{"id":17970,"date":"2021-09-11T19:58:25","date_gmt":"2021-09-11T19:58:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/?post_type=chapter&#038;p=17970"},"modified":"2021-09-11T20:16:11","modified_gmt":"2021-09-11T20:16:11","slug":"section-8-1-right-triangle-trigonometry-applications","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/section-8-1-right-triangle-trigonometry-applications\/","title":{"raw":"Section 8.1: Right Triangle Trigonometry; Applications","rendered":"Section 8.1: Right Triangle Trigonometry; Applications"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Use right triangles to evaluate trigonometric functions.<\/li>\r\n \t<li style=\"font-weight: 400;\">Find function values for\u200930\u00b0 (\u03c0\/6),\u2009\u2009\u200945\u00b0 (\u03c0\/4),\u2009and\u200960\u00b0 (\u03c0\/3).<\/li>\r\n \t<li style=\"font-weight: 400;\">Use cofunctions of complementary angles.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use the de\ufb01nitions of trigonometric functions of any angle.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use right triangle trigonometry to solve applied problems.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Using Right Triangles to Evaluate Trigonometric Functions<\/h2>\r\nIn earlier sections, we used a unit circle to define the <strong>trigonometric functions<\/strong>. In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of [latex]t[\/latex] is its value at [latex]t[\/latex] radians. First, we need to create our right triangle. Figure 1\u00a0shows a point on a <strong>unit circle<\/strong> of radius 1. If we drop a vertical line segment from the point [latex]\\left(x,y\\right)\\\\[\/latex] to the <em>x<\/em>-axis, we have a right triangle whose vertical side has length [latex]y[\/latex] and whose horizontal side has length [latex]x[\/latex]. We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.<span id=\"fs-id1165137602828\">\r\n<\/span>\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\/27003730\/CNX_Precalc_Figure_05_04_0012.jpg\" alt=\"Graph of quarter circle with radius of 1 and angle of t. Point of (x,y) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"208\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\n&nbsp;\r\n\r\nWe know\r\n<div style=\"text-align: center;\">[latex]\\cos t=\\frac{x}{1}=x[\/latex]<\/div>\r\nLikewise, we know\r\n<div style=\"text-align: center;\">[latex]\\sin t=\\frac{y}{1}=y[\/latex]<\/div>\r\nThese ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using [latex]\\left(x,y\\right)[\/latex] coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of [latex]x[\/latex], we will call the side between the given angle and the right angle the <strong>adjacent side<\/strong> to angle [latex]t[\/latex]. (Adjacent means \"next to.\") Instead of [latex]y[\/latex], we will call the side most distant from the given angle the <strong>opposite side<\/strong> from angle [latex]t[\/latex]. And instead of [latex]1[\/latex], we will call the side of a right triangle opposite the right angle the <strong>hypotenuse<\/strong>. These sides are labeled in Figure 2.\r\n\r\n<span id=\"fs-id1165137465030\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003732\/CNX_Precalc_Figure_05_04_0022.jpg\" alt=\"A right triangle with hypotenuse, opposite, and adjacent sides labeled.\" \/><\/span>\r\n<p style=\"text-align: center;\"><strong>Figure 2.\u00a0<\/strong>The sides of a right triangle in relation to angle [latex]t[\/latex].<\/p>\r\n\r\n<h2>Understanding Right Triangle Relationships<\/h2>\r\nGiven a right triangle with an acute angle of [latex]t[\/latex],\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;\\sin \\left(t\\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} &amp;\\csc \\left(t\\right)=\\frac{\\text{hypotenuse}}{\\text{opposite}}\\\\ &amp;\\cos \\left(t\\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} &amp;\\sec \\left(t\\right)=\\frac{\\text{hypotenuse}}{\\text{adjacent}}\\\\ &amp;\\tan \\left(t\\right)=\\frac{\\text{opposite}}{\\text{adjacent}} &amp;\\cot \\left(t\\right)=\\frac{\\text{adjacent}}{\\text{opposite}}\\end{align}[\/latex]<\/div>\r\nA common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of \"<u>S<\/u>ine is <u>o<\/u>pposite over <u>h<\/u>ypotenuse, <u>C<\/u>osine is <u>a<\/u>djacent over <u>h<\/u>ypotenuse, <u>T<\/u>angent is <u>o<\/u>pposite over <u>a<\/u>djacent.\"\r\n<div class=\"textbox\">\r\n<h3>How To: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.<\/h3>\r\n<ol>\r\n \t<li>Find the sine as the ratio of the opposite side to the hypotenuse<\/li>\r\n \t<li>Find the cosine as the ratio of the adjacent side to the hypotenuse.<\/li>\r\n \t<li>Find the tangent is the ratio of the opposite side to the adjacent side.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Evaluating a Trigonometric Function of a Right Triangle<\/h3>\r\nGiven the triangle shown in Figure 3, find the value of [latex]\\cos \\alpha[\/latex].<span id=\"fs-id1165137414609\">\r\n<\/span>\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\/27003734\/CNX_Precalc_Figure_05_04_0032.jpg\" alt=\"A right triangle with side lengths of 8, 15, and 17. Angle alpha also labeled.\" width=\"487\" height=\"188\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"880775\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"880775\"]\r\n\r\nThe side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(\\alpha \\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} =\\frac{15}{17} \\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\nGiven the triangle shown in Figure 4, find the value of [latex]\\text{sin}t[\/latex].<span id=\"fs-id1165135191134\">\r\n<\/span>\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\/27003736\/CNX_Precalc_Figure_05_04_0042.jpg\" alt=\"A right triangle with sides of 7, 24, and 25. Also labeled is angle t.\" width=\"487\" height=\"180\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"261765\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"261765\"]\r\n\r\n[latex]\\frac{7}{25}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Relating Angles and Their Functions<\/h2>\r\nWhen working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 5. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.\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\/27003739\/CNX_Precalc_Figure_05_04_0052.jpg\" alt=\"Right triangle with angles alpha and beta. Sides are labeled hypotenuse, adjacent to alpha\/opposite to beta, and adjacent to beta\/opposite alpha.\" width=\"487\" height=\"181\" \/> <b>Figure 5.<\/b> The side adjacent to one angle is opposite the other.[\/caption]\r\n\r\nWe will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.\r\n<div class=\"textbox\">\r\n<h3>How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.<\/h3>\r\n<ol>\r\n \t<li>If needed, draw the right triangle and label the angle provided.<\/li>\r\n \t<li>Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.<\/li>\r\n \t<li>Find the required function:\r\n<ul>\r\n \t<li>sine as the ratio of the opposite side to the hypotenuse<\/li>\r\n \t<li>cosine as the ratio of the adjacent side to the hypotenuse<\/li>\r\n \t<li>tangent as the ratio of the opposite side to the adjacent side<\/li>\r\n \t<li>secant as the ratio of the hypotenuse to the adjacent side<\/li>\r\n \t<li>cosecant as the ratio of the hypotenuse to the opposite side<\/li>\r\n \t<li>cotangent as the ratio of the adjacent side to the opposite side<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Evaluating Trigonometric Functions of Angles Not in Standard Position<\/h3>\r\nUsing the triangle shown in Figure 6, evaluate [latex]\\sin \\alpha[\/latex], [latex]\\cos \\alpha[\/latex], [latex]\\tan \\alpha[\/latex], [latex]\\sec \\alpha[\/latex], [latex]\\csc \\alpha [\/latex], and [latex]\\cot \\alpha[\/latex].<span id=\"fs-id1165137542988\">\r\n<\/span>\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\/27003741\/CNX_Precalc_Figure_05_04_0062.jpg\" alt=\"Right triangle with sides of 3, 4, and 5. Angle alpha is also labeled.\" width=\"487\" height=\"162\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"626810\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"626810\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\sin \\alpha =\\frac{\\text{opposite }\\alpha }{\\text{hypotenuse}}=\\frac{4}{5}\\\\ &amp;\\cos \\alpha =\\frac{\\text{adjacent to }\\alpha }{\\text{hypotenuse}}=\\frac{3}{5} \\\\ &amp;\\tan \\alpha =\\frac{\\text{opposite }\\alpha }{\\text{adjacent to }\\alpha }=\\frac{4}{3} \\\\ &amp;\\sec \\alpha =\\frac{\\text{hypotenuse}}{\\text{adjacent to }\\alpha }=\\frac{5}{3} \\\\ &amp;\\csc \\alpha =\\frac{\\text{hypotenuse}}{\\text{opposite }\\alpha }=\\frac{5}{4} \\\\ &amp;\\cot \\alpha =\\frac{\\text{adjacent to }\\alpha }{\\text{opposite }\\alpha }=\\frac{3}{4} \\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\nUsing the triangle shown in Figure 7, evaluate [latex]\\sin t[\/latex], [latex]\\cos t[\/latex], [latex]\\tan t[\/latex], [latex]\\sec t[\/latex], [latex]\\csc t[\/latex], and [latex]\\cot t[\/latex].\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\/27003744\/CNX_Precalc_Figure_05_04_0072.jpg\" alt=\"Right triangle with sides 33, 56, and 65. Angle t is also labeled.\" width=\"487\" height=\"204\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"252611\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"252611\"]\r\n\r\n[latex]\\begin{align}&amp;\\sin t=\\frac{33}{65},\\cos t=\\frac{56}{65},\\tan t=\\frac{33}{56}, \\\\ &amp;\\sec t=\\frac{65}{56},\\csc t=\\frac{65}{33},\\cot t=\\frac{56}{33} \\end{align}[\/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]155312[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding Trigonometric Functions of Special Angles Using Side Lengths<\/h2>\r\nWe have already discussed the trigonometric functions as they relate to the <strong>special angles<\/strong> on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of [latex]30^\\circ [\/latex], [latex]60^\\circ [\/latex], and [latex]45^\\circ[\/latex],\u00a0however, remember that when dealing with right triangles, we are limited to angles between [latex]0^\\circ \\text{ and } 90^\\circ[\/latex].\r\n\r\nSuppose we have a [latex]30^\\circ ,60^\\circ ,90^\\circ [\/latex] triangle, which can also be described as a [latex]\\frac{\\pi }{6}, \\frac{\\pi }{3},\\frac{\\pi }{2}[\/latex] triangle. The sides have lengths in the relation [latex]s,\\sqrt{3}s,2s[\/latex]. The sides of a [latex]45^\\circ ,45^\\circ ,90^\\circ [\/latex] triangle, which can also be described as a [latex]\\frac{\\pi }{4},\\frac{\\pi }{4},\\frac{\\pi }{2}[\/latex] triangle, have lengths in the relation [latex]s,s,\\sqrt{2}s[\/latex]. These relations are shown in Figure 8.\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\/27003746\/CNX_Precalc_Figure_05_04_0082.jpg\" alt=\"Two side by side graphs of circles with inscribed angles. First circle has angle of pi\/3 inscribed. Second circle has angle of pi\/4 inscribed.\" width=\"975\" height=\"371\" \/> <b>Figure 8.<\/b> Side lengths of special triangles[\/caption]\r\n\r\nWe can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.\r\n<div class=\"textbox\">\r\n<h3>How To: Given trigonometric functions of a special angle, evaluate using side lengths.<\/h3>\r\n<ol>\r\n \t<li>Use the side lengths shown in Figure 8\u00a0for the special angle you wish to evaluate.<\/li>\r\n \t<li>Use the ratio of side lengths appropriate to the function you wish to evaluate.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Evaluating Trigonometric Functions of Special Angles Using Side Lengths<\/h3>\r\nFind the exact value of the trigonometric functions of [latex]\\frac{\\pi }{3}[\/latex], using side lengths.\r\n\r\n[reveal-answer q=\"547484\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"547484\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\sin \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{opp}}{\\text{hyp}}=\\frac{\\sqrt{3}s}{2s}=\\frac{\\sqrt{3}}{2}\\\\ &amp;\\cos \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{adj}}{\\text{hyp}}=\\frac{s}{2s}=\\frac{1}{2}\\\\ &amp;\\tan \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{opp}}{\\text{adj}}=\\frac{\\sqrt{3}s}{s}=\\sqrt{3}\\\\ &amp;\\sec \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{hyp}}{\\text{adj}}=\\frac{2s}{s}=2\\\\ &amp;\\csc \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{hyp}}{\\text{opp}}=\\frac{2s}{\\sqrt{3}s}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3} \\\\ &amp;\\cot \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{adj}}{\\text{opp}}=\\frac{s}{\\sqrt{3}s}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3} \\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 exact value of the trigonometric functions of [latex]\\frac{\\pi }{4}[\/latex], using side lengths.\r\n\r\n[reveal-answer q=\"676808\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"676808\"]\r\n\r\n[latex]\\sin \\left(\\frac{\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\cos \\left(\\frac{\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex],\r\n[latex]\\sec \\left(\\frac{\\pi }{4}\\right)=\\sqrt{2},\\csc\\left(\\frac{\\pi }{4}\\right)=\\sqrt{2},\\cot \\left(\\frac{\\pi }{4}\\right)=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/Ujyl_zQw2zE\r\n\r\nBy looking at all six trig functions and their special angles, we can organize it all in one place.\r\n<table id=\"Table_05_03_01\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Angle<\/strong><\/td>\r\n<td><strong> [latex]0[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{6},\\text{ or }{30}^{\\circ}[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{4},\\text{ or } {45}^{\\circ }[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{3},\\text{ or }{60}^{\\circ }[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{2},\\text{ or }{90}^{\\circ }[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosine<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Sine<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Tangent<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>Undefined<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Secant<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>2<\/td>\r\n<td>Undefined<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosecant<\/strong><\/td>\r\n<td>Undefined<\/td>\r\n<td>2<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cotangent<\/strong><\/td>\r\n<td>Undefined<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that we have this table, we can use it to find the exact values of trigonometric expressions.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Evaluating Trigonometric Functions of Special Angles<\/h3>\r\nFind the exact value of the trigonometric function [latex]\\frac{\\sin\\left(\\frac{\\pi}{6}\\right)}{1+\\cos^{2}\\left(\\frac{\\pi}{6}\\right)}[\/latex] using the table above.\r\n\r\n[reveal-answer q=\"547480\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"547480\"]\r\n\r\n[latex]\\begin{gathered} &amp; \\frac{\\frac{1}{2}}{1+\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}} &amp;&amp; \\text{Put in the values from the table. Note the squaring of cosine.} \\\\ &amp;\\frac{\\frac{1}{2}}{1+\\frac{3}{4}} &amp;&amp; \\text{Square the fraction.} \\\\ &amp;\\frac{\\frac{1}{2}}{\\frac{7}{4}} &amp;&amp; \\text{Get common denominators.} \\\\ &amp;\\left(\\frac{1}{2}\\right)\\left({\\frac{4}{7}}\\right) &amp;&amp; \\text{Get common denominators.} \\\\ &amp;\\frac{2}{7} &amp;&amp; \\text{Simplify}\\end{gathered}[\/latex]\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 exact value of the trigonometric function [latex]\\sec^{2}\\left(45^\\circ\\right)+\\tan^{2}\\left(60^\\circ\\right)[\/latex] using the table above.\r\n\r\n[reveal-answer q=\"676800\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"676800\"]\r\n\r\n5\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using Equal Cofunction of Complements<\/h2>\r\nIf we look at the table above, we will notice a pattern. In a right triangle with angles of [latex]\\frac{\\pi }{6}[\/latex] and [latex]\\frac{\\pi }{3}[\/latex], we see that the sine of [latex]\\frac{\\pi }{3}[\/latex], namely [latex]\\frac{\\sqrt{3}}{2}[\/latex], is also the cosine of [latex]\\frac{\\pi }{6}[\/latex], while the sine of [latex]\\frac{\\pi }{6}[\/latex], namely [latex]\\frac{1}{2}[\/latex], is also the cosine of [latex]\\frac{\\pi }{3}[\/latex].\r\n<div>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;\\sin \\frac{\\pi }{3}=\\cos \\frac{\\pi }{6}=\\frac{\\sqrt{3}s}{2s}=\\frac{\\sqrt{3}}{2} \\\\ &amp;\\sin \\frac{\\pi }{6}=\\cos \\frac{\\pi }{3}=\\frac{s}{2s}=\\frac{1}{2} \\end{align}[\/latex]<span id=\"fs-id1165137409548\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003749\/CNX_Precalc_Figure_05_04_0092.jpg\" alt=\"A graph of circle with angle pi\/3 inscribed.\" \/><\/span><\/p>\r\n<p style=\"text-align: center;\"><strong>Figure 9.\u00a0<\/strong>The sine of [latex]\\frac{\\pi }{3}[\/latex] equals the cosine of [latex]\\frac{\\pi }{6}[\/latex] and vice versa.<\/p>\r\n\r\n<\/div>\r\nThis result should not be surprising because, as we see from Figure 9, the side opposite the angle of [latex]\\frac{\\pi }{3}[\/latex] is also the side adjacent to [latex]\\frac{\\pi }{6}[\/latex], so [latex]\\sin \\left(\\frac{\\pi }{3}\\right)[\/latex] and [latex]\\cos \\left(\\frac{\\pi }{6}\\right)[\/latex] are exactly the same ratio of the same two sides, [latex]\\sqrt{3}s[\/latex] and [latex]2s[\/latex]. Similarly, [latex]\\cos \\left(\\frac{\\pi }{3}\\right)[\/latex] and [latex]\\sin \\left(\\frac{\\pi }{6}\\right)[\/latex] are also the same ratio using the same two sides, [latex]s[\/latex] and [latex]2s[\/latex].\r\n\r\nThe interrelationship between the sines and cosines of [latex]\\frac{\\pi }{6}[\/latex] and [latex]\\frac{\\pi }{3}[\/latex] also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to [latex]\\pi [\/latex], and the right angle is [latex]\\frac{\\pi }{2}[\/latex], the remaining two angles must also add up to [latex]\\frac{\\pi }{2}[\/latex]. That means that a right triangle can be formed with any two angles that add to [latex]\\frac{\\pi }{2}[\/latex] \u2014in other words, any two complementary angles. So we may state a <em>cofunction identity<\/em>: If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 10.\r\n\r\n<span id=\"fs-id1165137655742\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003808\/CNX_Precalc_Figure_05_04_0102.jpg\" alt=\"Right triangle with angles alpha and beta. Equivalence between sin alpha and cos beta. Equivalence between sin beta and cos alpha.\" \/><\/span>\r\n<p style=\"text-align: center;\"><strong>Figure 10.\u00a0<\/strong>Cofunction identity of sine and cosine of complementary angles<\/p>\r\nUsing this identity, we can state without calculating, for instance, that the sine of [latex]\\frac{\\pi }{12}[\/latex] equals the cosine of [latex]\\frac{5\\pi }{12}[\/latex], and that the sine of [latex]\\frac{5\\pi }{12}[\/latex] equals the cosine of [latex]\\frac{\\pi }{12}[\/latex]. We can also state that if, for a certain angle [latex]t[\/latex], [latex]\\cos \\text{ }t=\\frac{5}{13}[\/latex], then [latex]\\sin \\left(\\frac{\\pi }{2}-t\\right)=\\frac{5}{13}[\/latex] as well.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Cofunction Identities<\/h3>\r\nThe <strong>cofunction identities<\/strong> in radians are listed in the table below.\r\n<table id=\"Table_05_04_01\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\cos t=\\sin \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\r\n<td>[latex]\\sin t=\\cos \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\tan t=\\cot \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\r\n<td>[latex]\\cot t=\\tan \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\sec t=\\csc \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\r\n<td>[latex]\\csc t=\\sec \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the sine and cosine of an angle, find the sine or cosine of its complement.<\/h3>\r\n<ol>\r\n \t<li>To find the sine of the complementary angle, find the cosine of the original angle.<\/li>\r\n \t<li>To find the cosine of the complementary angle, find the sine of the original angle.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Using Cofunction Identities<\/h3>\r\nWrite the following as an equivalent cosine expression: [latex]\\sin \\left(\\frac{5\\pi}{12}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"797038\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"797038\"]\r\n\r\nAccording to the cofunction identities for sine and cosine,\r\n<p style=\"text-align: center;\">[latex]\\sin t=\\cos \\left(\\frac{\\pi }{2}-t\\right)[\/latex].<\/p>\r\nSo\r\n<p style=\"text-align: center;\">[latex]\\cos \\left(\\frac{\\pi }{2}-\\frac{5\\pi}{12}\\right)=\\cos\\left(\\frac{\\pi}{12}\\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Using Cofunction Identities<\/h3>\r\nWrite the following as an equivalent tangent expression: [latex]\\cot\\left(25^\\circ\\right)[\/latex].\r\n\r\n[reveal-answer q=\"797040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"797040\"]\r\n\r\nAccording to the cofunction identities for tangent and cotangent,\r\n<p style=\"text-align: center;\">[latex]\\tan t=\\cot \\left(90^\\circ-t\\right)[\/latex].<\/p>\r\nSo\r\n<p style=\"text-align: center;\">[latex]\\tan \\left(90^\\circ-25^\\circ\\right)=\\tan\\left(65^\\circ\\right)[\/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\nWrite the following as an equivalent cosecant expression: [latex]\\sec\\left(68^\\circ\\right)[\/latex].\r\n\r\n[reveal-answer q=\"164220\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164220\"]\r\n\r\n[latex]\\csc\\left(22^\\circ\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/_gkuml--4_Q\r\n<h2>Using Trigonometric Functions<\/h2>\r\nIn previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.<\/h3>\r\n<ol>\r\n \t<li>For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.<\/li>\r\n \t<li>Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.<\/li>\r\n \t<li>Using the value of the trigonometric function and the known side length, solve for the missing side length.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Finding Missing Side Lengths Using Trigonometric Ratios<\/h3>\r\nFind the unknown sides of the triangle in Figure 11.\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\/27003813\/CNX_Precalc_Figure_05_04_0112.jpg\" alt=\"A right triangle with sides a, c, and 7. Angle of 30 degrees is also labeled.\" width=\"487\" height=\"250\" \/> <b>Figure 11<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"112724\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"112724\"]\r\n\r\nWe know the angle and the opposite side, so we can use the tangent to find the adjacent side.\r\n<p style=\"text-align: center;\">[latex]\\tan \\left(30^\\circ \\right)=\\frac{7}{a}[\/latex]<\/p>\r\nWe rearrange to solve for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}a&amp;=\\frac{7}{\\tan \\left(30^\\circ \\right)} \\\\ &amp;=12.1\\end{align}[\/latex]<\/p>\r\nWe can use the sine to find the hypotenuse.\r\n<p style=\"text-align: center;\">[latex]\\sin \\left(30^\\circ \\right)=\\frac{7}{c}[\/latex]<\/p>\r\nAgain, we rearrange to solve for [latex]c[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}c&amp;=\\frac{7}{\\sin \\left(30^\\circ \\right)} \\\\ &amp;=14 \\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\nA right triangle has one angle of [latex]\\frac{\\pi }{3}[\/latex]\u00a0and a hypotenuse of 20. Find the unknown sides and angle of the triangle.\r\n\r\n[reveal-answer q=\"869816\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"869816\"]\r\n\r\n[latex]\\text{adjacent}=10[\/latex]; [latex]\\text{opposite}=10\\sqrt{3}[\/latex] ; missing angle is [latex]\\frac{\\pi }{6}[\/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]121967[\/ohm_question]\r\n\r\n<\/div>\r\nRight-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye.\u00a0 In the figure below, [latex]\\alpha[\/latex] represents the <strong>angle of elevation<\/strong> and [latex]\\beta[\/latex] represents the <strong>angle of depression<\/strong>.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"733\"]<img src=\"http:\/\/www.hutchmath.com\/Images\/elevationdepression.JPG\" alt=\"diagram of an angle of elevation and depression. The angle of elevation is measured from teh horizontal and looks up at an object. The angle of depression is measured from the horizontal and looks down on something.\" width=\"733\" height=\"192\" \/> <strong>Figure 12<\/strong>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>How To: Given a tall object, measure its height indirectly.<\/h3>\r\n<ol>\r\n \t<li>Make a sketch of the problem situation to keep track of known and unknown information.<\/li>\r\n \t<li>Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.<\/li>\r\n \t<li>At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.<\/li>\r\n \t<li>Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.<\/li>\r\n \t<li>Solve the equation for the unknown height.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Measuring a Distance Indirectly<\/h3>\r\nTo find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures the angle of elevation to be [latex]57^\\circ [\/latex], as shown in Figure 13. Find the height of the tree.\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\/27003822\/CNX_Precalc_Figure_05_04_0122.jpg\" alt=\"A tree with angle of 57 degrees from vantage point. Vantage point is 30 feet from tree.\" width=\"487\" height=\"242\" \/> <b>Figure 13<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"91153\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"91153\"]\r\n\r\nWe know that the angle of elevation is [latex]57^\\circ [\/latex] and the adjacent side is 30 ft long. The opposite side is the unknown height.\r\n\r\nThe trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of [latex]57^\\circ [\/latex], letting [latex]h[\/latex] be the unknown height.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\tan \\theta =\\frac{\\text{opposite}}{\\text{adjacent}} \\\\ &amp;\\tan\\left(57^\\circ \\right)=\\frac{h}{30}&amp;&amp; \\text{Solve for }h. \\\\ &amp;h=30\\tan \\left(57^\\circ \\right)&amp;&amp; \\text{Multiply}.\\\\ &amp;h\\approx 46.2&amp;&amp; \\text{Use a calculator}. \\end{align}[\/latex]<\/p>\r\nThe tree is approximately 46 feet tall.\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\nHow long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building with an angle of elevation [latex]\\frac{5\\pi }{12}[\/latex] ? Round to the nearest foot.\r\n\r\n[reveal-answer q=\"798855\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"798855\"]\r\n\r\nAbout 52 ft\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]155310[\/ohm_question]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/8jU2R3BuR5E\r\n<h3>Inverse Trigonometric Function on a calculator<\/h3>\r\nIf two sides of a right triangle are given, an inverse trigonometric function can be used to find an acute angle in a triangle. There is an inverse trig button on a calculator that appears above the normal sin, cos, tan buttons on a calculator. A shift or second key is required to select these, and they are [latex]\\sin^{-1},\\cos^{-1},\\tan^{-1}[\/latex]. The calculator will return an angle in either radians or degrees, depending on what mode your calculator is in.\u00a0 We will need the inverse tangent function for bearing.\r\n<h2>Bearing<\/h2>\r\nBearing is the direction you are heading according to a compass. On most maps, N is up, S is down, W is to the left, and E is to the right. Bearings are written in this form: (N or S)(Acute Angle)(E or W). The angle is measured from either the north or south, whichever letter comes first in your bearing. For example, [latex]N20^\\circ E[\/latex] means that you will go north and then go 20 degrees to the east, or to the right. The drawing below gives an example of how bearings are drawn. As you can see, each angle is measure from either the N or S depending on the first letter of the bearing. Bearings are NOT drawn in standard position, which means the are NOT drawn from the positive x-axis.\r\n<img class=\"aligncenter\" src=\"http:\/\/www.hutchmath.com\/Images\/examplebearings.JPG\" alt=\"a picture drawn of N40E in the first quadrant, N60W in the second quadrant, S50W in the third quadrant, and S27E in the fourth quadrant\" width=\"295\" height=\"269\" \/>\r\n\r\nNow we will will look at some application problems that involve bearings and right triangles.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Find the bearing<\/h3>\r\nA semi travels east for 8 miles, makes a right turn, and then travels south for 11 more miles. What is the bearing from the starting point to the semi's current location?\r\n\r\n[reveal-answer q=\"797000\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"797000\"]\r\n\r\nFirst we need to draw a picture of the problem:\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/www.hutchmath.com\/Images\/semibearing.JPG\" alt=\"right triangle drawn in the fourth quadrant with adjacent side of 8 and opposite side of 11. The angle inside the triangle is beta, and the angle between the hypotenuse and the y-axis is beta.\" \/>\r\n\r\nWe want to find the angle [latex]\\alpha[\/latex] that is inside the triangle. To do this, we will set up a trig definition that relates the two given sides. We will choose tangent. Then we will take the inverse tangent to find the angle:\r\n[latex]\\begin{align} &amp;\\tan\\left(\\alpha\\right)=\\frac{11}{8} \\\\&amp; \\tan^{-1}\\left(\\tan\\left(\\alpha\\right)\\right)=\\tan^{-1}\\left(\\frac{11}{8}\\right) \\\\ &amp;\\alpha=\\tan^{-1}\\left(\\frac{11}{8}\\right) \\\\&amp; \\alpha\\approx 54^\\circ \\end{align}[\/latex]\r\n\r\nThe bearing must be measured from south since the triangle is drawn in the fourth quadrant. On the picture, we must find [latex]\\beta[\/latex]. Since [latex]\\alpha+\\beta=90^\\circ[\/latex], then [latex]\\beta=90^\\circ-\\alpha=36^\\circ[\/latex]. Therefore our bearing is [latex]S 36^\\circ E[\/latex].\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\nA cyclist travels west for 7 miles, makes a right turn and travels north for 4 miles. What is the bearing from the starting point to the cyclist's current position? Round your angle to one decimal place.\r\n\r\n[reveal-answer q=\"798809\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"798809\"]\r\n\r\n[latex]N 60.3^\\circ W[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Find the bearing<\/h3>\r\nA jeep travels on a bearing of [latex]N 40^\\circ W[\/latex] for 6 miles. How far north and how far west is the jeep from its starting point? Round your answers to one decimal place.\r\n\r\n[reveal-answer q=\"797005\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"797005\"]\r\n\r\nFirst we need to draw a picture of the problem:\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/www.hutchmath.com\/Images\/jeepbearing.JPG\" alt=\"right triangle drawn in the second quadrant. An angle of 40 degrees is indicated between the hypotenuse and the y-axis\" \/>\r\n\r\nWe need to find the angle [latex]\\alpha[\/latex] inside of the triangle before we set up a trig equation: [latex]\\alpha=90^\\circ-40^\\circ=50^\\circ[\/latex].\r\n\r\nNext we will set up a trig function to find the indicated sides of the triangle:\r\n\r\nFirst we will find the number of miles traveled North:\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp; \\sin\\left(50^\\circ\\right)=\\frac{N}{6} \\\\&amp; 6\\sin\\left(50^\\circ\\right)=N \\\\ &amp; N\\approx 4.6\\text{ miles} \\end{align}[\/latex]<\/p>\r\nNow we will find the number of miles traveled west:\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp; \\cos\\left(50^\\circ\\right)=\\frac{W}{6} \\\\&amp; 6\\cos\\left(50^\\circ\\right)=W \\\\ &amp; W\\approx 3.9\\text{ miles} \\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\nA jogger travels on a bearing of [latex]S 53.13^\\circ E[\/latex] for 5 miles. How far east and how far south is the jogger from their starting point? Round your answers to the nearest mile.\r\n\r\n[reveal-answer q=\"798800\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"798800\"]\r\n\r\n4 miles east, 3 miles south\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Convert Percent Grade Into Degrees<\/h2>\r\n<strong>Percent grade<\/strong> is often seen on roadways or trails, and it is a measure of steepness.\u00a0 For example, if a road has a 6% grade, this means the road rises 6 feet over a horizontal distance (run) of 100 feet.\u00a0 Percent grade is found by dividing the rise over the run as shown in Figure 14.\u00a0The rise and run are also the opposite and adjacent sides.\u00a0 We can find [latex]\\theta[\/latex] by taking the inverse tangent.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"230\"]<img src=\"http:\/\/www.hutchmath.com\/Images\/ElevationGrade.JPG\" alt=\"right triangle with height labeled as rise, and its base labeled as run\" width=\"230\" height=\"151\" \/> <b>Figure 14<\/b>[\/caption]\r\n\r\nIn order to calculate the degree measurement of a percent grade, first change the percent into a decimal by dividing by 100. Then take the inverse tangent of this decimal, and this will give the angle in degrees.\u00a0 This is actually the <strong>angle of elevation<\/strong>, which we studied earlier in this section!\r\n<div class=\"textbox\">\r\n<h3>How To: Convert Percent Grade inTo Degrees<\/h3>\r\nMake sure your calculator is in degree mode. Given the percent grade, the angle [latex]\\theta[\/latex] can be calculated using:\r\n<p style=\"text-align: center;\">[latex]\\theta=\\tan^{-1}\\left(\\dfrac{\\text{percent grade}}{100}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Percent Grade to Degrees<\/h3>\r\nFilbert Street, the steepest street in San Francisco, has a 31.5% grade.\u00a0 What is the angle the street makes with a horizontal line, rounded to one decimal place?\r\n\r\n[reveal-answer q=\"91150\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"91150\"]\r\n\r\n[latex]\\theta=\\tan^{-1}\\left(\\dfrac{31.5}{100}\\right)\\approx 17.5^\\circ[\/latex]\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\nA road has a grade of 10.9%.\u00a0 What is the angle the road makes with a horizontal line, rounded to one decimal place?\r\n\r\n[reveal-answer q=\"798850\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"798850\"]\r\n\r\n[latex]\\text{About }6.2^\\circ[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n&nbsp;\r\n<table id=\"eip-id1165137409421\" style=\"width: 422px;\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 128.494px;\">Cofunction Identities<\/td>\r\n<td style=\"text-align: center; width: 448.494px;\">[latex]\\begin{gathered} \\cos t=\\sin \\left(\\frac{\\pi }{2}-t\\right) \\\\ \\sin t=\\cos \\left(\\frac{\\pi }{2}-t\\right)\\\\ \\tan t=\\cot \\left(\\frac{\\pi }{2}-t\\right) \\\\ \\cot t=\\tan \\left(\\frac{\\pi }{2}-t\\right) \\\\ \\sec t=\\csc \\left(\\frac{\\pi }{2}-t\\right) \\\\ \\csc t=\\sec \\left(\\frac{\\pi }{2}-t\\right) \\end{gathered}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 128.494px;\">Percent Grade to Degrees<\/td>\r\n<td style=\"text-align: center; width: 448.494px;\">[latex]\\theta=\\tan^{-1}\\left(\\dfrac{\\text{percent grade}}{100}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<section id=\"fs-id1165137481899\" class=\"key-concepts\">\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165137415357\">\r\n \t<li>We can define trigonometric functions as ratios of the side lengths of a right triangle.<\/li>\r\n \t<li>The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle.<\/li>\r\n \t<li>We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur.<\/li>\r\n \t<li>Any two complementary angles could be the two acute angles of a right triangle.<\/li>\r\n \t<li>If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa.<\/li>\r\n \t<li>We can use trigonometric functions of an angle to find unknown side lengths.<\/li>\r\n \t<li>Select the trigonometric function representing the ratio of the unknown side to the known side.<\/li>\r\n \t<li>Right-triangle trigonometry permits the measurement of inaccessible heights and distances.<\/li>\r\n \t<li>The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known.<\/li>\r\n \t<li>Bearings are measured from either from N or S, depending on the first letter of the bearing.<\/li>\r\n \t<li>Percent grade can be expressed as a degree, which is the angle of elevation.<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137446119\" class=\"definition\">\r\n \t<dt>adjacent side<\/dt>\r\n \t<dd id=\"fs-id1165137446123\">in a right triangle, the side between a given angle and the right angle<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137465232\" class=\"definition\">\r\n \t<dt>angle of depression<\/dt>\r\n \t<dd id=\"fs-id1165135175026\">the angle between the horizontal and the line from the object to the observer\u2019s eye, assuming the object is positioned lower than the observer<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137447602\" class=\"definition\">\r\n \t<dt>angle of elevation<\/dt>\r\n \t<dd id=\"fs-id1165135185281\">the angle between the horizontal and the line from the object to the observer\u2019s eye, assuming the object is positioned higher than the observer<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137558543\" class=\"definition\">\r\n \t<dt>opposite side<\/dt>\r\n \t<dd id=\"fs-id1165137558547\">in a right triangle, the side most distant from a given angle<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135477488\" class=\"definition\">\r\n \t<dt>hypotenuse<\/dt>\r\n \t<dd id=\"fs-id1165137588091\">the side of a right triangle opposite the right angle<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135477488\" class=\"definition\">\r\n \t<dt>percent grade<\/dt>\r\n \t<dd id=\"fs-id1165137588091\">The ratio of the rise over the run, expressed as a percent.\u00a0 It is a measure of steepness.<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/section>&nbsp;\r\n<h2 style=\"text-align: center;\">Section 8.1 Homework Exercises<\/h2>\r\n1. For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003825\/CNX_Precalc_Figure_05_04_2012.jpg\" alt=\"A right triangle.\" \/>\r\n\r\n2. When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates?\r\n\r\n3. The tangent of an angle compares which sides of the right triangle?\r\n\r\n4.\u00a0What is the relationship between the two acute angles in a right triangle?\r\n\r\n5. Explain the cofunction identity.\r\n\r\nFor the following exercises, evaluate the expression.\u00a0 Rationalize the denominator if necessary.\r\n\r\n6. [latex]\\sin\\frac{\\pi}{4}\\cos\\frac{\\pi}{4}-\\tan\\frac{\\pi}{3}[\/latex]\r\n\r\n7.\u00a0[latex]\\sin\\frac{\\pi}{4}\\cos\\frac{\\pi}{3}-\\tan\\frac{\\pi}{4}[\/latex]\r\n\r\n8.\u00a0[latex]2\\sec^{2}\\frac{\\pi}{4}+\\cot^{2}\\frac{\\pi}{3}[\/latex]\r\n\r\n9. [latex]3\\csc^{2}\\frac{\\pi}{3}+\\cot^{2}\\frac{\\pi}{4}[\/latex]\r\n\r\nFor the following exercises, use cofunctions of complementary angles.\r\n\r\n10. [latex]\\cos \\left(34^\\circ\\right)=\\sin \\left(\\text{__}^\\circ\\right)[\/latex]\r\n\r\n11. [latex]\\cos \\left(\\frac{\\pi }{3}\\right)=\\sin \\text{(___)}[\/latex]\r\n\r\n12.\u00a0[latex]\\csc \\left(21^\\circ\\right)=\\sec \\left(\\text{___}^\\circ \\right)[\/latex]\r\n\r\n13. [latex]\\tan \\left(\\frac{\\pi }{4}\\right)=\\cot \\left(\\text{__}\\right)[\/latex]\r\n\r\nFor the following exercises, find the lengths of the missing sides if side [latex]a[\/latex] is opposite angle [latex]A[\/latex], side [latex]b[\/latex] is opposite angle [latex]B[\/latex], and side [latex]c[\/latex] is the hypotenuse.\r\n\r\n14. [latex]\\cos B=\\frac{4}{5},a=10[\/latex]\r\n\r\n15. [latex]\\sin B=\\frac{1}{2}, a=20[\/latex]\r\n\r\n16.\u00a0[latex]\\tan A=\\frac{5}{12},b=6[\/latex]\r\n\r\n17. [latex]\\tan A=100,b=100[\/latex]\r\n\r\n18.\u00a0[latex]\\sin B=\\frac{1}{\\sqrt{3}}, a=2[\/latex]\r\n\r\n19. [latex]a=5,\\measuredangle A={60}^{\\circ }[\/latex]\r\n\r\n20.\u00a0[latex]c=12,\\measuredangle A={45}^{\\circ }[\/latex]\r\n\r\nFor the following exercises, use Figure 14 to evaluate each trigonometric function of angle [latex]A[\/latex].\r\n<figure id=\"Figure_05_04_203\">\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\/27003827\/CNX_Precalc_Figure_05_04_2032.jpg\" alt=\"A right triangle with sides 4 and 10 and angle of A labeled.\" width=\"487\" height=\"202\" \/> <b>Figure 14<\/b>[\/caption]<\/figure>\r\n21. [latex]\\sin A[\/latex]\r\n\r\n22.\u00a0[latex]\\cos A[\/latex]\r\n\r\n23. [latex]\\tan A[\/latex]\r\n\r\n24.\u00a0[latex]\\csc A[\/latex]\r\n\r\n25. [latex]\\sec A[\/latex]\r\n\r\n26.\u00a0[latex]\\cot A[\/latex]\r\n\r\nFor the following exercises, use Figure 15 to evaluate each trigonometric function of angle [latex]A[\/latex].\r\n<figure id=\"Figure_05_04_204\">\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\/27003830\/CNX_Precalc_Figure_05_04_2042.jpg\" alt=\"A right triangle with sides of 10 and 8 and angle of A labeled.\" width=\"487\" height=\"454\" \/> <b>Figure 15<\/b>[\/caption]<\/figure>\r\n27. [latex]\\sin A[\/latex]\r\n\r\n28.\u00a0[latex]\\cos A[\/latex]\r\n\r\n29. [latex]\\tan A[\/latex]\r\n\r\n30.\u00a0[latex]\\csc A[\/latex]\r\n\r\n31. [latex]\\sec A[\/latex]\r\n\r\n32.\u00a0[latex]\\cot A[\/latex]\r\n\r\nFor the following exercises, solve for the unknown sides of the given triangle.\r\n\r\n33.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003832\/CNX_Precalc_Figure_05_04_2052.jpg\" alt=\"A right triangle with sides of 7, b, and c labeled. Angles of B and 30 degrees also labeled.\" \/>\r\n\r\n34.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003834\/CNX_Precalc_Figure_05_04_2062.jpg\" alt=\"A right triangle with sides of 10, a, and c. Angles of 60 degrees and A also labeled.\" \/>\r\n\r\n35.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003838\/CNX_Precalc_Figure_05_04_2072.jpg\" alt=\"A right triangle with corners labeled A, B, and C. Hypotenuse has length of 15 times square root of 2. Angle B is 45 degrees.\" \/>\r\n\r\nFor the following exercises, use a calculator to find the length of each side to four decimal places.\r\n\r\n36.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003840\/CNX_Precalc_Figure_05_04_2082.jpg\" alt=\"A right triangle with sides of 10, a, and c. Angles of A and 62 degrees are also labeled.\" \/>\r\n\r\n37.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003843\/CNX_Precalc_Figure_05_04_2092.jpg\" alt=\"A right triangle with sides of 7, b, and c. Angles of 35 degrees and B are also labeled.\" \/>\r\n\r\n38.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003845\/CNX_Precalc_Figure_05_04_2102.jpg\" alt=\"A right triangle with sides of a, b, and 10 labeled. Angles of 65 degrees and B are also labeled.\" \/>\r\n\r\n39.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003850\/CNX_Precalc_Figure_05_04_2112.jpg\" alt=\"A right triangle with sides a, b, and 12. Angles of 10 degrees and B are also labeled.\" \/>\r\n\r\n40.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003852\/CNX_Precalc_Figure_05_04_2122.jpg\" alt=\"A right triangle with corners labeled A, B, and C. Sides labeled b, c, and 16.5. Angle of 81 degrees also labeled.\" \/>\r\n\r\n41. [latex]b=15,\\measuredangle B={15}^{\\circ }[\/latex]\r\n\r\n42.\u00a0[latex]c=200,\\measuredangle B={5}^{\\circ }[\/latex]\r\n\r\n43. [latex]c=50,\\measuredangle B={21}^{\\circ }[\/latex]\r\n\r\n44.\u00a0[latex]a=30,\\measuredangle A={27}^{\\circ }[\/latex]\r\n\r\n45. [latex]b=3.5,\\measuredangle A={78}^{\\circ }[\/latex]\r\n\r\n46. Find [latex]x[\/latex].\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003854\/CNX_Precalc_Figure_05_04_2132.jpg\" alt=\"A triangle with angles of 63 degrees and 39 degrees and side x. Bisector in triangle with length of 82.\" \/>\r\n\r\n47. Find [latex]x[\/latex].\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003857\/CNX_Precalc_Figure_05_04_2142.jpg\" alt=\"A triangle with angles of 36 degrees and 50 degrees and side x. Bisector in triangle with length of 85.\" \/>\r\n\r\n48. Find [latex]x[\/latex].\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003900\/CNX_Precalc_Figure_05_04_2152.jpg\" alt=\"A right triangle with side of 115 and angle of 35 degrees. Within right triangle there is another right triangle with angle of 56 degrees. Side length difference between two triangles is x.\" \/>\r\n\r\n49. Find [latex]x[\/latex].\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003903\/CNX_Precalc_Figure_05_04_2162.jpg\" alt=\"A right triangle with side of 119 and angle of 26 degrees. Within right triangle there is another right triangle with angle of 70 degrees instead of 26 degrees. Difference in side length between two triangles is x.\" \/>\r\n\r\n50.\u00a0A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]36^\\circ [\/latex], and that the angle of depression to the bottom of the tower is [latex]23^\\circ [\/latex]. How tall is the tower?\r\n\r\n51. A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]43^\\circ [\/latex], and that the angle of depression to the bottom of the tower is [latex]31^\\circ [\/latex]. How tall is the tower?\r\n\r\n52.\u00a0A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]15^\\circ [\/latex], and that the angle of depression to the bottom of the tower is [latex]2^\\circ [\/latex]. How far is the person from the monument?\r\n\r\n53. A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]18^\\circ [\/latex], and that the angle of depression to the bottom of the tower is [latex]3^\\circ [\/latex]. How far is the person from the monument?\r\n\r\n54.\u00a0There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]40^\\circ [\/latex]. From the same location, the angle of elevation to the top of the antenna is measured to be [latex]43^\\circ [\/latex]. Find the height of the antenna.\r\n\r\n55. There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]36^\\circ [\/latex]. From the same location, the angle of elevation to the top of the lightning rod is measured to be [latex]38^\\circ [\/latex]. Find the height of the lightning rod.\r\n\r\n56. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\\circ [\/latex]. How high does the ladder reach up the side of the building?\r\n\r\n57. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\\circ [\/latex]. How high does the ladder reach up the side of the building?\r\n\r\n58.\u00a0The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.\r\n\r\n59. The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building.\r\n\r\n60.\u00a0Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be [latex]60^\\circ [\/latex], how far from the base of the tree am I?\r\n\r\n61. A car travels west for 5 miles, turns left, and then travels south for 9 miles.\u00a0 What is the bearing from car's starting position to its current position?\u00a0 Round your answer to two decimal places.\r\n\r\n62. A truck travels east for 4 miles, turns left, and then travels north for 6 miles.\u00a0 What is the bearing from the truck's starting position to its current position? Round your answer to two decimal places.\r\n\r\n63. An ant travels on a bearing of [latex]N 22^\\circ E[\/latex] for 36 inches.\u00a0 How far east and how far north is the ant from its starting position?\u00a0 Round your answers to two decimal places.\r\n\r\n64. A spider crawls on a bearing of [latex]S 34^\\circ W[\/latex] for 20 inches.\u00a0 How far west and how far south is the spider from its starting point? Round your answers to two decimal places.\r\n\r\n65.\u00a0Originally built in 1901 by Colonel J.W. Eddy, Angels Flight in Los Angeles is said to be the world's shortest incorporated railway.\u00a0 The counterbalanced cars, controlled by cables, travel a 33% grade for 315 feet.\u00a0 What is the angle the track makes with a horizontal line, rounded to one decimal place?\r\n\r\n66. The Saluda Grade is the steepest standard-gauge mainline railway grade in the United States.\u00a0 Between Melrose and Saluda, North Carolina, the maximum grade is 4.9% for about 300 feet.\u00a0 What is the angle the track makes with a horizontal line, rounded to one decimal place?","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"font-weight: 400;\">Use right triangles to evaluate trigonometric functions.<\/li>\n<li style=\"font-weight: 400;\">Find function values for\u200930\u00b0 (\u03c0\/6),\u2009\u2009\u200945\u00b0 (\u03c0\/4),\u2009and\u200960\u00b0 (\u03c0\/3).<\/li>\n<li style=\"font-weight: 400;\">Use cofunctions of complementary angles.<\/li>\n<li style=\"font-weight: 400;\">Use the de\ufb01nitions of trigonometric functions of any angle.<\/li>\n<li style=\"font-weight: 400;\">Use right triangle trigonometry to solve applied problems.<\/li>\n<\/ul>\n<\/div>\n<h2>Using Right Triangles to Evaluate Trigonometric Functions<\/h2>\n<p>In earlier sections, we used a unit circle to define the <strong>trigonometric functions<\/strong>. In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of [latex]t[\/latex] is its value at [latex]t[\/latex] radians. First, we need to create our right triangle. Figure 1\u00a0shows a point on a <strong>unit circle<\/strong> of radius 1. If we drop a vertical line segment from the point [latex]\\left(x,y\\right)\\\\[\/latex] to the <em>x<\/em>-axis, we have a right triangle whose vertical side has length [latex]y[\/latex] and whose horizontal side has length [latex]x[\/latex]. We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.<span id=\"fs-id1165137602828\"><br \/>\n<\/span><\/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\/27003730\/CNX_Precalc_Figure_05_04_0012.jpg\" alt=\"Graph of quarter circle with radius of 1 and angle of t. Point of (x,y) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"208\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>We know<\/p>\n<div style=\"text-align: center;\">[latex]\\cos t=\\frac{x}{1}=x[\/latex]<\/div>\n<p>Likewise, we know<\/p>\n<div style=\"text-align: center;\">[latex]\\sin t=\\frac{y}{1}=y[\/latex]<\/div>\n<p>These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using [latex]\\left(x,y\\right)[\/latex] coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of [latex]x[\/latex], we will call the side between the given angle and the right angle the <strong>adjacent side<\/strong> to angle [latex]t[\/latex]. (Adjacent means &#8220;next to.&#8221;) Instead of [latex]y[\/latex], we will call the side most distant from the given angle the <strong>opposite side<\/strong> from angle [latex]t[\/latex]. And instead of [latex]1[\/latex], we will call the side of a right triangle opposite the right angle the <strong>hypotenuse<\/strong>. These sides are labeled in Figure 2.<\/p>\n<p><span id=\"fs-id1165137465030\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003732\/CNX_Precalc_Figure_05_04_0022.jpg\" alt=\"A right triangle with hypotenuse, opposite, and adjacent sides labeled.\" \/><\/span><\/p>\n<p style=\"text-align: center;\"><strong>Figure 2.\u00a0<\/strong>The sides of a right triangle in relation to angle [latex]t[\/latex].<\/p>\n<h2>Understanding Right Triangle Relationships<\/h2>\n<p>Given a right triangle with an acute angle of [latex]t[\/latex],<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&\\sin \\left(t\\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} &\\csc \\left(t\\right)=\\frac{\\text{hypotenuse}}{\\text{opposite}}\\\\ &\\cos \\left(t\\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} &\\sec \\left(t\\right)=\\frac{\\text{hypotenuse}}{\\text{adjacent}}\\\\ &\\tan \\left(t\\right)=\\frac{\\text{opposite}}{\\text{adjacent}} &\\cot \\left(t\\right)=\\frac{\\text{adjacent}}{\\text{opposite}}\\end{align}[\/latex]<\/div>\n<p>A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of &#8220;<u>S<\/u>ine is <u>o<\/u>pposite over <u>h<\/u>ypotenuse, <u>C<\/u>osine is <u>a<\/u>djacent over <u>h<\/u>ypotenuse, <u>T<\/u>angent is <u>o<\/u>pposite over <u>a<\/u>djacent.&#8221;<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.<\/h3>\n<ol>\n<li>Find the sine as the ratio of the opposite side to the hypotenuse<\/li>\n<li>Find the cosine as the ratio of the adjacent side to the hypotenuse.<\/li>\n<li>Find the tangent is the ratio of the opposite side to the adjacent side.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Evaluating a Trigonometric Function of a Right Triangle<\/h3>\n<p>Given the triangle shown in Figure 3, find the value of [latex]\\cos \\alpha[\/latex].<span id=\"fs-id1165137414609\"><br \/>\n<\/span><\/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\/27003734\/CNX_Precalc_Figure_05_04_0032.jpg\" alt=\"A right triangle with side lengths of 8, 15, and 17. Angle alpha also labeled.\" width=\"487\" height=\"188\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q880775\">Show Solution<\/span><\/p>\n<div id=\"q880775\" class=\"hidden-answer\" style=\"display: none\">\n<p>The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(\\alpha \\right)=\\frac{\\text{adjacent}}{\\text{hypotenuse}} =\\frac{15}{17} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given the triangle shown in Figure 4, find the value of [latex]\\text{sin}t[\/latex].<span id=\"fs-id1165135191134\"><br \/>\n<\/span><\/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\/27003736\/CNX_Precalc_Figure_05_04_0042.jpg\" alt=\"A right triangle with sides of 7, 24, and 25. Also labeled is angle t.\" width=\"487\" height=\"180\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q261765\">Show Solution<\/span><\/p>\n<div id=\"q261765\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{7}{25}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Relating Angles and Their Functions<\/h2>\n<p>When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 5. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.<\/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\/27003739\/CNX_Precalc_Figure_05_04_0052.jpg\" alt=\"Right triangle with angles alpha and beta. Sides are labeled hypotenuse, adjacent to alpha\/opposite to beta, and adjacent to beta\/opposite alpha.\" width=\"487\" height=\"181\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5.<\/b> The side adjacent to one angle is opposite the other.<\/p>\n<\/div>\n<p>We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.<\/h3>\n<ol>\n<li>If needed, draw the right triangle and label the angle provided.<\/li>\n<li>Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.<\/li>\n<li>Find the required function:\n<ul>\n<li>sine as the ratio of the opposite side to the hypotenuse<\/li>\n<li>cosine as the ratio of the adjacent side to the hypotenuse<\/li>\n<li>tangent as the ratio of the opposite side to the adjacent side<\/li>\n<li>secant as the ratio of the hypotenuse to the adjacent side<\/li>\n<li>cosecant as the ratio of the hypotenuse to the opposite side<\/li>\n<li>cotangent as the ratio of the adjacent side to the opposite side<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Evaluating Trigonometric Functions of Angles Not in Standard Position<\/h3>\n<p>Using the triangle shown in Figure 6, evaluate [latex]\\sin \\alpha[\/latex], [latex]\\cos \\alpha[\/latex], [latex]\\tan \\alpha[\/latex], [latex]\\sec \\alpha[\/latex], [latex]\\csc \\alpha[\/latex], and [latex]\\cot \\alpha[\/latex].<span id=\"fs-id1165137542988\"><br \/>\n<\/span><\/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\/27003741\/CNX_Precalc_Figure_05_04_0062.jpg\" alt=\"Right triangle with sides of 3, 4, and 5. Angle alpha is also labeled.\" width=\"487\" height=\"162\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q626810\">Show Solution<\/span><\/p>\n<div id=\"q626810\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\sin \\alpha =\\frac{\\text{opposite }\\alpha }{\\text{hypotenuse}}=\\frac{4}{5}\\\\ &\\cos \\alpha =\\frac{\\text{adjacent to }\\alpha }{\\text{hypotenuse}}=\\frac{3}{5} \\\\ &\\tan \\alpha =\\frac{\\text{opposite }\\alpha }{\\text{adjacent to }\\alpha }=\\frac{4}{3} \\\\ &\\sec \\alpha =\\frac{\\text{hypotenuse}}{\\text{adjacent to }\\alpha }=\\frac{5}{3} \\\\ &\\csc \\alpha =\\frac{\\text{hypotenuse}}{\\text{opposite }\\alpha }=\\frac{5}{4} \\\\ &\\cot \\alpha =\\frac{\\text{adjacent to }\\alpha }{\\text{opposite }\\alpha }=\\frac{3}{4} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Using the triangle shown in Figure 7, evaluate [latex]\\sin t[\/latex], [latex]\\cos t[\/latex], [latex]\\tan t[\/latex], [latex]\\sec t[\/latex], [latex]\\csc t[\/latex], and [latex]\\cot t[\/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\/27003744\/CNX_Precalc_Figure_05_04_0072.jpg\" alt=\"Right triangle with sides 33, 56, and 65. Angle t is also labeled.\" width=\"487\" height=\"204\" \/><\/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=\"q252611\">Show Solution<\/span><\/p>\n<div id=\"q252611\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}&\\sin t=\\frac{33}{65},\\cos t=\\frac{56}{65},\\tan t=\\frac{33}{56}, \\\\ &\\sec t=\\frac{65}{56},\\csc t=\\frac{65}{33},\\cot t=\\frac{56}{33} \\end{align}[\/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=\"ohm155312\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=155312&theme=oea&iframe_resize_id=ohm155312\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding Trigonometric Functions of Special Angles Using Side Lengths<\/h2>\n<p>We have already discussed the trigonometric functions as they relate to the <strong>special angles<\/strong> on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of [latex]30^\\circ[\/latex], [latex]60^\\circ[\/latex], and [latex]45^\\circ[\/latex],\u00a0however, remember that when dealing with right triangles, we are limited to angles between [latex]0^\\circ \\text{ and } 90^\\circ[\/latex].<\/p>\n<p>Suppose we have a [latex]30^\\circ ,60^\\circ ,90^\\circ[\/latex] triangle, which can also be described as a [latex]\\frac{\\pi }{6}, \\frac{\\pi }{3},\\frac{\\pi }{2}[\/latex] triangle. The sides have lengths in the relation [latex]s,\\sqrt{3}s,2s[\/latex]. The sides of a [latex]45^\\circ ,45^\\circ ,90^\\circ[\/latex] triangle, which can also be described as a [latex]\\frac{\\pi }{4},\\frac{\\pi }{4},\\frac{\\pi }{2}[\/latex] triangle, have lengths in the relation [latex]s,s,\\sqrt{2}s[\/latex]. These relations are shown in Figure 8.<\/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\/27003746\/CNX_Precalc_Figure_05_04_0082.jpg\" alt=\"Two side by side graphs of circles with inscribed angles. First circle has angle of pi\/3 inscribed. Second circle has angle of pi\/4 inscribed.\" width=\"975\" height=\"371\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8.<\/b> Side lengths of special triangles<\/p>\n<\/div>\n<p>We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given trigonometric functions of a special angle, evaluate using side lengths.<\/h3>\n<ol>\n<li>Use the side lengths shown in Figure 8\u00a0for the special angle you wish to evaluate.<\/li>\n<li>Use the ratio of side lengths appropriate to the function you wish to evaluate.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Evaluating Trigonometric Functions of Special Angles Using Side Lengths<\/h3>\n<p>Find the exact value of the trigonometric functions of [latex]\\frac{\\pi }{3}[\/latex], using side lengths.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q547484\">Show Solution<\/span><\/p>\n<div id=\"q547484\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\sin \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{opp}}{\\text{hyp}}=\\frac{\\sqrt{3}s}{2s}=\\frac{\\sqrt{3}}{2}\\\\ &\\cos \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{adj}}{\\text{hyp}}=\\frac{s}{2s}=\\frac{1}{2}\\\\ &\\tan \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{opp}}{\\text{adj}}=\\frac{\\sqrt{3}s}{s}=\\sqrt{3}\\\\ &\\sec \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{hyp}}{\\text{adj}}=\\frac{2s}{s}=2\\\\ &\\csc \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{hyp}}{\\text{opp}}=\\frac{2s}{\\sqrt{3}s}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3} \\\\ &\\cot \\left(\\frac{\\pi }{3}\\right)=\\frac{\\text{adj}}{\\text{opp}}=\\frac{s}{\\sqrt{3}s}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3} \\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 exact value of the trigonometric functions of [latex]\\frac{\\pi }{4}[\/latex], using side lengths.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q676808\">Show Solution<\/span><\/p>\n<div id=\"q676808\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sin \\left(\\frac{\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\cos \\left(\\frac{\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex],<br \/>\n[latex]\\sec \\left(\\frac{\\pi }{4}\\right)=\\sqrt{2},\\csc\\left(\\frac{\\pi }{4}\\right)=\\sqrt{2},\\cot \\left(\\frac{\\pi }{4}\\right)=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Introduction to Trigonometric Functions Using Triangles\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/Ujyl_zQw2zE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>By looking at all six trig functions and their special angles, we can organize it all in one place.<\/p>\n<table id=\"Table_05_03_01\" summary=\"..\">\n<tbody>\n<tr>\n<td><strong>Angle<\/strong><\/td>\n<td><strong> [latex]0[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{6},\\text{ or }{30}^{\\circ}[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{4},\\text{ or } {45}^{\\circ }[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{3},\\text{ or }{60}^{\\circ }[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{2},\\text{ or }{90}^{\\circ }[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Cosine<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td><strong>Sine<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td><strong>Tangent<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>Undefined<\/td>\n<\/tr>\n<tr>\n<td><strong>Secant<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>2<\/td>\n<td>Undefined<\/td>\n<\/tr>\n<tr>\n<td><strong>Cosecant<\/strong><\/td>\n<td>Undefined<\/td>\n<td>2<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td><strong>Cotangent<\/strong><\/td>\n<td>Undefined<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have this table, we can use it to find the exact values of trigonometric expressions.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 4: Evaluating Trigonometric Functions of Special Angles<\/h3>\n<p>Find the exact value of the trigonometric function [latex]\\frac{\\sin\\left(\\frac{\\pi}{6}\\right)}{1+\\cos^{2}\\left(\\frac{\\pi}{6}\\right)}[\/latex] using the table above.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q547480\">Show Solution<\/span><\/p>\n<div id=\"q547480\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{gathered} & \\frac{\\frac{1}{2}}{1+\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}} && \\text{Put in the values from the table. Note the squaring of cosine.} \\\\ &\\frac{\\frac{1}{2}}{1+\\frac{3}{4}} && \\text{Square the fraction.} \\\\ &\\frac{\\frac{1}{2}}{\\frac{7}{4}} && \\text{Get common denominators.} \\\\ &\\left(\\frac{1}{2}\\right)\\left({\\frac{4}{7}}\\right) && \\text{Get common denominators.} \\\\ &\\frac{2}{7} && \\text{Simplify}\\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the exact value of the trigonometric function [latex]\\sec^{2}\\left(45^\\circ\\right)+\\tan^{2}\\left(60^\\circ\\right)[\/latex] using the table above.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q676800\">Show Solution<\/span><\/p>\n<div id=\"q676800\" class=\"hidden-answer\" style=\"display: none\">\n<p>5<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using Equal Cofunction of Complements<\/h2>\n<p>If we look at the table above, we will notice a pattern. In a right triangle with angles of [latex]\\frac{\\pi }{6}[\/latex] and [latex]\\frac{\\pi }{3}[\/latex], we see that the sine of [latex]\\frac{\\pi }{3}[\/latex], namely [latex]\\frac{\\sqrt{3}}{2}[\/latex], is also the cosine of [latex]\\frac{\\pi }{6}[\/latex], while the sine of [latex]\\frac{\\pi }{6}[\/latex], namely [latex]\\frac{1}{2}[\/latex], is also the cosine of [latex]\\frac{\\pi }{3}[\/latex].<\/p>\n<div>\n<p style=\"text-align: center;\">[latex]\\begin{align} &\\sin \\frac{\\pi }{3}=\\cos \\frac{\\pi }{6}=\\frac{\\sqrt{3}s}{2s}=\\frac{\\sqrt{3}}{2} \\\\ &\\sin \\frac{\\pi }{6}=\\cos \\frac{\\pi }{3}=\\frac{s}{2s}=\\frac{1}{2} \\end{align}[\/latex]<span id=\"fs-id1165137409548\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003749\/CNX_Precalc_Figure_05_04_0092.jpg\" alt=\"A graph of circle with angle pi\/3 inscribed.\" \/><\/span><\/p>\n<p style=\"text-align: center;\"><strong>Figure 9.\u00a0<\/strong>The sine of [latex]\\frac{\\pi }{3}[\/latex] equals the cosine of [latex]\\frac{\\pi }{6}[\/latex] and vice versa.<\/p>\n<\/div>\n<p>This result should not be surprising because, as we see from Figure 9, the side opposite the angle of [latex]\\frac{\\pi }{3}[\/latex] is also the side adjacent to [latex]\\frac{\\pi }{6}[\/latex], so [latex]\\sin \\left(\\frac{\\pi }{3}\\right)[\/latex] and [latex]\\cos \\left(\\frac{\\pi }{6}\\right)[\/latex] are exactly the same ratio of the same two sides, [latex]\\sqrt{3}s[\/latex] and [latex]2s[\/latex]. Similarly, [latex]\\cos \\left(\\frac{\\pi }{3}\\right)[\/latex] and [latex]\\sin \\left(\\frac{\\pi }{6}\\right)[\/latex] are also the same ratio using the same two sides, [latex]s[\/latex] and [latex]2s[\/latex].<\/p>\n<p>The interrelationship between the sines and cosines of [latex]\\frac{\\pi }{6}[\/latex] and [latex]\\frac{\\pi }{3}[\/latex] also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to [latex]\\pi[\/latex], and the right angle is [latex]\\frac{\\pi }{2}[\/latex], the remaining two angles must also add up to [latex]\\frac{\\pi }{2}[\/latex]. That means that a right triangle can be formed with any two angles that add to [latex]\\frac{\\pi }{2}[\/latex] \u2014in other words, any two complementary angles. So we may state a <em>cofunction identity<\/em>: If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 10.<\/p>\n<p><span id=\"fs-id1165137655742\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003808\/CNX_Precalc_Figure_05_04_0102.jpg\" alt=\"Right triangle with angles alpha and beta. Equivalence between sin alpha and cos beta. Equivalence between sin beta and cos alpha.\" \/><\/span><\/p>\n<p style=\"text-align: center;\"><strong>Figure 10.\u00a0<\/strong>Cofunction identity of sine and cosine of complementary angles<\/p>\n<p>Using this identity, we can state without calculating, for instance, that the sine of [latex]\\frac{\\pi }{12}[\/latex] equals the cosine of [latex]\\frac{5\\pi }{12}[\/latex], and that the sine of [latex]\\frac{5\\pi }{12}[\/latex] equals the cosine of [latex]\\frac{\\pi }{12}[\/latex]. We can also state that if, for a certain angle [latex]t[\/latex], [latex]\\cos \\text{ }t=\\frac{5}{13}[\/latex], then [latex]\\sin \\left(\\frac{\\pi }{2}-t\\right)=\\frac{5}{13}[\/latex] as well.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Cofunction Identities<\/h3>\n<p>The <strong>cofunction identities<\/strong> in radians are listed in the table below.<\/p>\n<table id=\"Table_05_04_01\" summary=\"..\">\n<tbody>\n<tr>\n<td>[latex]\\cos t=\\sin \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\n<td>[latex]\\sin t=\\cos \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\tan t=\\cot \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\n<td>[latex]\\cot t=\\tan \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\sec t=\\csc \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\n<td>[latex]\\csc t=\\sec \\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the sine and cosine of an angle, find the sine or cosine of its complement.<\/h3>\n<ol>\n<li>To find the sine of the complementary angle, find the cosine of the original angle.<\/li>\n<li>To find the cosine of the complementary angle, find the sine of the original angle.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Using Cofunction Identities<\/h3>\n<p>Write the following as an equivalent cosine expression: [latex]\\sin \\left(\\frac{5\\pi}{12}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q797038\">Show Solution<\/span><\/p>\n<div id=\"q797038\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the cofunction identities for sine and cosine,<\/p>\n<p style=\"text-align: center;\">[latex]\\sin t=\\cos \\left(\\frac{\\pi }{2}-t\\right)[\/latex].<\/p>\n<p>So<\/p>\n<p style=\"text-align: center;\">[latex]\\cos \\left(\\frac{\\pi }{2}-\\frac{5\\pi}{12}\\right)=\\cos\\left(\\frac{\\pi}{12}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Using Cofunction Identities<\/h3>\n<p>Write the following as an equivalent tangent expression: [latex]\\cot\\left(25^\\circ\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q797040\">Show Solution<\/span><\/p>\n<div id=\"q797040\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the cofunction identities for tangent and cotangent,<\/p>\n<p style=\"text-align: center;\">[latex]\\tan t=\\cot \\left(90^\\circ-t\\right)[\/latex].<\/p>\n<p>So<\/p>\n<p style=\"text-align: center;\">[latex]\\tan \\left(90^\\circ-25^\\circ\\right)=\\tan\\left(65^\\circ\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write the following as an equivalent cosecant expression: [latex]\\sec\\left(68^\\circ\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164220\">Show Solution<\/span><\/p>\n<div id=\"q164220\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\csc\\left(22^\\circ\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Cofunction Identities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/_gkuml--4_Q?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Using Trigonometric Functions<\/h2>\n<p>In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.<\/h3>\n<ol>\n<li>For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.<\/li>\n<li>Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.<\/li>\n<li>Using the value of the trigonometric function and the known side length, solve for the missing side length.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Finding Missing Side Lengths Using Trigonometric Ratios<\/h3>\n<p>Find the unknown sides of the triangle in Figure 11.<\/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\/27003813\/CNX_Precalc_Figure_05_04_0112.jpg\" alt=\"A right triangle with sides a, c, and 7. Angle of 30 degrees is also labeled.\" width=\"487\" height=\"250\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q112724\">Show Solution<\/span><\/p>\n<div id=\"q112724\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know the angle and the opposite side, so we can use the tangent to find the adjacent side.<\/p>\n<p style=\"text-align: center;\">[latex]\\tan \\left(30^\\circ \\right)=\\frac{7}{a}[\/latex]<\/p>\n<p>We rearrange to solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}a&=\\frac{7}{\\tan \\left(30^\\circ \\right)} \\\\ &=12.1\\end{align}[\/latex]<\/p>\n<p>We can use the sine to find the hypotenuse.<\/p>\n<p style=\"text-align: center;\">[latex]\\sin \\left(30^\\circ \\right)=\\frac{7}{c}[\/latex]<\/p>\n<p>Again, we rearrange to solve for [latex]c[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}c&=\\frac{7}{\\sin \\left(30^\\circ \\right)} \\\\ &=14 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A right triangle has one angle of [latex]\\frac{\\pi }{3}[\/latex]\u00a0and a hypotenuse of 20. Find the unknown sides and angle of the triangle.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q869816\">Show Solution<\/span><\/p>\n<div id=\"q869816\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\text{adjacent}=10[\/latex]; [latex]\\text{opposite}=10\\sqrt{3}[\/latex] ; missing angle is [latex]\\frac{\\pi }{6}[\/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=\"ohm121967\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=121967&theme=oea&iframe_resize_id=ohm121967\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer&#8217;s eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer&#8217;s eye.\u00a0 In the figure below, [latex]\\alpha[\/latex] represents the <strong>angle of elevation<\/strong> and [latex]\\beta[\/latex] represents the <strong>angle of depression<\/strong>.<\/p>\n<div style=\"width: 743px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/www.hutchmath.com\/Images\/elevationdepression.JPG\" alt=\"diagram of an angle of elevation and depression. The angle of elevation is measured from teh horizontal and looks up at an object. The angle of depression is measured from the horizontal and looks down on something.\" width=\"733\" height=\"192\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 12<\/strong><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a tall object, measure its height indirectly.<\/h3>\n<ol>\n<li>Make a sketch of the problem situation to keep track of known and unknown information.<\/li>\n<li>Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.<\/li>\n<li>At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.<\/li>\n<li>Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.<\/li>\n<li>Solve the equation for the unknown height.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Measuring a Distance Indirectly<\/h3>\n<p>To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures the angle of elevation to be [latex]57^\\circ[\/latex], as shown in Figure 13. Find the height of the tree.<\/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\/27003822\/CNX_Precalc_Figure_05_04_0122.jpg\" alt=\"A tree with angle of 57 degrees from vantage point. Vantage point is 30 feet from tree.\" width=\"487\" height=\"242\" \/><\/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=\"q91153\">Show Solution<\/span><\/p>\n<div id=\"q91153\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know that the angle of elevation is [latex]57^\\circ[\/latex] and the adjacent side is 30 ft long. The opposite side is the unknown height.<\/p>\n<p>The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of [latex]57^\\circ[\/latex], letting [latex]h[\/latex] be the unknown height.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\tan \\theta =\\frac{\\text{opposite}}{\\text{adjacent}} \\\\ &\\tan\\left(57^\\circ \\right)=\\frac{h}{30}&& \\text{Solve for }h. \\\\ &h=30\\tan \\left(57^\\circ \\right)&& \\text{Multiply}.\\\\ &h\\approx 46.2&& \\text{Use a calculator}. \\end{align}[\/latex]<\/p>\n<p>The tree is approximately 46 feet tall.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building with an angle of elevation [latex]\\frac{5\\pi }{12}[\/latex] ? Round to the nearest foot.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q798855\">Show Solution<\/span><\/p>\n<div id=\"q798855\" class=\"hidden-answer\" style=\"display: none\">\n<p>About 52 ft<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm155310\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=155310&theme=oea&iframe_resize_id=ohm155310\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Example:  Determine the Length of a Side of a Right Triangle Using a Trig Equation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/8jU2R3BuR5E?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Inverse Trigonometric Function on a calculator<\/h3>\n<p>If two sides of a right triangle are given, an inverse trigonometric function can be used to find an acute angle in a triangle. There is an inverse trig button on a calculator that appears above the normal sin, cos, tan buttons on a calculator. A shift or second key is required to select these, and they are [latex]\\sin^{-1},\\cos^{-1},\\tan^{-1}[\/latex]. The calculator will return an angle in either radians or degrees, depending on what mode your calculator is in.\u00a0 We will need the inverse tangent function for bearing.<\/p>\n<h2>Bearing<\/h2>\n<p>Bearing is the direction you are heading according to a compass. On most maps, N is up, S is down, W is to the left, and E is to the right. Bearings are written in this form: (N or S)(Acute Angle)(E or W). The angle is measured from either the north or south, whichever letter comes first in your bearing. For example, [latex]N20^\\circ E[\/latex] means that you will go north and then go 20 degrees to the east, or to the right. The drawing below gives an example of how bearings are drawn. As you can see, each angle is measure from either the N or S depending on the first letter of the bearing. Bearings are NOT drawn in standard position, which means the are NOT drawn from the positive x-axis.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/www.hutchmath.com\/Images\/examplebearings.JPG\" alt=\"a picture drawn of N40E in the first quadrant, N60W in the second quadrant, S50W in the third quadrant, and S27E in the fourth quadrant\" width=\"295\" height=\"269\" \/><\/p>\n<p>Now we will will look at some application problems that involve bearings and right triangles.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 9: Find the bearing<\/h3>\n<p>A semi travels east for 8 miles, makes a right turn, and then travels south for 11 more miles. What is the bearing from the starting point to the semi&#8217;s current location?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q797000\">Show Solution<\/span><\/p>\n<div id=\"q797000\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we need to draw a picture of the problem:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/www.hutchmath.com\/Images\/semibearing.JPG\" alt=\"right triangle drawn in the fourth quadrant with adjacent side of 8 and opposite side of 11. The angle inside the triangle is beta, and the angle between the hypotenuse and the y-axis is beta.\" \/><\/p>\n<p>We want to find the angle [latex]\\alpha[\/latex] that is inside the triangle. To do this, we will set up a trig definition that relates the two given sides. We will choose tangent. Then we will take the inverse tangent to find the angle:<br \/>\n[latex]\\begin{align} &\\tan\\left(\\alpha\\right)=\\frac{11}{8} \\\\& \\tan^{-1}\\left(\\tan\\left(\\alpha\\right)\\right)=\\tan^{-1}\\left(\\frac{11}{8}\\right) \\\\ &\\alpha=\\tan^{-1}\\left(\\frac{11}{8}\\right) \\\\& \\alpha\\approx 54^\\circ \\end{align}[\/latex]<\/p>\n<p>The bearing must be measured from south since the triangle is drawn in the fourth quadrant. On the picture, we must find [latex]\\beta[\/latex]. Since [latex]\\alpha+\\beta=90^\\circ[\/latex], then [latex]\\beta=90^\\circ-\\alpha=36^\\circ[\/latex]. Therefore our bearing is [latex]S 36^\\circ E[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A cyclist travels west for 7 miles, makes a right turn and travels north for 4 miles. What is the bearing from the starting point to the cyclist&#8217;s current position? Round your angle to one decimal place.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q798809\">Show Solution<\/span><\/p>\n<div id=\"q798809\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]N 60.3^\\circ W[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 10: Find the bearing<\/h3>\n<p>A jeep travels on a bearing of [latex]N 40^\\circ W[\/latex] for 6 miles. How far north and how far west is the jeep from its starting point? Round your answers to one decimal place.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q797005\">Show Solution<\/span><\/p>\n<div id=\"q797005\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we need to draw a picture of the problem:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/www.hutchmath.com\/Images\/jeepbearing.JPG\" alt=\"right triangle drawn in the second quadrant. An angle of 40 degrees is indicated between the hypotenuse and the y-axis\" \/><\/p>\n<p>We need to find the angle [latex]\\alpha[\/latex] inside of the triangle before we set up a trig equation: [latex]\\alpha=90^\\circ-40^\\circ=50^\\circ[\/latex].<\/p>\n<p>Next we will set up a trig function to find the indicated sides of the triangle:<\/p>\n<p>First we will find the number of miles traveled North:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} & \\sin\\left(50^\\circ\\right)=\\frac{N}{6} \\\\& 6\\sin\\left(50^\\circ\\right)=N \\\\ & N\\approx 4.6\\text{ miles} \\end{align}[\/latex]<\/p>\n<p>Now we will find the number of miles traveled west:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} & \\cos\\left(50^\\circ\\right)=\\frac{W}{6} \\\\& 6\\cos\\left(50^\\circ\\right)=W \\\\ & W\\approx 3.9\\text{ miles} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A jogger travels on a bearing of [latex]S 53.13^\\circ E[\/latex] for 5 miles. How far east and how far south is the jogger from their starting point? Round your answers to the nearest mile.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q798800\">Show Solution<\/span><\/p>\n<div id=\"q798800\" class=\"hidden-answer\" style=\"display: none\">\n<p>4 miles east, 3 miles south<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Convert Percent Grade Into Degrees<\/h2>\n<p><strong>Percent grade<\/strong> is often seen on roadways or trails, and it is a measure of steepness.\u00a0 For example, if a road has a 6% grade, this means the road rises 6 feet over a horizontal distance (run) of 100 feet.\u00a0 Percent grade is found by dividing the rise over the run as shown in Figure 14.\u00a0The rise and run are also the opposite and adjacent sides.\u00a0 We can find [latex]\\theta[\/latex] by taking the inverse tangent.<\/p>\n<div style=\"width: 240px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/www.hutchmath.com\/Images\/ElevationGrade.JPG\" alt=\"right triangle with height labeled as rise, and its base labeled as run\" width=\"230\" height=\"151\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14<\/b><\/p>\n<\/div>\n<p>In order to calculate the degree measurement of a percent grade, first change the percent into a decimal by dividing by 100. Then take the inverse tangent of this decimal, and this will give the angle in degrees.\u00a0 This is actually the <strong>angle of elevation<\/strong>, which we studied earlier in this section!<\/p>\n<div class=\"textbox\">\n<h3>How To: Convert Percent Grade inTo Degrees<\/h3>\n<p>Make sure your calculator is in degree mode. Given the percent grade, the angle [latex]\\theta[\/latex] can be calculated using:<\/p>\n<p style=\"text-align: center;\">[latex]\\theta=\\tan^{-1}\\left(\\dfrac{\\text{percent grade}}{100}\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 11: Percent Grade to Degrees<\/h3>\n<p>Filbert Street, the steepest street in San Francisco, has a 31.5% grade.\u00a0 What is the angle the street makes with a horizontal line, rounded to one decimal place?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q91150\">Show Solution<\/span><\/p>\n<div id=\"q91150\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\theta=\\tan^{-1}\\left(\\dfrac{31.5}{100}\\right)\\approx 17.5^\\circ[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A road has a grade of 10.9%.\u00a0 What is the angle the road makes with a horizontal line, rounded to one decimal place?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q798850\">Show Solution<\/span><\/p>\n<div id=\"q798850\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\text{About }6.2^\\circ[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Equations<\/h2>\n<p>&nbsp;<\/p>\n<table id=\"eip-id1165137409421\" style=\"width: 422px;\" summary=\"..\">\n<tbody>\n<tr>\n<td style=\"width: 128.494px;\">Cofunction Identities<\/td>\n<td style=\"text-align: center; width: 448.494px;\">[latex]\\begin{gathered} \\cos t=\\sin \\left(\\frac{\\pi }{2}-t\\right) \\\\ \\sin t=\\cos \\left(\\frac{\\pi }{2}-t\\right)\\\\ \\tan t=\\cot \\left(\\frac{\\pi }{2}-t\\right) \\\\ \\cot t=\\tan \\left(\\frac{\\pi }{2}-t\\right) \\\\ \\sec t=\\csc \\left(\\frac{\\pi }{2}-t\\right) \\\\ \\csc t=\\sec \\left(\\frac{\\pi }{2}-t\\right) \\end{gathered}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 128.494px;\">Percent Grade to Degrees<\/td>\n<td style=\"text-align: center; width: 448.494px;\">[latex]\\theta=\\tan^{-1}\\left(\\dfrac{\\text{percent grade}}{100}\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<section id=\"fs-id1165137481899\" class=\"key-concepts\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137415357\">\n<li>We can define trigonometric functions as ratios of the side lengths of a right triangle.<\/li>\n<li>The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle.<\/li>\n<li>We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur.<\/li>\n<li>Any two complementary angles could be the two acute angles of a right triangle.<\/li>\n<li>If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa.<\/li>\n<li>We can use trigonometric functions of an angle to find unknown side lengths.<\/li>\n<li>Select the trigonometric function representing the ratio of the unknown side to the known side.<\/li>\n<li>Right-triangle trigonometry permits the measurement of inaccessible heights and distances.<\/li>\n<li>The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known.<\/li>\n<li>Bearings are measured from either from N or S, depending on the first letter of the bearing.<\/li>\n<li>Percent grade can be expressed as a degree, which is the angle of elevation.<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137446119\" class=\"definition\">\n<dt>adjacent side<\/dt>\n<dd id=\"fs-id1165137446123\">in a right triangle, the side between a given angle and the right angle<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137465232\" class=\"definition\">\n<dt>angle of depression<\/dt>\n<dd id=\"fs-id1165135175026\">the angle between the horizontal and the line from the object to the observer\u2019s eye, assuming the object is positioned lower than the observer<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137447602\" class=\"definition\">\n<dt>angle of elevation<\/dt>\n<dd id=\"fs-id1165135185281\">the angle between the horizontal and the line from the object to the observer\u2019s eye, assuming the object is positioned higher than the observer<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137558543\" class=\"definition\">\n<dt>opposite side<\/dt>\n<dd id=\"fs-id1165137558547\">in a right triangle, the side most distant from a given angle<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135477488\" class=\"definition\">\n<dt>hypotenuse<\/dt>\n<dd id=\"fs-id1165137588091\">the side of a right triangle opposite the right angle<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135477488\" class=\"definition\">\n<dt>percent grade<\/dt>\n<dd id=\"fs-id1165137588091\">The ratio of the rise over the run, expressed as a percent.\u00a0 It is a measure of steepness.<\/dd>\n<\/dl>\n<\/div>\n<\/section>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: center;\">Section 8.1 Homework Exercises<\/h2>\n<p>1. For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003825\/CNX_Precalc_Figure_05_04_2012.jpg\" alt=\"A right triangle.\" \/><\/p>\n<p>2. When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates?<\/p>\n<p>3. The tangent of an angle compares which sides of the right triangle?<\/p>\n<p>4.\u00a0What is the relationship between the two acute angles in a right triangle?<\/p>\n<p>5. Explain the cofunction identity.<\/p>\n<p>For the following exercises, evaluate the expression.\u00a0 Rationalize the denominator if necessary.<\/p>\n<p>6. [latex]\\sin\\frac{\\pi}{4}\\cos\\frac{\\pi}{4}-\\tan\\frac{\\pi}{3}[\/latex]<\/p>\n<p>7.\u00a0[latex]\\sin\\frac{\\pi}{4}\\cos\\frac{\\pi}{3}-\\tan\\frac{\\pi}{4}[\/latex]<\/p>\n<p>8.\u00a0[latex]2\\sec^{2}\\frac{\\pi}{4}+\\cot^{2}\\frac{\\pi}{3}[\/latex]<\/p>\n<p>9. [latex]3\\csc^{2}\\frac{\\pi}{3}+\\cot^{2}\\frac{\\pi}{4}[\/latex]<\/p>\n<p>For the following exercises, use cofunctions of complementary angles.<\/p>\n<p>10. [latex]\\cos \\left(34^\\circ\\right)=\\sin \\left(\\text{__}^\\circ\\right)[\/latex]<\/p>\n<p>11. [latex]\\cos \\left(\\frac{\\pi }{3}\\right)=\\sin \\text{(___)}[\/latex]<\/p>\n<p>12.\u00a0[latex]\\csc \\left(21^\\circ\\right)=\\sec \\left(\\text{___}^\\circ \\right)[\/latex]<\/p>\n<p>13. [latex]\\tan \\left(\\frac{\\pi }{4}\\right)=\\cot \\left(\\text{__}\\right)[\/latex]<\/p>\n<p>For the following exercises, find the lengths of the missing sides if side [latex]a[\/latex] is opposite angle [latex]A[\/latex], side [latex]b[\/latex] is opposite angle [latex]B[\/latex], and side [latex]c[\/latex] is the hypotenuse.<\/p>\n<p>14. [latex]\\cos B=\\frac{4}{5},a=10[\/latex]<\/p>\n<p>15. [latex]\\sin B=\\frac{1}{2}, a=20[\/latex]<\/p>\n<p>16.\u00a0[latex]\\tan A=\\frac{5}{12},b=6[\/latex]<\/p>\n<p>17. [latex]\\tan A=100,b=100[\/latex]<\/p>\n<p>18.\u00a0[latex]\\sin B=\\frac{1}{\\sqrt{3}}, a=2[\/latex]<\/p>\n<p>19. [latex]a=5,\\measuredangle A={60}^{\\circ }[\/latex]<\/p>\n<p>20.\u00a0[latex]c=12,\\measuredangle A={45}^{\\circ }[\/latex]<\/p>\n<p>For the following exercises, use Figure 14 to evaluate each trigonometric function of angle [latex]A[\/latex].<\/p>\n<figure id=\"Figure_05_04_203\">\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\/27003827\/CNX_Precalc_Figure_05_04_2032.jpg\" alt=\"A right triangle with sides 4 and 10 and angle of A labeled.\" width=\"487\" height=\"202\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14<\/b><\/p>\n<\/div>\n<\/figure>\n<p>21. [latex]\\sin A[\/latex]<\/p>\n<p>22.\u00a0[latex]\\cos A[\/latex]<\/p>\n<p>23. [latex]\\tan A[\/latex]<\/p>\n<p>24.\u00a0[latex]\\csc A[\/latex]<\/p>\n<p>25. [latex]\\sec A[\/latex]<\/p>\n<p>26.\u00a0[latex]\\cot A[\/latex]<\/p>\n<p>For the following exercises, use Figure 15 to evaluate each trigonometric function of angle [latex]A[\/latex].<\/p>\n<figure id=\"Figure_05_04_204\">\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\/27003830\/CNX_Precalc_Figure_05_04_2042.jpg\" alt=\"A right triangle with sides of 10 and 8 and angle of A labeled.\" width=\"487\" height=\"454\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 15<\/b><\/p>\n<\/div>\n<\/figure>\n<p>27. [latex]\\sin A[\/latex]<\/p>\n<p>28.\u00a0[latex]\\cos A[\/latex]<\/p>\n<p>29. [latex]\\tan A[\/latex]<\/p>\n<p>30.\u00a0[latex]\\csc A[\/latex]<\/p>\n<p>31. [latex]\\sec A[\/latex]<\/p>\n<p>32.\u00a0[latex]\\cot A[\/latex]<\/p>\n<p>For the following exercises, solve for the unknown sides of the given triangle.<\/p>\n<p>33.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003832\/CNX_Precalc_Figure_05_04_2052.jpg\" alt=\"A right triangle with sides of 7, b, and c labeled. Angles of B and 30 degrees also labeled.\" \/><\/p>\n<p>34.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003834\/CNX_Precalc_Figure_05_04_2062.jpg\" alt=\"A right triangle with sides of 10, a, and c. Angles of 60 degrees and A also labeled.\" \/><\/p>\n<p>35.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003838\/CNX_Precalc_Figure_05_04_2072.jpg\" alt=\"A right triangle with corners labeled A, B, and C. Hypotenuse has length of 15 times square root of 2. Angle B is 45 degrees.\" \/><\/p>\n<p>For the following exercises, use a calculator to find the length of each side to four decimal places.<\/p>\n<p>36.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003840\/CNX_Precalc_Figure_05_04_2082.jpg\" alt=\"A right triangle with sides of 10, a, and c. Angles of A and 62 degrees are also labeled.\" \/><\/p>\n<p>37.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003843\/CNX_Precalc_Figure_05_04_2092.jpg\" alt=\"A right triangle with sides of 7, b, and c. Angles of 35 degrees and B are also labeled.\" \/><\/p>\n<p>38.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003845\/CNX_Precalc_Figure_05_04_2102.jpg\" alt=\"A right triangle with sides of a, b, and 10 labeled. Angles of 65 degrees and B are also labeled.\" \/><\/p>\n<p>39.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003850\/CNX_Precalc_Figure_05_04_2112.jpg\" alt=\"A right triangle with sides a, b, and 12. Angles of 10 degrees and B are also labeled.\" \/><\/p>\n<p>40.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003852\/CNX_Precalc_Figure_05_04_2122.jpg\" alt=\"A right triangle with corners labeled A, B, and C. Sides labeled b, c, and 16.5. Angle of 81 degrees also labeled.\" \/><\/p>\n<p>41. [latex]b=15,\\measuredangle B={15}^{\\circ }[\/latex]<\/p>\n<p>42.\u00a0[latex]c=200,\\measuredangle B={5}^{\\circ }[\/latex]<\/p>\n<p>43. [latex]c=50,\\measuredangle B={21}^{\\circ }[\/latex]<\/p>\n<p>44.\u00a0[latex]a=30,\\measuredangle A={27}^{\\circ }[\/latex]<\/p>\n<p>45. [latex]b=3.5,\\measuredangle A={78}^{\\circ }[\/latex]<\/p>\n<p>46. Find [latex]x[\/latex].<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003854\/CNX_Precalc_Figure_05_04_2132.jpg\" alt=\"A triangle with angles of 63 degrees and 39 degrees and side x. Bisector in triangle with length of 82.\" \/><\/p>\n<p>47. Find [latex]x[\/latex].<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003857\/CNX_Precalc_Figure_05_04_2142.jpg\" alt=\"A triangle with angles of 36 degrees and 50 degrees and side x. Bisector in triangle with length of 85.\" \/><\/p>\n<p>48. Find [latex]x[\/latex].<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003900\/CNX_Precalc_Figure_05_04_2152.jpg\" alt=\"A right triangle with side of 115 and angle of 35 degrees. Within right triangle there is another right triangle with angle of 56 degrees. Side length difference between two triangles is x.\" \/><\/p>\n<p>49. Find [latex]x[\/latex].<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003903\/CNX_Precalc_Figure_05_04_2162.jpg\" alt=\"A right triangle with side of 119 and angle of 26 degrees. Within right triangle there is another right triangle with angle of 70 degrees instead of 26 degrees. Difference in side length between two triangles is x.\" \/><\/p>\n<p>50.\u00a0A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]36^\\circ[\/latex], and that the angle of depression to the bottom of the tower is [latex]23^\\circ[\/latex]. How tall is the tower?<\/p>\n<p>51. A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]43^\\circ[\/latex], and that the angle of depression to the bottom of the tower is [latex]31^\\circ[\/latex]. How tall is the tower?<\/p>\n<p>52.\u00a0A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]15^\\circ[\/latex], and that the angle of depression to the bottom of the tower is [latex]2^\\circ[\/latex]. How far is the person from the monument?<\/p>\n<p>53. A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]18^\\circ[\/latex], and that the angle of depression to the bottom of the tower is [latex]3^\\circ[\/latex]. How far is the person from the monument?<\/p>\n<p>54.\u00a0There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]40^\\circ[\/latex]. From the same location, the angle of elevation to the top of the antenna is measured to be [latex]43^\\circ[\/latex]. Find the height of the antenna.<\/p>\n<p>55. There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]36^\\circ[\/latex]. From the same location, the angle of elevation to the top of the lightning rod is measured to be [latex]38^\\circ[\/latex]. Find the height of the lightning rod.<\/p>\n<p>56. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\\circ[\/latex]. How high does the ladder reach up the side of the building?<\/p>\n<p>57. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\\circ[\/latex]. How high does the ladder reach up the side of the building?<\/p>\n<p>58.\u00a0The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.<\/p>\n<p>59. The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building.<\/p>\n<p>60.\u00a0Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be [latex]60^\\circ[\/latex], how far from the base of the tree am I?<\/p>\n<p>61. A car travels west for 5 miles, turns left, and then travels south for 9 miles.\u00a0 What is the bearing from car&#8217;s starting position to its current position?\u00a0 Round your answer to two decimal places.<\/p>\n<p>62. A truck travels east for 4 miles, turns left, and then travels north for 6 miles.\u00a0 What is the bearing from the truck&#8217;s starting position to its current position? Round your answer to two decimal places.<\/p>\n<p>63. An ant travels on a bearing of [latex]N 22^\\circ E[\/latex] for 36 inches.\u00a0 How far east and how far north is the ant from its starting position?\u00a0 Round your answers to two decimal places.<\/p>\n<p>64. A spider crawls on a bearing of [latex]S 34^\\circ W[\/latex] for 20 inches.\u00a0 How far west and how far south is the spider from its starting point? Round your answers to two decimal places.<\/p>\n<p>65.\u00a0Originally built in 1901 by Colonel J.W. Eddy, Angels Flight in Los Angeles is said to be the world&#8217;s shortest incorporated railway.\u00a0 The counterbalanced cars, controlled by cables, travel a 33% grade for 315 feet.\u00a0 What is the angle the track makes with a horizontal line, rounded to one decimal place?<\/p>\n<p>66. The Saluda Grade is the steepest standard-gauge mainline railway grade in the United States.\u00a0 Between Melrose and Saluda, North Carolina, the maximum grade is 4.9% for about 300 feet.\u00a0 What is the angle the track makes with a horizontal line, rounded to one decimal place?<\/p>\n","protected":false},"author":264444,"menu_order":1,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-17970","chapter","type-chapter","status-publish","hentry"],"part":14256,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17970","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17970\/revisions"}],"predecessor-version":[{"id":17987,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17970\/revisions\/17987"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/14256"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17970\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=17970"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=17970"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=17970"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=17970"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}