{"id":14311,"date":"2018-09-27T16:52:11","date_gmt":"2018-09-27T16:52:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-law-of-sines\/"},"modified":"2021-02-04T22:47:49","modified_gmt":"2021-02-04T22:47:49","slug":"non-right-triangles-law-of-sines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/chapter\/non-right-triangles-law-of-sines\/","title":{"raw":"Walkthrough of Unit 8: Law of Sines and Law of Cosines","rendered":"Walkthrough of Unit 8: Law of Sines and Law of Cosines"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Use the Law of Sines to solve oblique triangles.<\/li>\r\n \t<li>Find the area of an oblique triangle using the sine function.<\/li>\r\n \t<li>Solve applied problems using the Law of Sines.<\/li>\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&nbsp;\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165008\/CNX_Precalc_Figure_08_01_0012.jpg\" alt=\"A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted line perpendicular to the ground side connecting the airplane vertex with the ground - an altitude line.\" width=\"487\" height=\"134\" \/> <b>Figure 1<\/b>[\/caption]\r\n<h2>Using the Law of Sines to Solve Obliques Triangles<\/h2>\r\nIn any triangle, we can draw an <strong>altitude<\/strong>, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.\r\n\r\nAny triangle that is not a right triangle is an <strong>oblique triangle<\/strong>. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:\r\n<ol>\r\n \t<li><strong>ASA (angle-side-angle)<\/strong> We know the measurements of two angles and the included side. See Figure 2.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165011\/CNX_Precalc_Figure_08_01_0022.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma's values are known, as is the side opposite beta, between alpha and gamma.\" width=\"487\" height=\"141\" \/> <b>Figure 2<\/b>[\/caption]<\/li>\r\n \t<li><strong>AAS (angle-angle-side)<\/strong> We know the measurements of two angles and a side that is not between the known angles. See Figure 3.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165013\/CNX_Precalc_Figure_08_01_0032.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma are known, as is the side opposite alpha, between beta and gamma.\" width=\"487\" height=\"141\" \/> <b>Figure 3<\/b>[\/caption]<\/li>\r\n \t<li><strong>SSA (side-side-angle)<\/strong> We know the measurements of two sides and an angle that is not between the known sides. See Figure 4.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165016\/CNX_Precalc_Figure_08_01_0042.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha is the only angle known. Two sides are known. The first is opposite alpha, between beta and gamma, and the second is opposite gamma, between alpha and beta.\" width=\"487\" height=\"141\" \/> <b>Figure 4<\/b>[\/caption]<\/li>\r\n<\/ol>\r\nKnowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let\u2019s see how this statement is derived by considering the triangle shown in Figure 5.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165018\/CNX_Precalc_Figure_08_01_0052.jpg\" alt=\"An oblique triangle consisting of sides a, b, and c, and angles alpha, beta, and gamma. Side c is opposide angle gamma and is the horizontal base of the triangle. Side b is opposite angle beta, and side a is opposite angle alpha. There is a dotted perpendicular line - an altitude - from the gamma angle to the horizontal base c.\" width=\"487\" height=\"143\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\nUsing the right triangle relationships, we know that [latex]\\sin \\alpha =\\frac{h}{b}[\/latex] and [latex]\\sin \\beta =\\frac{h}{a}[\/latex]. Solving both equations for [latex]h[\/latex] gives two different expressions for [latex]h[\/latex].\r\n<div style=\"text-align: center\">[latex]h=b\\sin \\alpha \\text{ and }h=a\\sin \\beta [\/latex]<\/div>\r\nWe then set the expressions equal to each other.\r\n<div style=\"text-align: center\">[latex]\\begin{align}b\\sin \\alpha &amp;=a\\sin \\beta \\\\ \\left(\\frac{1}{ab}\\right)\\left(b\\sin \\alpha \\right)&amp;=\\left(a\\sin \\beta \\right)\\left(\\frac{1}{ab}\\right) &amp;&amp; \\text{Multiply both sides by}\\frac{1}{ab}. \\\\ \\frac{\\sin \\alpha }{a}&amp;=\\frac{\\sin \\beta }{b} \\end{align}[\/latex]<\/div>\r\nSimilarly, we can compare the other ratios.\r\n<div style=\"text-align: center\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\gamma }{c}\\text{ and }\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/div>\r\nCollectively, these relationships are called the <strong>Law of Sines<\/strong>.\r\n<div style=\"text-align: center\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\lambda }{c}[\/latex]<\/div>\r\nNote the standard way of labeling triangles: angle [latex]\\alpha [\/latex] (alpha) is opposite side [latex]a[\/latex]; angle [latex]\\beta [\/latex] (beta) is opposite side [latex]b[\/latex]; and angle [latex]\\gamma [\/latex] (gamma) is opposite side [latex]c[\/latex]. See Figure 6.\r\n\r\nWhile calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165020\/CNX_Precalc_Figure_08_01_0062.jpg\" alt=\"A triangle with standard labels.\" width=\"487\" height=\"197\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Law of Sines<\/h3>\r\nGiven a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The <strong>Law of Sines<\/strong> is based on proportions and is presented symbolically two ways.\r\n<p style=\"text-align: center\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\frac{a}{\\sin \\alpha }=\\frac{b}{\\sin \\beta }=\\frac{c}{\\sin \\gamma }[\/latex]<\/p>\r\nTo solve an oblique triangle, use any pair of applicable ratios.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Solving for Two Unknown Sides and Angle of an AAS Triangle<\/h3>\r\nSolve the triangle shown in Figure 7\u00a0to the nearest tenth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165022\/CNX_Precalc_Figure_08_01_0072.jpg\" alt=\"An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.\" width=\"487\" height=\"200\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"604625\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"604625\"]\r\n\r\nThe three angles must add up to 180 degrees. From this, we can determine that\r\n<p style=\"text-align: center\">[latex]\\begin{align} \\beta =180^\\circ -50^\\circ -30^\\circ =100^\\circ \\end{align}[\/latex]<\/p>\r\nTo find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle [latex]\\alpha =50^\\circ [\/latex] and its corresponding side [latex]a=10[\/latex]. We can use the following proportion from the Law of Sines to find the length of [latex]c[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\frac{\\sin \\left(30^\\circ \\right)}{c} \\\\ &amp;c\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\sin \\left(30^\\circ \\right) &amp;&amp; \\text{Multiply both sides by }c. \\\\ &amp;c=\\sin \\left(30^\\circ \\right)\\frac{10}{\\sin \\left(50^\\circ \\right)} &amp;&amp; \\text{Multiply by the reciprocal to isolate }c. \\\\ &amp;c\\approx 6.5 \\end{align}[\/latex]<\/p>\r\nSimilarly, to solve for [latex]b[\/latex], we set up another proportion.\r\n<p style=\"text-align: center\">[latex]\\begin{align} &amp;\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\frac{\\sin \\left(100^\\circ \\right)}{b} \\\\ &amp;b\\sin \\left(50^\\circ \\right)=10\\sin \\left(100^\\circ \\right) &amp;&amp; \\text{Multiply both sides by }b. \\\\ &amp;b=\\frac{10\\sin \\left(100^\\circ \\right)}{\\sin \\left(50^\\circ \\right)} &amp;&amp; \\text{Multiply by the reciprocal to isolate }b. \\\\ &amp;b\\approx 12.9\\end{align}[\/latex]<\/p>\r\nTherefore, the complete set of angles and sides is\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\alpha =50^\\circ,\\beta =100^\\circ,\\gamma =30^\\circ \\\\ a=10,b\\approx 12.9,c\\approx 6.5 \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve the triangle shown in Figure 8\u00a0to the nearest tenth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165024\/CNX_Precalc_Figure_08_01_0082.jpg\" alt=\"An oblique triangle with standard labels. Angle alpha is 98 degrees, angle gamma is 43 degrees, and side b is of length 22. Side b is the horizontal base.\" width=\"487\" height=\"247\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"933392\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"933392\"]\r\n\r\n[latex]\\begin{gathered}\\alpha ={98}^{\\circ },\\beta ={39}^{\\circ },\\gamma ={43}^{\\circ } \\\\ a=34.6, b=22, c=23.8\\end{gathered}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149230[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using The Law of Sines to Solve SSA Triangles<\/h2>\r\nWe can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an <strong>ambiguous case<\/strong>. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Possible Outcomes for SSA Triangles<\/h3>\r\nOblique triangles in the category SSA may have four different outcomes.\u00a0Figure 9\u00a0illustrates the solutions with the known sides [latex]a[\/latex] and [latex]b[\/latex] and known angle [latex]\\alpha [\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165027\/CNX_Precalc_Figure_08_01_009n2.jpg\" alt=\"Four attempted oblique triangles are in a row, all with standard labels. Side c is the horizontal base. In the first attempted triangle, side a is less than the altitude height. Since side a cannot reach side c, there is no triangle. In the second attempted triangle, side a is equal to the length of the altitude height, so side a forms a right angle with side c. In the third attempted triangle, side a is greater than the altitude height and less than side b, so side a can form either an acute or obtuse angle with side c. In the fourth attempted triangle, side a is greater than or equal to side b, so side a forms an acute angle with side c.\" width=\"731\" height=\"483\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Solving an Oblique SSA Triangle<\/h3>\r\nSolve the triangle in Figure 10\u00a0for the missing side and find the missing angle measures to the nearest tenth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165029\/CNX_Precalc_Figure_08_01_0102.jpg\" alt=\"An oblique triangle with standard labels where side a is of length 6, side b is of length 8, and angle alpha is 35 degrees.\" width=\"487\" height=\"225\" \/> <b>Figure 10<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"588123\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"588123\"]\r\n\r\nUse the Law of Sines to find angle [latex]\\beta [\/latex] and angle [latex]\\gamma [\/latex], and then side [latex]c[\/latex]. Solving for [latex]\\beta [\/latex], we have the proportion\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b} \\\\ \\frac{\\sin \\left(35^\\circ \\right)}{6}=\\frac{\\sin \\beta }{8}\\\\ \\frac{8\\sin \\left(35^\\circ \\right)}{6}=\\sin \\beta \\\\ 0.7648\\approx \\sin \\beta \\\\ {\\sin }^{-1}\\left(0.7648\\right)\\approx 49.9^\\circ \\\\ \\beta \\approx 49.9^\\circ \\end{gathered}[\/latex]<\/p>\r\nHowever, in the diagram, angle [latex]\\beta [\/latex] appears to be an obtuse angle and may be greater than 90\u00b0. How did we get an acute angle, and how do we find the measurement of [latex]\\beta ?[\/latex] Let\u2019s investigate further. Dropping a perpendicular from [latex]\\gamma [\/latex] and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165031\/CNX_Precalc_Figure_08_01_0112.jpg\" alt=\"An oblique triangle built from the previous with standard prime labels. Side a is of length 6, side b is of length 8, and angle alpha prime is 35 degrees. An isosceles triangle is attached, using side a as one of its congruent legs and the angle supplementary to angle beta as one of its congruent base angles. The other congruent angle is called beta prime, and the entire new horizontal base, which extends from the original side c, is called c prime. There is a dotted altitude line from angle gamma prime to side c prime.\" width=\"487\" height=\"248\" \/> <b>Figure 11<\/b>[\/caption]\r\n\r\nThe angle supplementary to [latex]\\beta [\/latex] is approximately equal to 49.9\u00b0, which means that [latex]\\beta =180^\\circ -49.9^\\circ =130.1^\\circ [\/latex]. (Remember that the sine function is positive in both the first and second quadrants.) Solving for [latex]\\gamma [\/latex], we have\r\n<p style=\"text-align: center\">[latex]\\gamma =180^\\circ -35^\\circ -130.1^\\circ \\approx 14.9^\\circ [\/latex]<\/p>\r\nWe can then use these measurements to solve the other triangle. Since [latex]{\\gamma }^{\\prime }[\/latex] is supplementary to [latex]\\gamma [\/latex], we have\r\n<p style=\"text-align: center\">[latex]{\\gamma }^{\\prime }=180^\\circ -35^\\circ -49.9^\\circ \\approx 95.1^\\circ [\/latex]<\/p>\r\nNow we need to find [latex]c[\/latex] and [latex]{c}^{\\prime }[\/latex].\r\n\r\nWe have\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\frac{c}{\\sin \\left(14.9^\\circ \\right)}=\\frac{6}{\\sin \\left(35^\\circ \\right)} \\\\ c=\\frac{6\\sin \\left(14.9^\\circ \\right)}{\\sin \\left(35^\\circ \\right)}\\approx 2.7 \\end{gathered}[\/latex]<\/p>\r\nFinally,\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\frac{{c}^{\\prime }}{\\sin \\left(95.1^\\circ \\right)}=\\frac{6}{\\sin \\left(35^\\circ \\right)} \\\\ {c}^{\\prime }=\\frac{6\\sin \\left(95.1^\\circ \\right)}{\\sin \\left(35^\\circ \\right)}\\approx 10.4 \\end{gathered}[\/latex]<\/p>\r\nTo summarize, there are two triangles with an angle of 35\u00b0, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165033\/CNX_Precalc_Figure_08_01_012ab2.jpg\" alt=\"There are two triangles with standard labels. Triangle a is the orginal triangle. It has angles alpha of 35 degrees, beta of 130.1 degrees, and gamma of 14.9 degrees. It has sides a = 6, b = 8, and c is approximately 2.7. Triangle b is the extended triangle. It has angles alpha prime = 35 degrees, angle beta prime = 49.9 degrees, and angle gamma prime = 95.1 degrees. It has side a prime = 6, side b prime = 8, and side c prime is approximately 10.4.\" width=\"731\" height=\"280\" \/> <b>Figure 12<\/b>[\/caption]\r\n\r\nHowever, we were looking for the values for the triangle with an obtuse angle [latex]\\beta [\/latex]. We can see them in the first triangle (a) in Figure 12.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]\\alpha =80^\\circ ,a=120[\/latex], and [latex]b=121[\/latex], find the missing side and angles. If there is more than one possible solution, show both.\r\n\r\n[reveal-answer q=\"65799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"65799\"]\r\n\r\n<strong>Solution 1<\/strong>\r\n[latex]\\begin{align}&amp;\\alpha =80^\\circ &amp;&amp; a=120\\hfill \\\\ &amp;\\beta \\approx 83.2^\\circ &amp;&amp; b=121 \\\\ &amp;\\gamma \\approx 16.8^\\circ &amp;&amp; c\\approx 35.2 \\end{align}[\/latex]\r\n<strong>Solution 2<\/strong>\r\n[latex]\\begin{align}&amp;{\\alpha }^{\\prime }=80^\\circ &amp;&amp;{a}^{\\prime }=120 \\\\ &amp;{\\beta }^{\\prime }\\approx 96.8^\\circ &amp;&amp;{b}^{\\prime }=121 \\\\ &amp;{\\gamma }^{\\prime }\\approx 3.2^\\circ &amp;&amp;{c}^{\\prime }\\approx 6.8 \\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Solving for the Unknown Sides and Angles of a SSA Triangle<\/h3>\r\nIn the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165035\/CNX_Precalc_Figure_08_01_0142.jpg\" alt=\"An oblique triangle with standard labels. Side b is 9, side c is 12, and angle gamma is 85. Angle alpha, angle beta, and side a are unknown.\" width=\"487\" height=\"212\" \/> <b>Figure 13<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"357327\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"357327\"]\r\n\r\nIn choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle [latex]\\gamma =85^\\circ [\/latex], and its corresponding side [latex]c=12[\/latex], and we know side [latex]b=9[\/latex]. We will use this proportion to solve for [latex]\\beta [\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\sin \\left(85^\\circ \\right)}{12}&amp;=\\frac{\\sin \\beta }{9} &amp;&amp; \\text{Isolate the unknown}.\\\\ \\frac{9\\sin \\left(85^\\circ \\right)}{12}&amp;=\\sin \\beta \\end{align}[\/latex]<\/p>\r\nTo find [latex]\\beta [\/latex], apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for [latex]\\beta [\/latex]. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\beta &amp;={\\sin }^{-1}\\left(\\frac{9\\sin \\left(85^\\circ \\right)}{12}\\right) \\\\ \\beta &amp;\\approx {\\sin }^{-1}\\left(0.7471\\right) \\\\ \\beta &amp;\\approx 48.3^\\circ \\end{align}[\/latex]<\/p>\r\nIn this case, if we subtract [latex]\\beta [\/latex] from 180\u00b0, we find that there may be a second possible solution. Thus, [latex]\\beta =180^\\circ -48.3^\\circ \\approx 131.7^\\circ [\/latex]. To check the solution, subtract both angles, 131.7\u00b0 and 85\u00b0, from 180\u00b0. This gives\r\n<p style=\"text-align: center\">[latex]\\alpha =180^\\circ -85^\\circ -131.7^\\circ \\approx -36.7^\\circ [\/latex],<\/p>\r\nwhich is impossible, and so [latex]\\beta \\approx 48.3^\\circ [\/latex].\r\n\r\nTo find the remaining missing values, we calculate [latex]\\alpha =180^\\circ -85^\\circ -48.3^\\circ \\approx 46.7^\\circ [\/latex]. Now, only side [latex]a[\/latex] is needed. Use the Law of Sines to solve for [latex]a[\/latex] by one of the proportions.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\frac{\\sin \\left(85^\\circ \\right)}{12}=\\frac{\\sin \\left(46.7^\\circ \\right)}{a} \\\\ a\\frac{\\sin \\left(85^\\circ \\right)}{12}=\\sin \\left(46.7^\\circ \\right) \\\\ a=\\frac{12\\sin \\left(46.7^\\circ \\right)}{\\sin \\left(85^\\circ \\right)}\\approx 8.8 \\end{gathered}[\/latex]<\/p>\r\nThe complete set of solutions for the given triangle is\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\alpha \\approx 46.7^\\circ \\text{, }a\\approx 8.8 \\\\ \\beta \\approx 48.3^\\circ \\text{, }b=9 \\\\ \\gamma =85^\\circ \\text{, }c=12\\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]\\alpha =80^\\circ ,a=100,b=10[\/latex], find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.\r\n\r\n[reveal-answer q=\"396947\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"396947\"]\r\n\r\n[latex]\\beta \\approx 5.7^\\circ ,\\gamma \\approx 94.3^\\circ ,c\\approx 101.3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Finding the Triangles That Meet the Given Criteria<\/h3>\r\nFind all possible triangles if one side has length 4 opposite an angle of 50\u00b0, and a second side has length 10.\r\n\r\n[reveal-answer q=\"979773\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979773\"]\r\n\r\nUsing the given information, we can solve for the angle opposite the side of length 10.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\frac{\\sin \\alpha }{10}=\\frac{\\sin \\left(50^\\circ \\right)}{4} \\\\ \\sin \\alpha =\\frac{10\\sin \\left(50^\\circ \\right)}{4} \\\\ \\sin \\alpha \\approx 1.915 \\end{gathered}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165038\/CNX_Precalc_Figure_08_01_0152.jpg\" alt=\"An incomplete triangle. One side has length 4 opposite a 50 degree angle, and a second side has length 10 opposite angle a. The side of length 4 is too short to reach the side of length 10, so there is no third angle.\" width=\"487\" height=\"220\" \/> <b>Figure 14<\/b>[\/caption]\r\n\r\nWe can stop here without finding the value of [latex]\\alpha [\/latex]. Because the range of the sine function is [latex]\\left[-1,1\\right][\/latex], it is impossible for the sine value to be 1.915. In fact, inputting [latex]{\\sin }^{-1}\\left(1.915\\right)[\/latex] in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nDetermine the number of triangles possible given [latex]a=31,b=26,\\beta =48^\\circ [\/latex].\r\n\r\n[reveal-answer q=\"996706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"996706\"]\r\n\r\nTwo\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149233[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding the Area of an Oblique Triangle Using the Sine Function<\/h2>\r\nNow that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as [latex]\\text{Area}=\\frac{1}{2}bh[\/latex], where [latex]b[\/latex] is base and [latex]h[\/latex] is height. For oblique triangles, we must find [latex]h[\/latex] before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property [latex]\\sin \\alpha =\\frac{\\text{opposite}}{\\text{hypotenuse}}[\/latex] to write an equation for area in oblique triangles. In the acute triangle, we have [latex]\\sin \\alpha =\\frac{h}{c}[\/latex] or [latex]c\\sin \\alpha =h[\/latex]. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base [latex]b[\/latex] to form a right triangle. The angle used in calculation is [latex]{\\alpha }^{\\prime }[\/latex], or [latex]180-\\alpha [\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165040\/CNX_Precalc_Figure_08_01_016.jpg\" alt=\"Two oblique triangles with standard labels. Both have a dotted altitude line h extended from angle beta to the horizontal base side b. In the first, which is an acute triangle, the altitude is within the triangle. In the second, which is an obtuse triangle, the altitude h is outside of the triangle. \" width=\"975\" height=\"235\" \/> <b>Figure 15<\/b>[\/caption]\r\n\r\nThus,\r\n<div style=\"text-align: center\">[latex]\\text{Area}=\\frac{1}{2}\\left(\\text{base}\\right)\\left(\\text{height}\\right)=\\frac{1}{2}b\\left(c\\sin \\alpha \\right)[\/latex]<\/div>\r\nSimilarly,\r\n<div style=\"text-align: center\">[latex]\\text{Area}=\\frac{1}{2}a\\left(b\\sin \\gamma \\right)=\\frac{1}{2}a\\left(c\\sin \\beta \\right)[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Area of an Oblique Triangle<\/h3>\r\nThe formula for the area of an oblique triangle is given by\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\text{Area}&amp;=\\frac{1}{2}bc\\sin \\alpha \\\\ &amp;=\\frac{1}{2}ac\\sin \\beta \\\\ &amp;=\\frac{1}{2}ab\\sin \\gamma \\end{align}[\/latex]<\/p>\r\nThis is equivalent to one-half of the product of two sides and the sine of their included angle.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Finding the Area of an Oblique Triangle<\/h3>\r\nFind the area of a triangle with sides [latex]a=90,b=52[\/latex], and angle [latex]\\gamma =102^\\circ [\/latex]. Round the area to the nearest integer.\r\n\r\n[reveal-answer q=\"99495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"99495\"]\r\n\r\nUsing the formula, we have\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\text{Area}&amp;=\\frac{1}{2}ab\\sin \\gamma \\\\ \\text{Area}&amp;=\\frac{1}{2}\\left(90\\right)\\left(52\\right)\\sin \\left(102^\\circ \\right) \\\\ \\text{Area}&amp;\\approx 2289\\text{square units} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the area of the triangle given [latex]\\beta =42^\\circ ,a=7.2\\text{ft},c=3.4\\text{ft}[\/latex]. Round the area to the nearest tenth.\r\n\r\n[reveal-answer q=\"426440\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426440\"]\r\n\r\nabout\u00a08.2 square feet\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]97465[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Applied Problems Using the Law of Sines<\/h2>\r\nThe more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Finding an Altitude<\/h3>\r\nFind the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165043\/CNX_Precalc_Figure_08_01_017.jpg\" alt=\"A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted altitude line perpendicular to the ground side connecting the airplane vertex with the ground.\" width=\"487\" height=\"134\" \/> <b>Figure 16<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"861681\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"861681\"]\r\n\r\nTo find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side [latex]a[\/latex], and then use right triangle relationships to find the height of the aircraft, [latex]h[\/latex].\r\n\r\nBecause the angles in the triangle add up to 180 degrees, the unknown angle must be 180\u00b0\u221215\u00b0\u221235\u00b0=130\u00b0. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\frac{\\sin \\left(130^\\circ \\right)}{20}=\\frac{\\sin \\left(35^\\circ \\right)}{a} \\\\ a\\sin \\left(130^\\circ \\right)=20\\sin \\left(35^\\circ \\right) \\\\ a=\\frac{20\\sin \\left(35^\\circ \\right)}{\\sin \\left(130^\\circ \\right)} \\\\ a\\approx 14.98 \\end{gathered}[\/latex]<\/p>\r\nThe distance from one station to the aircraft is about 14.98 miles.\r\n\r\nNow that we know [latex]a[\/latex], we can use right triangle relationships to solve for [latex]h[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin \\left(15^\\circ \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ \\sin \\left(15^\\circ \\right)=\\frac{h}{a} \\\\ \\sin \\left(15^\\circ \\right)=\\frac{h}{14.98} \\\\ h=14.98\\sin \\left(15^\\circ \\right) \\\\ h\\approx 3.88 \\end{gathered}[\/latex]<\/p>\r\nThe aircraft is at an altitude of approximately 3.9 miles.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nThe diagram shown in Figure 17 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70\u00b0, the angle of elevation from the northern end zone, point [latex]B[\/latex], is 62\u00b0, and the distance between the viewing points of the two end zones is 145 yards.\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165046\/CNX_Precalc_Figure_08_01_018.jpg\" alt=\"An oblique triangle formed from three vertices A, B, and C. Verticies A and B are points on the ground, and vertex C is the blimp in the air between them. The distance between A and B is 145 yards. The angle at vertex A is 70 degrees, and the angle at vertex B is 62 degrees.\" width=\"487\" height=\"535\" \/> <b>Figure 17<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"826578\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"826578\"]\r\n\r\n161.9 yd\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149308[\/ohm_question]\r\n\r\n<\/div>\r\nSuppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 1.\u00a0How far from port is the boat?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165212\/CNX_Precalc_Figure_08_02_0012.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\" \/> <b>Figure 18<\/b>[\/caption]\r\n\r\nUnfortunately, 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 <strong>SAS (side-angle-side) triangle<\/strong>, or when all three sides are known, but no angles are known, a <strong>SSS (side-side-side) triangle<\/strong>. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.\r\n<h2>Using the Law of Cosines to Solve Oblique Triangles<\/h2>\r\nThe 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.\r\n\r\nUnderstanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the <strong>Generalized Pythagorean Theorem<\/strong>, which is an extension of the <strong>Pythagorean Theorem<\/strong> 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 Figure 2. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165215\/CNX_Precalc_Figure_08_02_0022.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\" \/> <b>Figure 19<\/b>[\/caption]\r\n\r\nWe can drop a perpendicular from [latex]C[\/latex] to the <em>x-<\/em>axis (this is the altitude or height). Recalling the basic <strong>trigonometric identities<\/strong>, we know that\r\n<div style=\"text-align: center\">[latex]\\cos \\theta =\\frac{x\\text{(adjacent)}}{b\\text{(hypotenuse)}}\\text{ and }\\sin \\theta =\\frac{y\\text{(opposite)}}{b\\text{(hypotenuse)}}[\/latex]<\/div>\r\nIn terms of [latex]\\theta ,\\text{ }x=b\\cos \\theta [\/latex] and [latex]y=b\\sin \\theta .\\text{ }[\/latex] The [latex]\\left(x,y\\right)[\/latex] point located at [latex]C[\/latex] has coordinates [latex]\\left(b\\cos \\theta ,b\\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,\r\n<div style=\"text-align: center\">[latex]\\begin{align} {a}^{2}&amp;={\\left(x-c\\right)}^{2}+{y}^{2} \\\\ &amp;={\\left(b\\cos \\theta -c\\right)}^{2}+{\\left(b\\sin \\theta \\right)}^{2} &amp;&amp; \\text{Substitute }\\left(b\\cos \\theta \\right)\\text{ for}x\\text{and }\\left(b\\sin \\theta \\right)\\text{for }y. \\\\ &amp;=\\left({b}^{2}{\\cos }^{2}\\theta -2bc\\cos \\theta +{c}^{2}\\right)+{b}^{2}{\\sin }^{2}\\theta &amp;&amp; \\text{Expand the perfect square}. \\\\ &amp;={b}^{2}{\\cos }^{2}\\theta +{b}^{2}{\\sin }^{2}\\theta +{c}^{2}-2bc\\cos \\theta &amp;&amp; \\text{Group terms noting that }{\\cos }^{2}\\theta +{\\sin }^{2}\\theta =1. \\\\ &amp;={b}^{2}\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)+{c}^{2}-2bc\\cos \\theta &amp;&amp; \\text{Factor out }{b}^{2}. \\\\ {a}^{2}&amp;={b}^{2}+{c}^{2}-2bc\\cos \\theta &amp;&amp;{\\cos }^{2}\\theta +{\\sin }^{2}\\theta = 1 \\end{align}[\/latex]<\/div>\r\nThe formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.\r\n\r\nKeep 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.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Law of Cosines<\/h3>\r\nThe <strong>Law of Cosines<\/strong> 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 Figure 3, 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.\r\n<p style=\"text-align: center\">[latex]{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\alpha[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]{b}^{2}={a}^{2}+{c}^{2}-2ac\\cos \\beta[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]{c}^{2}={a}^{2}+{b}^{2}-2ab\\cos \\gamma [\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165217\/CNX_Precalc_Figure_08_02_003n2.jpg\" alt=\"A triangle with standard labels: angles alpha, beta, and gamma with opposite sides a, b, and c respectively.\" width=\"487\" height=\"239\" \/> <b>Figure 20<\/b>[\/caption]\r\n\r\nTo solve for a missing side measurement, the corresponding opposite angle measure is needed.\r\n\r\nWhen 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.\r\n<p style=\"text-align: center\">[latex]\\cos \\alpha =\\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\cos \\beta =\\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\cos\\gamma =\\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To:\u00a0Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.<\/h3>\r\n<ol id=\"fs-id1165132949849\">\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 <strong>Law of Sines<\/strong> 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 class=\"textbox shaded\">\r\n<h3>Example 7: Finding the Unknown Side and Angles of a SAS Triangle<\/h3>\r\nFind the unknown side and angles of the triangle in Figure 4.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165219\/CNX_Precalc_Figure_08_02_0042.jpg\" alt=\"A triangle with standard labels. Side a = 10, side c = 12, and angle beta = 30 degrees.\" width=\"487\" height=\"189\" \/> <b>Figure 21<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"359338\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"359338\"]\r\n\r\nFirst, 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.\r\n\r\nEach 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].\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;{b}^{2}={a}^{2}+{c}^{2}-2ac\\cos \\beta \\\\ &amp;{b}^{2}={10}^{2}+{12}^{2}-2\\left(10\\right)\\left(12\\right)\\cos \\left({30}^{\\circ }\\right) &amp;&amp; \\text{Substitute the measurements for the known quantities}. \\\\ &amp;{b}^{2}=100+144 - 240\\left(\\frac{\\sqrt{3}}{2}\\right)&amp;&amp; \\text{Evaluate the cosine and begin to simplify}. \\\\ &amp;{b}^{2}=244 - 120\\sqrt{3} \\\\ &amp;b=\\sqrt{244 - 120\\sqrt{3}}&amp;&amp; \\text{Use the square root property}. \\\\ &amp;b\\approx 6.013 \\end{align}[\/latex]<\/p>\r\nBecause 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\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b} \\\\ &amp;\\frac{\\sin \\alpha }{10}=\\frac{\\sin \\left(30^\\circ \\right)}{6.013} \\\\ &amp;\\sin \\alpha =\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}&amp;&amp; \\text{Multiply both sides of the equation by 10}. \\\\ &amp;\\alpha ={\\sin }^{-1}\\left(\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}\\right)&amp;&amp; \\text{Find the inverse sine of }\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}. \\\\ &amp;\\alpha \\approx 56.3^\\circ \\end{align}[\/latex]<\/p>\r\nThe other possibility for [latex]\\alpha [\/latex] would be [latex]\\alpha =180^\\circ -56.3^\\circ \\approx 123.7^\\circ [\/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^\\circ [\/latex] does not make sense. Notice that if we choose to apply the <strong>Law of Cosines<\/strong>, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between [latex]0^\\circ [\/latex] and [latex]180^\\circ [\/latex]. Proceeding with [latex]\\alpha \\approx 56.3^\\circ [\/latex], we can then find the third angle of the triangle.\r\n<p style=\"text-align: center\">[latex]\\gamma =180^\\circ -30^\\circ -56.3^\\circ \\approx 93.7^\\circ [\/latex]<\/p>\r\nThe complete set of angles and sides is\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;\\alpha \\approx 56.3^\\circ &amp;&amp; a=10\\\\ &amp;\\beta =30^\\circ &amp;&amp; b\\approx 6.013\\\\ &amp;\\gamma \\approx 93.7^\\circ &amp;&amp; c=12 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the missing side and angles of the given triangle: [latex]\\alpha =30^\\circ ,b=12,c=24[\/latex].\r\n\r\n[reveal-answer q=\"885435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"885435\"]\r\n\r\n[latex]a\\approx 14.9,\\beta \\approx 23.8^\\circ ,\\gamma \\approx 126.2^\\circ [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174837[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Solving for an Angle of a SSS Triangle<\/h3>\r\nFind 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].\r\n\r\n[reveal-answer q=\"211238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"211238\"]\r\n\r\nFor 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\r\n<p style=\"text-align: center\">[latex]\\begin{align} &amp;{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\alpha \\\\ &amp;{20}^{2}={25}^{2}+{18}^{2}-2\\left(25\\right)\\left(18\\right)\\cos \\alpha&amp;&amp; \\text{Substitute the appropriate measurements}. \\\\ &amp;400=625+324 - 900\\cos \\alpha&amp;&amp; \\text{Simplify in each step}. \\\\ &amp;400=949 - 900\\cos \\alpha \\\\ &amp;-549=-900\\cos \\alpha&amp;&amp; \\text{Isolate cos }\\alpha . \\\\ &amp;\\frac{-549}{-900}=\\cos \\alpha \\\\ &amp;0.61\\approx \\cos \\alpha \\\\ &amp;{\\cos }^{-1}\\left(0.61\\right)\\approx \\alpha&amp;&amp; \\text{Find the inverse cosine}. \\\\ &amp;\\alpha \\approx 52.4^\\circ \\end{align}[\/latex]<\/p>\r\nSee Figure 5.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165221\/CNX_Precalc_Figure_08_02_0052.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\" \/> <b>Figure 22<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nBecause the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]a=5,b=7[\/latex], and [latex]c=10[\/latex], find the missing angles.\r\n\r\n[reveal-answer q=\"488277\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"488277\"]\r\n\r\n[latex]\\alpha \\approx 27.7^\\circ ,\\beta \\approx 40.5^\\circ ,\\gamma \\approx 111.8^\\circ [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173789[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Applied Problems Using the Law of Cosines<\/h2>\r\nJust 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.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Using the Law of Cosines to Solve a Communication Problem<\/h3>\r\nOn 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.\r\n\r\n[reveal-answer q=\"960347\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"960347\"]\r\n\r\nFor simplicity, we start by drawing a diagram similar to Figure 6\u00a0and labeling our given information.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165224\/CNX_Precalc_Figure_08_02_0062.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\" \/> <b>Figure 23<\/b>[\/caption]\r\n\r\nUsing 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].\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\theta \\\\ {\\left(2420\\right)}^{2}={\\left(5050\\right)}^{2}+{\\left(6000\\right)}^{2}-2\\left(5050\\right)\\left(6000\\right)\\cos \\theta \\\\ {\\left(2420\\right)}^{2}-{\\left(5050\\right)}^{2}-{\\left(6000\\right)}^{2}=-2\\left(5050\\right)\\left(6000\\right)\\cos \\theta \\\\ \\frac{{\\left(2420\\right)}^{2}-{\\left(5050\\right)}^{2}-{\\left(6000\\right)}^{2}}{-2\\left(5050\\right)\\left(6000\\right)}=\\cos \\theta \\\\ \\cos \\theta \\approx 0.9183 \\\\ \\theta \\approx {\\cos }^{-1}\\left(0.9183\\right) \\\\ \\theta \\approx 23.3^\\circ \\end{gathered}[\/latex]<\/p>\r\nTo 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 Figure 7.\u00a0This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165226\/CNX_Precalc_Figure_08_02_0072.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\" \/> <b>Figure 24<\/b>[\/caption]\r\n\r\nUsing the angle [latex]\\theta =23.3^\\circ [\/latex] and the basic trigonometric identities, we can find the solutions. Thus\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\cos \\left(23.3^\\circ \\right)=\\frac{x}{5050} \\\\ x=5050\\cos \\left(23.3^\\circ \\right) \\\\ x\\approx 4638.15\\text{feet} \\\\ \\sin \\left(23.3^\\circ \\right)=\\frac{y}{5050} \\\\ y=5050\\sin \\left(23.3^\\circ \\right) \\\\ y\\approx 1997.5\\text{feet} \\end{gathered}[\/latex]<\/p>\r\nThe cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Calculating Distance Traveled Using a SAS Triangle<\/h3>\r\nReturning 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 Figure 8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165228\/CNX_Precalc_Figure_08_02_0092.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\" \/> <b>Figure 25<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"569516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"569516\"]\r\n\r\nThe boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, [latex]180^\\circ -20^\\circ =160^\\circ [\/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.\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;{x}^{2}={8}^{2}+{10}^{2}-2\\left(8\\right)\\left(10\\right)\\cos \\left(160^\\circ \\right) \\\\ &amp;{x}^{2}=314.35 \\\\ &amp;x=\\sqrt{314.35} \\\\ &amp;x\\approx 17.7\\text{miles} \\end{align}[\/latex]<\/p>\r\nThe boat is about 17.7 miles from port.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using Heron's Formula to Find the Area of a Triangle<\/h2>\r\nWe 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 <strong>Heron\u2019s formula<\/strong> instead of finding the height. <strong>Heron of Alexandria<\/strong> 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.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Heron\u2019s Formula<\/h3>\r\nHeron\u2019s formula finds the area of oblique triangles in which sides [latex]a,b[\/latex], and [latex]c[\/latex] are known.\r\n<p style=\"text-align: center\">[latex]\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex]<\/p>\r\nwhere [latex]s=\\frac{\\left(a+b+c\\right)}{2}[\/latex] is one half of the perimeter of the triangle, sometimes called the semi-perimeter.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Using Heron\u2019s Formula to Find the Area of a Given Triangle<\/h3>\r\nFind the area of the triangle in Figure 9\u00a0using Heron\u2019s formula.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165230\/CNX_Precalc_Figure_08_02_0102.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\" \/> <b>Figure 26<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"256710\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"256710\"]\r\n\r\nFirst, we calculate [latex]s[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{align} s&amp;=\\frac{\\left(a+b+c\\right)}{2} \\\\ s&amp;=\\frac{\\left(10+15+7\\right)}{2}=16 \\end{align}[\/latex]<\/p>\r\nThen we apply the formula.\r\n<p style=\"text-align: center\">[latex]\\begin{align} \\text{Area}&amp;=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)} \\\\ \\text{Area}&amp;=\\sqrt{16\\left(16 - 10\\right)\\left(16 - 15\\right)\\left(16 - 7\\right)} \\\\ \\text{Area}&amp;\\approx 29.4 \\end{align}[\/latex]<\/p>\r\nThe area is approximately 29.4 square units.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nUse 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].\r\n\r\n[reveal-answer q=\"159200\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"159200\"]\r\n\r\nArea = 552 square feet\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149312[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 12: Applying Heron\u2019s Formula to a Real-World Problem<\/h3>\r\nA 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 Figure 10\u00a0for a view of the city property.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165233\/CNX_Precalc_Figure_08_02_0112.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\" \/> <b>Figure 27<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"63725\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"63725\"]\r\n\r\nFind the measurement for [latex]s[\/latex], which is one-half of the perimeter.\r\n<p style=\"text-align: center\">[latex]\\begin{align}s&amp;=\\frac{\\left(62.4+43.5+34.1\\right)}{2} \\\\ s&amp;=70\\text{m} \\end{align}[\/latex]<\/p>\r\nApply Heron\u2019s formula.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\text{Area}&amp;=\\sqrt{70\\left(70 - 62.4\\right)\\left(70 - 43.5\\right)\\left(70 - 34.1\\right)} \\\\ \\text{Area}&amp;=\\sqrt{506,118.2} \\\\ \\text{Area}&amp;\\approx 711.4 \\end{align}[\/latex]<\/p>\r\nThe developer has about 711.4 square meters.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind 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]\r\n\r\n[reveal-answer q=\"942985\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"942985\"]\r\n\r\nabout 8.15 square feet\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]8473[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table id=\"eip-id2728866\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Law of Sines<\/td>\r\n<td>\r\n<p style=\"text-align: center\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\frac{a}{\\sin \\alpha }=\\frac{b}{\\sin \\beta }=\\frac{c}{\\sin \\gamma }[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Area for oblique triangles<\/td>\r\n<td>[latex]\\begin{align}\\text{Area}&amp;=\\frac{1}{2}bc\\sin \\alpha \\\\ &amp;=\\frac{1}{2}ac\\sin \\beta \\\\ &amp;=\\frac{1}{2}ab\\sin \\gamma \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table id=\"eip-id1956425\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Law of Cosines<\/td>\r\n<td>\r\n<p style=\"text-align: left\">[latex]{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\alpha[\/latex]<\/p>\r\n<p style=\"text-align: left\">[latex]{b}^{2}={a}^{2}+{c}^{2}-2ac\\cos \\beta[\/latex]<\/p>\r\n<p style=\"text-align: left\">[latex]{c}^{2}={a}^{2}+{b}^{2}-2ab\\cos \\gamma [\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Heron\u2019s formula<\/td>\r\n<td>[latex]\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex] where [latex]s=\\frac{\\left(a+b+c\\right)}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>The Law of Sines can be used to solve oblique triangles, which are non-right triangles.<\/li>\r\n \t<li>According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.<\/li>\r\n \t<li>There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution.<\/li>\r\n \t<li>The ambiguous case arises when an oblique triangle can have different outcomes.<\/li>\r\n \t<li>There are three possible cases that arise from SSA arrangement\u2014a single solution, two possible solutions, and no solution.<\/li>\r\n \t<li>The Law of Sines can be used to solve triangles with given criteria.<\/li>\r\n \t<li>The general area formula for triangles translates to oblique triangles by first finding the appropriate height value.<\/li>\r\n \t<li>There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation.<\/li>\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.<\/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.<\/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.<\/li>\r\n<\/ul>\r\n<div>\r\n<dl id=\"fs-id1165133234516\" class=\"definition\">\r\n \t<dd>\r\n<dl class=\"definition\">\r\n \t<dd>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135646148\" class=\"definition\">\r\n \t<dt>altitude<\/dt>\r\n \t<dd id=\"fs-id1165135646153\">a perpendicular line from one vertex of a triangle to the opposite side, or in the case of an obtuse triangle, to the line containing the opposite side, forming two right triangles<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134357482\" class=\"definition\">\r\n \t<dt>ambiguous case<\/dt>\r\n \t<dd id=\"fs-id1165134357487\">a scenario in which more than one triangle is a valid solution for a given oblique SSA triangle<\/dd>\r\n<\/dl>\r\n<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133184256\" class=\"definition\">\r\n \t<dt>\u00a0 \u00a0 \u00a0 Generalized Pythagorean Theorem<\/dt>\r\n \t<dd id=\"fs-id1165133184261\">\u00a0 \u00a0 \u00a0 \u00a0an extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0SSS triangles<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dd>\r\n<dl id=\"fs-id1165134357482\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165135690077\" class=\"definition\">\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<\/dt>\r\n<\/dl>\r\n<dl id=\"fs-id1165134357493\" class=\"definition\">\r\n \t<dt>Law of Sines<\/dt>\r\n \t<dd id=\"fs-id1165133234509\">states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133234516\" class=\"definition\">\r\n \t<dt>oblique triangle<\/dt>\r\n \t<dd id=\"fs-id1165135551173\">any triangle that is not a right triangle<\/dd>\r\n<\/dl>\r\n<\/dd>\r\n<\/dl>\r\n&nbsp;\r\n\r\n&nbsp;\r\n<dl id=\"fs-id1165133234516\" class=\"definition\">\r\n \t<dd id=\"fs-id1165135551173\"><\/dd>\r\n<\/dl>\r\n<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Use the Law of Sines to solve oblique triangles.<\/li>\n<li>Find the area of an oblique triangle using the sine function.<\/li>\n<li>Solve applied problems using the Law of Sines.<\/li>\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>&nbsp;<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165008\/CNX_Precalc_Figure_08_01_0012.jpg\" alt=\"A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted line perpendicular to the ground side connecting the airplane vertex with the ground - an altitude line.\" width=\"487\" height=\"134\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<h2>Using the Law of Sines to Solve Obliques Triangles<\/h2>\n<p>In any triangle, we can draw an <strong>altitude<\/strong>, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.<\/p>\n<p>Any triangle that is not a right triangle is an <strong>oblique triangle<\/strong>. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:<\/p>\n<ol>\n<li><strong>ASA (angle-side-angle)<\/strong> We know the measurements of two angles and the included side. See Figure 2.\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165011\/CNX_Precalc_Figure_08_01_0022.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma's values are known, as is the side opposite beta, between alpha and gamma.\" width=\"487\" height=\"141\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<\/li>\n<li><strong>AAS (angle-angle-side)<\/strong> We know the measurements of two angles and a side that is not between the known angles. See Figure 3.\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165013\/CNX_Precalc_Figure_08_01_0032.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma are known, as is the side opposite alpha, between beta and gamma.\" width=\"487\" height=\"141\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/li>\n<li><strong>SSA (side-side-angle)<\/strong> We know the measurements of two sides and an angle that is not between the known sides. See Figure 4.\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165016\/CNX_Precalc_Figure_08_01_0042.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha is the only angle known. Two sides are known. The first is opposite alpha, between beta and gamma, and the second is opposite gamma, between alpha and beta.\" width=\"487\" height=\"141\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<p>Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let\u2019s see how this statement is derived by considering the triangle shown in Figure 5.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165018\/CNX_Precalc_Figure_08_01_0052.jpg\" alt=\"An oblique triangle consisting of sides a, b, and c, and angles alpha, beta, and gamma. Side c is opposide angle gamma and is the horizontal base of the triangle. Side b is opposite angle beta, and side a is opposite angle alpha. There is a dotted perpendicular line - an altitude - from the gamma angle to the horizontal base c.\" width=\"487\" height=\"143\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<p>Using the right triangle relationships, we know that [latex]\\sin \\alpha =\\frac{h}{b}[\/latex] and [latex]\\sin \\beta =\\frac{h}{a}[\/latex]. Solving both equations for [latex]h[\/latex] gives two different expressions for [latex]h[\/latex].<\/p>\n<div style=\"text-align: center\">[latex]h=b\\sin \\alpha \\text{ and }h=a\\sin \\beta[\/latex]<\/div>\n<p>We then set the expressions equal to each other.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}b\\sin \\alpha &=a\\sin \\beta \\\\ \\left(\\frac{1}{ab}\\right)\\left(b\\sin \\alpha \\right)&=\\left(a\\sin \\beta \\right)\\left(\\frac{1}{ab}\\right) && \\text{Multiply both sides by}\\frac{1}{ab}. \\\\ \\frac{\\sin \\alpha }{a}&=\\frac{\\sin \\beta }{b} \\end{align}[\/latex]<\/div>\n<p>Similarly, we can compare the other ratios.<\/p>\n<div style=\"text-align: center\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\gamma }{c}\\text{ and }\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/div>\n<p>Collectively, these relationships are called the <strong>Law of Sines<\/strong>.<\/p>\n<div style=\"text-align: center\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\lambda }{c}[\/latex]<\/div>\n<p>Note the standard way of labeling triangles: angle [latex]\\alpha[\/latex] (alpha) is opposite side [latex]a[\/latex]; angle [latex]\\beta[\/latex] (beta) is opposite side [latex]b[\/latex]; and angle [latex]\\gamma[\/latex] (gamma) is opposite side [latex]c[\/latex]. See Figure 6.<\/p>\n<p>While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165020\/CNX_Precalc_Figure_08_01_0062.jpg\" alt=\"A triangle with standard labels.\" width=\"487\" height=\"197\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Law of Sines<\/h3>\n<p>Given a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The <strong>Law of Sines<\/strong> is based on proportions and is presented symbolically two ways.<\/p>\n<p style=\"text-align: center\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\frac{a}{\\sin \\alpha }=\\frac{b}{\\sin \\beta }=\\frac{c}{\\sin \\gamma }[\/latex]<\/p>\n<p>To solve an oblique triangle, use any pair of applicable ratios.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Solving for Two Unknown Sides and Angle of an AAS Triangle<\/h3>\n<p>Solve the triangle shown in Figure 7\u00a0to the nearest tenth.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165022\/CNX_Precalc_Figure_08_01_0072.jpg\" alt=\"An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.\" width=\"487\" height=\"200\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604625\">Show Solution<\/span><\/p>\n<div id=\"q604625\" class=\"hidden-answer\" style=\"display: none\">\n<p>The three angles must add up to 180 degrees. From this, we can determine that<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\beta =180^\\circ -50^\\circ -30^\\circ =100^\\circ \\end{align}[\/latex]<\/p>\n<p>To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle [latex]\\alpha =50^\\circ[\/latex] and its corresponding side [latex]a=10[\/latex]. We can use the following proportion from the Law of Sines to find the length of [latex]c[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\frac{\\sin \\left(30^\\circ \\right)}{c} \\\\ &c\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\sin \\left(30^\\circ \\right) && \\text{Multiply both sides by }c. \\\\ &c=\\sin \\left(30^\\circ \\right)\\frac{10}{\\sin \\left(50^\\circ \\right)} && \\text{Multiply by the reciprocal to isolate }c. \\\\ &c\\approx 6.5 \\end{align}[\/latex]<\/p>\n<p>Similarly, to solve for [latex]b[\/latex], we set up another proportion.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} &\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\frac{\\sin \\left(100^\\circ \\right)}{b} \\\\ &b\\sin \\left(50^\\circ \\right)=10\\sin \\left(100^\\circ \\right) && \\text{Multiply both sides by }b. \\\\ &b=\\frac{10\\sin \\left(100^\\circ \\right)}{\\sin \\left(50^\\circ \\right)} && \\text{Multiply by the reciprocal to isolate }b. \\\\ &b\\approx 12.9\\end{align}[\/latex]<\/p>\n<p>Therefore, the complete set of angles and sides is<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\alpha =50^\\circ,\\beta =100^\\circ,\\gamma =30^\\circ \\\\ a=10,b\\approx 12.9,c\\approx 6.5 \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve the triangle shown in Figure 8\u00a0to the nearest tenth.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165024\/CNX_Precalc_Figure_08_01_0082.jpg\" alt=\"An oblique triangle with standard labels. Angle alpha is 98 degrees, angle gamma is 43 degrees, and side b is of length 22. Side b is the horizontal base.\" width=\"487\" height=\"247\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q933392\">Show Solution<\/span><\/p>\n<div id=\"q933392\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{gathered}\\alpha ={98}^{\\circ },\\beta ={39}^{\\circ },\\gamma ={43}^{\\circ } \\\\ a=34.6, b=22, c=23.8\\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149230\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149230&theme=oea&iframe_resize_id=ohm149230\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using The Law of Sines to Solve SSA Triangles<\/h2>\n<p>We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an <strong>ambiguous case<\/strong>. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Possible Outcomes for SSA Triangles<\/h3>\n<p>Oblique triangles in the category SSA may have four different outcomes.\u00a0Figure 9\u00a0illustrates the solutions with the known sides [latex]a[\/latex] and [latex]b[\/latex] and known angle [latex]\\alpha[\/latex].<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165027\/CNX_Precalc_Figure_08_01_009n2.jpg\" alt=\"Four attempted oblique triangles are in a row, all with standard labels. Side c is the horizontal base. In the first attempted triangle, side a is less than the altitude height. Since side a cannot reach side c, there is no triangle. In the second attempted triangle, side a is equal to the length of the altitude height, so side a forms a right angle with side c. In the third attempted triangle, side a is greater than the altitude height and less than side b, so side a can form either an acute or obtuse angle with side c. In the fourth attempted triangle, side a is greater than or equal to side b, so side a forms an acute angle with side c.\" width=\"731\" height=\"483\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Solving an Oblique SSA Triangle<\/h3>\n<p>Solve the triangle in Figure 10\u00a0for the missing side and find the missing angle measures to the nearest tenth.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165029\/CNX_Precalc_Figure_08_01_0102.jpg\" alt=\"An oblique triangle with standard labels where side a is of length 6, side b is of length 8, and angle alpha is 35 degrees.\" width=\"487\" height=\"225\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q588123\">Show Solution<\/span><\/p>\n<div id=\"q588123\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the Law of Sines to find angle [latex]\\beta[\/latex] and angle [latex]\\gamma[\/latex], and then side [latex]c[\/latex]. Solving for [latex]\\beta[\/latex], we have the proportion<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b} \\\\ \\frac{\\sin \\left(35^\\circ \\right)}{6}=\\frac{\\sin \\beta }{8}\\\\ \\frac{8\\sin \\left(35^\\circ \\right)}{6}=\\sin \\beta \\\\ 0.7648\\approx \\sin \\beta \\\\ {\\sin }^{-1}\\left(0.7648\\right)\\approx 49.9^\\circ \\\\ \\beta \\approx 49.9^\\circ \\end{gathered}[\/latex]<\/p>\n<p>However, in the diagram, angle [latex]\\beta[\/latex] appears to be an obtuse angle and may be greater than 90\u00b0. How did we get an acute angle, and how do we find the measurement of [latex]\\beta ?[\/latex] Let\u2019s investigate further. Dropping a perpendicular from [latex]\\gamma[\/latex] and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165031\/CNX_Precalc_Figure_08_01_0112.jpg\" alt=\"An oblique triangle built from the previous with standard prime labels. Side a is of length 6, side b is of length 8, and angle alpha prime is 35 degrees. An isosceles triangle is attached, using side a as one of its congruent legs and the angle supplementary to angle beta as one of its congruent base angles. The other congruent angle is called beta prime, and the entire new horizontal base, which extends from the original side c, is called c prime. There is a dotted altitude line from angle gamma prime to side c prime.\" width=\"487\" height=\"248\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<p>The angle supplementary to [latex]\\beta[\/latex] is approximately equal to 49.9\u00b0, which means that [latex]\\beta =180^\\circ -49.9^\\circ =130.1^\\circ[\/latex]. (Remember that the sine function is positive in both the first and second quadrants.) Solving for [latex]\\gamma[\/latex], we have<\/p>\n<p style=\"text-align: center\">[latex]\\gamma =180^\\circ -35^\\circ -130.1^\\circ \\approx 14.9^\\circ[\/latex]<\/p>\n<p>We can then use these measurements to solve the other triangle. Since [latex]{\\gamma }^{\\prime }[\/latex] is supplementary to [latex]\\gamma[\/latex], we have<\/p>\n<p style=\"text-align: center\">[latex]{\\gamma }^{\\prime }=180^\\circ -35^\\circ -49.9^\\circ \\approx 95.1^\\circ[\/latex]<\/p>\n<p>Now we need to find [latex]c[\/latex] and [latex]{c}^{\\prime }[\/latex].<\/p>\n<p>We have<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\frac{c}{\\sin \\left(14.9^\\circ \\right)}=\\frac{6}{\\sin \\left(35^\\circ \\right)} \\\\ c=\\frac{6\\sin \\left(14.9^\\circ \\right)}{\\sin \\left(35^\\circ \\right)}\\approx 2.7 \\end{gathered}[\/latex]<\/p>\n<p>Finally,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\frac{{c}^{\\prime }}{\\sin \\left(95.1^\\circ \\right)}=\\frac{6}{\\sin \\left(35^\\circ \\right)} \\\\ {c}^{\\prime }=\\frac{6\\sin \\left(95.1^\\circ \\right)}{\\sin \\left(35^\\circ \\right)}\\approx 10.4 \\end{gathered}[\/latex]<\/p>\n<p>To summarize, there are two triangles with an angle of 35\u00b0, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165033\/CNX_Precalc_Figure_08_01_012ab2.jpg\" alt=\"There are two triangles with standard labels. Triangle a is the orginal triangle. It has angles alpha of 35 degrees, beta of 130.1 degrees, and gamma of 14.9 degrees. It has sides a = 6, b = 8, and c is approximately 2.7. Triangle b is the extended triangle. It has angles alpha prime = 35 degrees, angle beta prime = 49.9 degrees, and angle gamma prime = 95.1 degrees. It has side a prime = 6, side b prime = 8, and side c prime is approximately 10.4.\" width=\"731\" height=\"280\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12<\/b><\/p>\n<\/div>\n<p>However, we were looking for the values for the triangle with an obtuse angle [latex]\\beta[\/latex]. We can see them in the first triangle (a) in Figure 12.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given [latex]\\alpha =80^\\circ ,a=120[\/latex], and [latex]b=121[\/latex], find the missing side and angles. If there is more than one possible solution, show both.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q65799\">Show Solution<\/span><\/p>\n<div id=\"q65799\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Solution 1<\/strong><br \/>\n[latex]\\begin{align}&\\alpha =80^\\circ && a=120\\hfill \\\\ &\\beta \\approx 83.2^\\circ && b=121 \\\\ &\\gamma \\approx 16.8^\\circ && c\\approx 35.2 \\end{align}[\/latex]<br \/>\n<strong>Solution 2<\/strong><br \/>\n[latex]\\begin{align}&{\\alpha }^{\\prime }=80^\\circ &&{a}^{\\prime }=120 \\\\ &{\\beta }^{\\prime }\\approx 96.8^\\circ &&{b}^{\\prime }=121 \\\\ &{\\gamma }^{\\prime }\\approx 3.2^\\circ &&{c}^{\\prime }\\approx 6.8 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Solving for the Unknown Sides and Angles of a SSA Triangle<\/h3>\n<p>In the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165035\/CNX_Precalc_Figure_08_01_0142.jpg\" alt=\"An oblique triangle with standard labels. Side b is 9, side c is 12, and angle gamma is 85. Angle alpha, angle beta, and side a are unknown.\" width=\"487\" height=\"212\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 13<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q357327\">Show Solution<\/span><\/p>\n<div id=\"q357327\" class=\"hidden-answer\" style=\"display: none\">\n<p>In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle [latex]\\gamma =85^\\circ[\/latex], and its corresponding side [latex]c=12[\/latex], and we know side [latex]b=9[\/latex]. We will use this proportion to solve for [latex]\\beta[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\sin \\left(85^\\circ \\right)}{12}&=\\frac{\\sin \\beta }{9} && \\text{Isolate the unknown}.\\\\ \\frac{9\\sin \\left(85^\\circ \\right)}{12}&=\\sin \\beta \\end{align}[\/latex]<\/p>\n<p>To find [latex]\\beta[\/latex], apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for [latex]\\beta[\/latex]. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\beta &={\\sin }^{-1}\\left(\\frac{9\\sin \\left(85^\\circ \\right)}{12}\\right) \\\\ \\beta &\\approx {\\sin }^{-1}\\left(0.7471\\right) \\\\ \\beta &\\approx 48.3^\\circ \\end{align}[\/latex]<\/p>\n<p>In this case, if we subtract [latex]\\beta[\/latex] from 180\u00b0, we find that there may be a second possible solution. Thus, [latex]\\beta =180^\\circ -48.3^\\circ \\approx 131.7^\\circ[\/latex]. To check the solution, subtract both angles, 131.7\u00b0 and 85\u00b0, from 180\u00b0. This gives<\/p>\n<p style=\"text-align: center\">[latex]\\alpha =180^\\circ -85^\\circ -131.7^\\circ \\approx -36.7^\\circ[\/latex],<\/p>\n<p>which is impossible, and so [latex]\\beta \\approx 48.3^\\circ[\/latex].<\/p>\n<p>To find the remaining missing values, we calculate [latex]\\alpha =180^\\circ -85^\\circ -48.3^\\circ \\approx 46.7^\\circ[\/latex]. Now, only side [latex]a[\/latex] is needed. Use the Law of Sines to solve for [latex]a[\/latex] by one of the proportions.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\frac{\\sin \\left(85^\\circ \\right)}{12}=\\frac{\\sin \\left(46.7^\\circ \\right)}{a} \\\\ a\\frac{\\sin \\left(85^\\circ \\right)}{12}=\\sin \\left(46.7^\\circ \\right) \\\\ a=\\frac{12\\sin \\left(46.7^\\circ \\right)}{\\sin \\left(85^\\circ \\right)}\\approx 8.8 \\end{gathered}[\/latex]<\/p>\n<p>The complete set of solutions for the given triangle is<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\alpha \\approx 46.7^\\circ \\text{, }a\\approx 8.8 \\\\ \\beta \\approx 48.3^\\circ \\text{, }b=9 \\\\ \\gamma =85^\\circ \\text{, }c=12\\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given [latex]\\alpha =80^\\circ ,a=100,b=10[\/latex], find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q396947\">Show Solution<\/span><\/p>\n<div id=\"q396947\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\beta \\approx 5.7^\\circ ,\\gamma \\approx 94.3^\\circ ,c\\approx 101.3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Finding the Triangles That Meet the Given Criteria<\/h3>\n<p>Find all possible triangles if one side has length 4 opposite an angle of 50\u00b0, and a second side has length 10.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q979773\">Show Solution<\/span><\/p>\n<div id=\"q979773\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the given information, we can solve for the angle opposite the side of length 10.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\frac{\\sin \\alpha }{10}=\\frac{\\sin \\left(50^\\circ \\right)}{4} \\\\ \\sin \\alpha =\\frac{10\\sin \\left(50^\\circ \\right)}{4} \\\\ \\sin \\alpha \\approx 1.915 \\end{gathered}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165038\/CNX_Precalc_Figure_08_01_0152.jpg\" alt=\"An incomplete triangle. One side has length 4 opposite a 50 degree angle, and a second side has length 10 opposite angle a. The side of length 4 is too short to reach the side of length 10, so there is no third angle.\" width=\"487\" height=\"220\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14<\/b><\/p>\n<\/div>\n<p>We can stop here without finding the value of [latex]\\alpha[\/latex]. Because the range of the sine function is [latex]\\left[-1,1\\right][\/latex], it is impossible for the sine value to be 1.915. In fact, inputting [latex]{\\sin }^{-1}\\left(1.915\\right)[\/latex] in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Determine the number of triangles possible given [latex]a=31,b=26,\\beta =48^\\circ[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q996706\">Show Solution<\/span><\/p>\n<div id=\"q996706\" class=\"hidden-answer\" style=\"display: none\">\n<p>Two<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149233\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149233&theme=oea&iframe_resize_id=ohm149233\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding the Area of an Oblique Triangle Using the Sine Function<\/h2>\n<p>Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as [latex]\\text{Area}=\\frac{1}{2}bh[\/latex], where [latex]b[\/latex] is base and [latex]h[\/latex] is height. For oblique triangles, we must find [latex]h[\/latex] before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property [latex]\\sin \\alpha =\\frac{\\text{opposite}}{\\text{hypotenuse}}[\/latex] to write an equation for area in oblique triangles. In the acute triangle, we have [latex]\\sin \\alpha =\\frac{h}{c}[\/latex] or [latex]c\\sin \\alpha =h[\/latex]. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base [latex]b[\/latex] to form a right triangle. The angle used in calculation is [latex]{\\alpha }^{\\prime }[\/latex], or [latex]180-\\alpha[\/latex].<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165040\/CNX_Precalc_Figure_08_01_016.jpg\" alt=\"Two oblique triangles with standard labels. Both have a dotted altitude line h extended from angle beta to the horizontal base side b. In the first, which is an acute triangle, the altitude is within the triangle. In the second, which is an obtuse triangle, the altitude h is outside of the triangle.\" width=\"975\" height=\"235\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 15<\/b><\/p>\n<\/div>\n<p>Thus,<\/p>\n<div style=\"text-align: center\">[latex]\\text{Area}=\\frac{1}{2}\\left(\\text{base}\\right)\\left(\\text{height}\\right)=\\frac{1}{2}b\\left(c\\sin \\alpha \\right)[\/latex]<\/div>\n<p>Similarly,<\/p>\n<div style=\"text-align: center\">[latex]\\text{Area}=\\frac{1}{2}a\\left(b\\sin \\gamma \\right)=\\frac{1}{2}a\\left(c\\sin \\beta \\right)[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Area of an Oblique Triangle<\/h3>\n<p>The formula for the area of an oblique triangle is given by<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\text{Area}&=\\frac{1}{2}bc\\sin \\alpha \\\\ &=\\frac{1}{2}ac\\sin \\beta \\\\ &=\\frac{1}{2}ab\\sin \\gamma \\end{align}[\/latex]<\/p>\n<p>This is equivalent to one-half of the product of two sides and the sine of their included angle.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Finding the Area of an Oblique Triangle<\/h3>\n<p>Find the area of a triangle with sides [latex]a=90,b=52[\/latex], and angle [latex]\\gamma =102^\\circ[\/latex]. Round the area to the nearest integer.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q99495\">Show Solution<\/span><\/p>\n<div id=\"q99495\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the formula, we have<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\text{Area}&=\\frac{1}{2}ab\\sin \\gamma \\\\ \\text{Area}&=\\frac{1}{2}\\left(90\\right)\\left(52\\right)\\sin \\left(102^\\circ \\right) \\\\ \\text{Area}&\\approx 2289\\text{square units} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the area of the triangle given [latex]\\beta =42^\\circ ,a=7.2\\text{ft},c=3.4\\text{ft}[\/latex]. Round the area to the nearest tenth.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q426440\">Show Solution<\/span><\/p>\n<div id=\"q426440\" class=\"hidden-answer\" style=\"display: none\">\n<p>about\u00a08.2 square feet<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm97465\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=97465&theme=oea&iframe_resize_id=ohm97465\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Applied Problems Using the Law of Sines<\/h2>\n<p>The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 6: Finding an Altitude<\/h3>\n<p>Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165043\/CNX_Precalc_Figure_08_01_017.jpg\" alt=\"A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted altitude line perpendicular to the ground side connecting the airplane vertex with the ground.\" width=\"487\" height=\"134\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 16<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q861681\">Show Solution<\/span><\/p>\n<div id=\"q861681\" class=\"hidden-answer\" style=\"display: none\">\n<p>To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side [latex]a[\/latex], and then use right triangle relationships to find the height of the aircraft, [latex]h[\/latex].<\/p>\n<p>Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180\u00b0\u221215\u00b0\u221235\u00b0=130\u00b0. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\frac{\\sin \\left(130^\\circ \\right)}{20}=\\frac{\\sin \\left(35^\\circ \\right)}{a} \\\\ a\\sin \\left(130^\\circ \\right)=20\\sin \\left(35^\\circ \\right) \\\\ a=\\frac{20\\sin \\left(35^\\circ \\right)}{\\sin \\left(130^\\circ \\right)} \\\\ a\\approx 14.98 \\end{gathered}[\/latex]<\/p>\n<p>The distance from one station to the aircraft is about 14.98 miles.<\/p>\n<p>Now that we know [latex]a[\/latex], we can use right triangle relationships to solve for [latex]h[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin \\left(15^\\circ \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ \\sin \\left(15^\\circ \\right)=\\frac{h}{a} \\\\ \\sin \\left(15^\\circ \\right)=\\frac{h}{14.98} \\\\ h=14.98\\sin \\left(15^\\circ \\right) \\\\ h\\approx 3.88 \\end{gathered}[\/latex]<\/p>\n<p>The aircraft is at an altitude of approximately 3.9 miles.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>The diagram shown in Figure 17 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70\u00b0, the angle of elevation from the northern end zone, point [latex]B[\/latex], is 62\u00b0, and the distance between the viewing points of the two end zones is 145 yards.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165046\/CNX_Precalc_Figure_08_01_018.jpg\" alt=\"An oblique triangle formed from three vertices A, B, and C. Verticies A and B are points on the ground, and vertex C is the blimp in the air between them. The distance between A and B is 145 yards. The angle at vertex A is 70 degrees, and the angle at vertex B is 62 degrees.\" width=\"487\" height=\"535\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 17<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q826578\">Show Solution<\/span><\/p>\n<div id=\"q826578\" class=\"hidden-answer\" style=\"display: none\">\n<p>161.9 yd<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149308\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149308&theme=oea&iframe_resize_id=ohm149308\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 1.\u00a0How far from port is the boat?<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165212\/CNX_Precalc_Figure_08_02_0012.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\"><b>Figure 18<\/b><\/p>\n<\/div>\n<p>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 <strong>SAS (side-angle-side) triangle<\/strong>, or when all three sides are known, but no angles are known, a <strong>SSS (side-side-side) triangle<\/strong>. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.<\/p>\n<h2>Using the Law of Cosines to Solve Oblique Triangles<\/h2>\n<p>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>Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the <strong>Generalized Pythagorean Theorem<\/strong>, which is an extension of the <strong>Pythagorean Theorem<\/strong> 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 Figure 2. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165215\/CNX_Precalc_Figure_08_02_0022.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\"><b>Figure 19<\/b><\/p>\n<\/div>\n<p>We can drop a perpendicular from [latex]C[\/latex] to the <em>x-<\/em>axis (this is the altitude or height). Recalling the basic <strong>trigonometric identities<\/strong>, we know that<\/p>\n<div style=\"text-align: center\">[latex]\\cos \\theta =\\frac{x\\text{(adjacent)}}{b\\text{(hypotenuse)}}\\text{ and }\\sin \\theta =\\frac{y\\text{(opposite)}}{b\\text{(hypotenuse)}}[\/latex]<\/div>\n<p>In terms of [latex]\\theta ,\\text{ }x=b\\cos \\theta[\/latex] and [latex]y=b\\sin \\theta .\\text{ }[\/latex] The [latex]\\left(x,y\\right)[\/latex] point located at [latex]C[\/latex] has coordinates [latex]\\left(b\\cos \\theta ,b\\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 style=\"text-align: center\">[latex]\\begin{align} {a}^{2}&={\\left(x-c\\right)}^{2}+{y}^{2} \\\\ &={\\left(b\\cos \\theta -c\\right)}^{2}+{\\left(b\\sin \\theta \\right)}^{2} && \\text{Substitute }\\left(b\\cos \\theta \\right)\\text{ for}x\\text{and }\\left(b\\sin \\theta \\right)\\text{for }y. \\\\ &=\\left({b}^{2}{\\cos }^{2}\\theta -2bc\\cos \\theta +{c}^{2}\\right)+{b}^{2}{\\sin }^{2}\\theta && \\text{Expand the perfect square}. \\\\ &={b}^{2}{\\cos }^{2}\\theta +{b}^{2}{\\sin }^{2}\\theta +{c}^{2}-2bc\\cos \\theta && \\text{Group terms noting that }{\\cos }^{2}\\theta +{\\sin }^{2}\\theta =1. \\\\ &={b}^{2}\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)+{c}^{2}-2bc\\cos \\theta && \\text{Factor out }{b}^{2}. \\\\ {a}^{2}&={b}^{2}+{c}^{2}-2bc\\cos \\theta &&{\\cos }^{2}\\theta +{\\sin }^{2}\\theta = 1 \\end{align}[\/latex]<\/div>\n<p>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>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\">\n<h3>A General Note: Law of Cosines<\/h3>\n<p>The <strong>Law of Cosines<\/strong> 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 Figure 3, 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<p style=\"text-align: center\">[latex]{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\alpha[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]{b}^{2}={a}^{2}+{c}^{2}-2ac\\cos \\beta[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]{c}^{2}={a}^{2}+{b}^{2}-2ab\\cos \\gamma[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165217\/CNX_Precalc_Figure_08_02_003n2.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\"><b>Figure 20<\/b><\/p>\n<\/div>\n<p>To solve for a missing side measurement, the corresponding opposite angle measure is needed.<\/p>\n<p>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<p style=\"text-align: center\">[latex]\\cos \\alpha =\\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\cos \\beta =\\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\cos\\gamma =\\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To:\u00a0Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.<\/h3>\n<ol id=\"fs-id1165132949849\">\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 <strong>Law of Sines<\/strong> 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 class=\"textbox shaded\">\n<h3>Example 7: Finding the Unknown Side and Angles of a SAS Triangle<\/h3>\n<p>Find the unknown side and angles of the triangle in Figure 4.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165219\/CNX_Precalc_Figure_08_02_0042.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\"><b>Figure 21<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q359338\">Show Solution<\/span><\/p>\n<div id=\"q359338\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>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<p style=\"text-align: center\">[latex]\\begin{align}&{b}^{2}={a}^{2}+{c}^{2}-2ac\\cos \\beta \\\\ &{b}^{2}={10}^{2}+{12}^{2}-2\\left(10\\right)\\left(12\\right)\\cos \\left({30}^{\\circ }\\right) && \\text{Substitute the measurements for the known quantities}. \\\\ &{b}^{2}=100+144 - 240\\left(\\frac{\\sqrt{3}}{2}\\right)&& \\text{Evaluate the cosine and begin to simplify}. \\\\ &{b}^{2}=244 - 120\\sqrt{3} \\\\ &b=\\sqrt{244 - 120\\sqrt{3}}&& \\text{Use the square root property}. \\\\ &b\\approx 6.013 \\end{align}[\/latex]<\/p>\n<p>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<p style=\"text-align: center\">[latex]\\begin{align}&\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b} \\\\ &\\frac{\\sin \\alpha }{10}=\\frac{\\sin \\left(30^\\circ \\right)}{6.013} \\\\ &\\sin \\alpha =\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}&& \\text{Multiply both sides of the equation by 10}. \\\\ &\\alpha ={\\sin }^{-1}\\left(\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}\\right)&& \\text{Find the inverse sine of }\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}. \\\\ &\\alpha \\approx 56.3^\\circ \\end{align}[\/latex]<\/p>\n<p>The other possibility for [latex]\\alpha[\/latex] would be [latex]\\alpha =180^\\circ -56.3^\\circ \\approx 123.7^\\circ[\/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^\\circ[\/latex] does not make sense. Notice that if we choose to apply the <strong>Law of Cosines<\/strong>, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between [latex]0^\\circ[\/latex] and [latex]180^\\circ[\/latex]. Proceeding with [latex]\\alpha \\approx 56.3^\\circ[\/latex], we can then find the third angle of the triangle.<\/p>\n<p style=\"text-align: center\">[latex]\\gamma =180^\\circ -30^\\circ -56.3^\\circ \\approx 93.7^\\circ[\/latex]<\/p>\n<p>The complete set of angles and sides is<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&\\alpha \\approx 56.3^\\circ && a=10\\\\ &\\beta =30^\\circ && b\\approx 6.013\\\\ &\\gamma \\approx 93.7^\\circ && c=12 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the missing side and angles of the given triangle: [latex]\\alpha =30^\\circ ,b=12,c=24[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q885435\">Show Solution<\/span><\/p>\n<div id=\"q885435\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]a\\approx 14.9,\\beta \\approx 23.8^\\circ ,\\gamma \\approx 126.2^\\circ[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174837\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174837&theme=oea&iframe_resize_id=ohm174837\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Solving for an Angle of a SSS Triangle<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q211238\">Show Solution<\/span><\/p>\n<div id=\"q211238\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p style=\"text-align: center\">[latex]\\begin{align} &{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\alpha \\\\ &{20}^{2}={25}^{2}+{18}^{2}-2\\left(25\\right)\\left(18\\right)\\cos \\alpha&& \\text{Substitute the appropriate measurements}. \\\\ &400=625+324 - 900\\cos \\alpha&& \\text{Simplify in each step}. \\\\ &400=949 - 900\\cos \\alpha \\\\ &-549=-900\\cos \\alpha&& \\text{Isolate cos }\\alpha . \\\\ &\\frac{-549}{-900}=\\cos \\alpha \\\\ &0.61\\approx \\cos \\alpha \\\\ &{\\cos }^{-1}\\left(0.61\\right)\\approx \\alpha&& \\text{Find the inverse cosine}. \\\\ &\\alpha \\approx 52.4^\\circ \\end{align}[\/latex]<\/p>\n<p>See Figure 5.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165221\/CNX_Precalc_Figure_08_02_0052.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\"><b>Figure 22<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>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 class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given [latex]a=5,b=7[\/latex], and [latex]c=10[\/latex], find the missing angles.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q488277\">Show Solution<\/span><\/p>\n<div id=\"q488277\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\alpha \\approx 27.7^\\circ ,\\beta \\approx 40.5^\\circ ,\\gamma \\approx 111.8^\\circ[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173789\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173789&theme=oea&iframe_resize_id=ohm173789\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Applied Problems Using the Law of Cosines<\/h2>\n<p>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 class=\"textbox shaded\">\n<h3>Example 9: Using the Law of Cosines to Solve a Communication Problem<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q960347\">Show Solution<\/span><\/p>\n<div id=\"q960347\" class=\"hidden-answer\" style=\"display: none\">\n<p>For simplicity, we start by drawing a diagram similar to Figure 6\u00a0and labeling our given information.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165224\/CNX_Precalc_Figure_08_02_0062.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\"><b>Figure 23<\/b><\/p>\n<\/div>\n<p>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<p style=\"text-align: center\">[latex]\\begin{gathered}{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\theta \\\\ {\\left(2420\\right)}^{2}={\\left(5050\\right)}^{2}+{\\left(6000\\right)}^{2}-2\\left(5050\\right)\\left(6000\\right)\\cos \\theta \\\\ {\\left(2420\\right)}^{2}-{\\left(5050\\right)}^{2}-{\\left(6000\\right)}^{2}=-2\\left(5050\\right)\\left(6000\\right)\\cos \\theta \\\\ \\frac{{\\left(2420\\right)}^{2}-{\\left(5050\\right)}^{2}-{\\left(6000\\right)}^{2}}{-2\\left(5050\\right)\\left(6000\\right)}=\\cos \\theta \\\\ \\cos \\theta \\approx 0.9183 \\\\ \\theta \\approx {\\cos }^{-1}\\left(0.9183\\right) \\\\ \\theta \\approx 23.3^\\circ \\end{gathered}[\/latex]<\/p>\n<p>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 Figure 7.\u00a0This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165226\/CNX_Precalc_Figure_08_02_0072.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\"><b>Figure 24<\/b><\/p>\n<\/div>\n<p>Using the angle [latex]\\theta =23.3^\\circ[\/latex] and the basic trigonometric identities, we can find the solutions. Thus<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\cos \\left(23.3^\\circ \\right)=\\frac{x}{5050} \\\\ x=5050\\cos \\left(23.3^\\circ \\right) \\\\ x\\approx 4638.15\\text{feet} \\\\ \\sin \\left(23.3^\\circ \\right)=\\frac{y}{5050} \\\\ y=5050\\sin \\left(23.3^\\circ \\right) \\\\ y\\approx 1997.5\\text{feet} \\end{gathered}[\/latex]<\/p>\n<p>The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 10: Calculating Distance Traveled Using a SAS Triangle<\/h3>\n<p>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 Figure 8.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165228\/CNX_Precalc_Figure_08_02_0092.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\"><b>Figure 25<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q569516\">Show Solution<\/span><\/p>\n<div id=\"q569516\" class=\"hidden-answer\" style=\"display: none\">\n<p>The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, [latex]180^\\circ -20^\\circ =160^\\circ[\/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<p style=\"text-align: center\">[latex]\\begin{align}&{x}^{2}={8}^{2}+{10}^{2}-2\\left(8\\right)\\left(10\\right)\\cos \\left(160^\\circ \\right) \\\\ &{x}^{2}=314.35 \\\\ &x=\\sqrt{314.35} \\\\ &x\\approx 17.7\\text{miles} \\end{align}[\/latex]<\/p>\n<p>The boat is about 17.7 miles from port.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using Heron&#8217;s Formula to Find the Area of a Triangle<\/h2>\n<p>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 <strong>Heron\u2019s formula<\/strong> instead of finding the height. <strong>Heron of Alexandria<\/strong> 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 class=\"textbox\">\n<h3>A General Note: Heron\u2019s Formula<\/h3>\n<p>Heron\u2019s formula finds the area of oblique triangles in which sides [latex]a,b[\/latex], and [latex]c[\/latex] are known.<\/p>\n<p style=\"text-align: center\">[latex]\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex]<\/p>\n<p>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 class=\"textbox shaded\">\n<h3>Example 11: Using Heron\u2019s Formula to Find the Area of a Given Triangle<\/h3>\n<p>Find the area of the triangle in Figure 9\u00a0using Heron\u2019s formula.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165230\/CNX_Precalc_Figure_08_02_0102.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\"><b>Figure 26<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q256710\">Show Solution<\/span><\/p>\n<div id=\"q256710\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we calculate [latex]s[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} s&=\\frac{\\left(a+b+c\\right)}{2} \\\\ s&=\\frac{\\left(10+15+7\\right)}{2}=16 \\end{align}[\/latex]<\/p>\n<p>Then we apply the formula.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\text{Area}&=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)} \\\\ \\text{Area}&=\\sqrt{16\\left(16 - 10\\right)\\left(16 - 15\\right)\\left(16 - 7\\right)} \\\\ \\text{Area}&\\approx 29.4 \\end{align}[\/latex]<\/p>\n<p>The area is approximately 29.4 square units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q159200\">Show Solution<\/span><\/p>\n<div id=\"q159200\" class=\"hidden-answer\" style=\"display: none\">\n<p>Area = 552 square feet<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149312\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149312&theme=oea&iframe_resize_id=ohm149312\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 12: Applying Heron\u2019s Formula to a Real-World Problem<\/h3>\n<p>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 Figure 10\u00a0for a view of the city property.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165233\/CNX_Precalc_Figure_08_02_0112.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\"><b>Figure 27<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q63725\">Show Solution<\/span><\/p>\n<div id=\"q63725\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the measurement for [latex]s[\/latex], which is one-half of the perimeter.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}s&=\\frac{\\left(62.4+43.5+34.1\\right)}{2} \\\\ s&=70\\text{m} \\end{align}[\/latex]<\/p>\n<p>Apply Heron\u2019s formula.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\text{Area}&=\\sqrt{70\\left(70 - 62.4\\right)\\left(70 - 43.5\\right)\\left(70 - 34.1\\right)} \\\\ \\text{Area}&=\\sqrt{506,118.2} \\\\ \\text{Area}&\\approx 711.4 \\end{align}[\/latex]<\/p>\n<p>The developer has about 711.4 square meters.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q942985\">Show Solution<\/span><\/p>\n<div id=\"q942985\" class=\"hidden-answer\" style=\"display: none\">\n<p>about 8.15 square feet<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm8473\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=8473&theme=oea&iframe_resize_id=ohm8473\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Key Equations<\/h2>\n<table id=\"eip-id2728866\" summary=\"..\">\n<tbody>\n<tr>\n<td>Law of Sines<\/td>\n<td>\n<p style=\"text-align: center\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\frac{a}{\\sin \\alpha }=\\frac{b}{\\sin \\beta }=\\frac{c}{\\sin \\gamma }[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>Area for oblique triangles<\/td>\n<td>[latex]\\begin{align}\\text{Area}&=\\frac{1}{2}bc\\sin \\alpha \\\\ &=\\frac{1}{2}ac\\sin \\beta \\\\ &=\\frac{1}{2}ab\\sin \\gamma \\end{align}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table id=\"eip-id1956425\" summary=\"..\">\n<tbody>\n<tr>\n<td>Law of Cosines<\/td>\n<td>\n<p style=\"text-align: left\">[latex]{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\alpha[\/latex]<\/p>\n<p style=\"text-align: left\">[latex]{b}^{2}={a}^{2}+{c}^{2}-2ac\\cos \\beta[\/latex]<\/p>\n<p style=\"text-align: left\">[latex]{c}^{2}={a}^{2}+{b}^{2}-2ab\\cos \\gamma[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>Heron\u2019s formula<\/td>\n<td>[latex]\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex] where [latex]s=\\frac{\\left(a+b+c\\right)}{2}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>The Law of Sines can be used to solve oblique triangles, which are non-right triangles.<\/li>\n<li>According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.<\/li>\n<li>There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution.<\/li>\n<li>The ambiguous case arises when an oblique triangle can have different outcomes.<\/li>\n<li>There are three possible cases that arise from SSA arrangement\u2014a single solution, two possible solutions, and no solution.<\/li>\n<li>The Law of Sines can be used to solve triangles with given criteria.<\/li>\n<li>The general area formula for triangles translates to oblique triangles by first finding the appropriate height value.<\/li>\n<li>There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation.<\/li>\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.<\/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.<\/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.<\/li>\n<\/ul>\n<div>\n<dl id=\"fs-id1165133234516\" class=\"definition\">\n<dd>\n<dl class=\"definition\">\n<dd>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135646148\" class=\"definition\">\n<dt>altitude<\/dt>\n<dd id=\"fs-id1165135646153\">a perpendicular line from one vertex of a triangle to the opposite side, or in the case of an obtuse triangle, to the line containing the opposite side, forming two right triangles<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134357482\" class=\"definition\">\n<dt>ambiguous case<\/dt>\n<dd id=\"fs-id1165134357487\">a scenario in which more than one triangle is a valid solution for a given oblique SSA triangle<\/dd>\n<\/dl>\n<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133184256\" class=\"definition\">\n<dt>\u00a0 \u00a0 \u00a0 Generalized Pythagorean Theorem<\/dt>\n<dd id=\"fs-id1165133184261\">\u00a0 \u00a0 \u00a0 \u00a0an extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0SSS triangles<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dd>\n<dl id=\"fs-id1165134357482\" class=\"definition\">\n<dt>\n<\/dt>\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<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134357493\" class=\"definition\">\n<dt>Law of Sines<\/dt>\n<dd id=\"fs-id1165133234509\">states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133234516\" class=\"definition\">\n<dt>oblique triangle<\/dt>\n<dd id=\"fs-id1165135551173\">any triangle that is not a right triangle<\/dd>\n<\/dl>\n<\/dd>\n<\/dl>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<dl id=\"fs-id1165133234516\" class=\"definition\">\n<dd id=\"fs-id1165135551173\"><\/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-14311\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/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":17533,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-14311","chapter","type-chapter","status-publish","hentry"],"part":16037,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14311","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14311\/revisions"}],"predecessor-version":[{"id":16084,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14311\/revisions\/16084"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/parts\/16037"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapters\/14311\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/media?parent=14311"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=14311"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/contributor?post=14311"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/wp-json\/wp\/v2\/license?post=14311"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}