_08_02_231.jpg\" alt=\"Insert figure(table) alt text: A graph of two rays, which represent the paths of the two boats. Both rays start at the origin. The first goes into the first quadrant at a 60 degree angle at 4 mph. The second goes into the fourth quadrant at a 327 degree angle from the origin. The second travels at 18 mph.\" width=\"487\" height=\"404\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 17.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135695099\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135695099\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135695099\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135695102\">37.3 miles<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133340980\">\n<div id=\"fs-id1165133340982\">\n<p id=\"fs-id1165133340985\">A triangular swimming pool measures 40 feet on one side and 65 feet on another side. These sides form an angle that measures 50\u00b0. How long is the third side (to the nearest tenth)?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135453145\">\n<div id=\"fs-id1165135453147\">\n<p id=\"fs-id1165135453149\">A pilot flies in a straight path for 1 hour 30 min. She then makes a course correction, heading 10\u00b0 to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position?<\/p>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135453158\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135453158\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135453158\">2371 miles<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134331114\">\n<div id=\"fs-id1165134331116\">\n<p id=\"fs-id1165134331118\">Los Angeles is 1,744 miles from Chicago, Chicago is 714 miles from New York, and New York is 2,451 miles from Los Angeles. Draw a triangle connecting these three cities, and find the angles in the triangle.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135200372\">\n<div id=\"fs-id1165135200373\">\n<p id=\"fs-id1165135200374\">Philadelphia is 140 miles from Washington, D.C., Washington, D.C. is 442 miles from Boston, and Boston is 315 miles from Philadelphia. Draw a triangle connecting these three cities and find the angles in the triangle.<\/p>\n<\/div>\n<div id=\"fs-id1165135200381\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135200381\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135200381\" class=\"hidden-answer\" style=\"display: none\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19150009\/CNX_Precalc_Figure_08_02_233.jpg\" alt=\"Angle BO is 9.1 degrees, angle PH is 150.2 degrees, and angle DC is 20.7 degrees.\" \/><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135659166\">\n<div id=\"fs-id1165135659168\">\n<p id=\"fs-id1165134033219\">Two planes leave the same airport at the same time. One flies at 20\u00b0 east of north at 500 miles per hour. The second flies at 30\u00b0 east of south at 600 miles per hour. How far apart are the planes after 2 hours?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134033227\">\n<div id=\"fs-id1165134033229\">\n<p id=\"fs-id1165134033231\">Two airplanes take off in different directions. One travels 300 mph due west and the other travels 25\u00b0 north of west at 420 mph. After 90 minutes, how far apart are they, assuming they are flying at the same altitude?<\/p>\n<\/div>\n<div id=\"fs-id1165135694395\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135694395\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135694395\" class=\"hidden-answer\" style=\"display: none\">599.8 miles<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135694402\">\n<div id=\"fs-id1165134408478\">\n<p id=\"fs-id1165134408480\">A parallelogram has sides of length 15.4 units and 9.8 units. Its area is 72.9 square units. Find the measure of the longer diagonal.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134408487\">\n<div id=\"fs-id1165134408488\">\n<p id=\"fs-id1165134408489\">The four sequential sides of a quadrilateral have lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117\u00b0. What is the area of this quadrilateral?<\/p>\n<\/div>\n<div id=\"fs-id1165134408492\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134408492\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134408492\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137810097\">65.4 cm2<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137810105\">\n<div id=\"fs-id1165137810107\">\n<p id=\"fs-id1165133359474\">The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106\u00b0. What is the area of this quadrilateral?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133359483\">\n<div id=\"fs-id1165133359485\">\n<p id=\"fs-id1165133359487\">Find the area of a triangular piece of land that measures 30 feet on one side and 42 feet on another; the included angle measures 132\u00b0. Round to the nearest whole square foot.<\/p>\n<\/div>\n<div id=\"fs-id1165135506390\" class=\"solution textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135506390\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135506390\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135506393\">468 ft2<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131991396\">\n<div id=\"fs-id1165131991398\">\n<p id=\"fs-id1165131991400\">Find the area of a triangular piece of land that measures 110 feet on one side and 250 feet on another; the included angle measures 85\u00b0. Round to the nearest whole square foot.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165135690077\">\n<dt>Law of Cosines<\/dt>\n<dd id=\"fs-id1165135690083\">states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133184256\">\n<dt>Generalized Pythagorean Theorem<\/dt>\n<dd id=\"fs-id1165133184261\">an extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and SSS triangles<\/dd>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3104\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Algebra and Trigonometry. <strong>Authored by<\/strong>: Jay Abramson, et. al. <strong>Provided by<\/strong>: OpenStax CNX. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1\">http:\/\/cnx.org\/contents\/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":53384,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Algebra and Trigonometry\",\"author\":\"Jay Abramson, et. al\",\"organization\":\"OpenStax CNX\",\"url\":\"http:\/\/cnx.org\/contents\/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3104","chapter","type-chapter","status-publish","hentry"],"part":2978,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/pressbooks\/v2\/chapters\/3104","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/wp\/v2\/users\/53384"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/pressbooks\/v2\/chapters\/3104\/revisions"}],"predecessor-version":[{"id":3699,"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/pressbooks\/v2\/chapters\/3104\/revisions\/3699"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/pressbooks\/v2\/parts\/2978"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/pressbooks\/v2\/chapters\/3104\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/wp\/v2\/media?parent=3104"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/pressbooks\/v2\/chapter-type?post=3104"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/wp\/v2\/contributor?post=3104"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/wp-json\/wp\/v2\/license?post=3104"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}