{"id":222,"date":"2025-02-13T22:45:13","date_gmt":"2025-02-13T22:45:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-law-of-sines\/"},"modified":"2025-10-17T21:26:50","modified_gmt":"2025-10-17T21:26:50","slug":"non-right-triangles-law-of-sines","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-law-of-sines\/","title":{"raw":"Non-right Triangles with Law of Sines: Learn It 1","rendered":"Non-right Triangles with Law of Sines: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the Law of Sines to solve oblique triangles.<\/li>\r\n \t<li>Find the area of an oblique triangle using the sine function.<\/li>\r\n \t<li>Solve applied problems using the Law of Sines.<\/li>\r\n<\/ul>\r\n<\/section>Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the <strong>angle of elevation<\/strong> measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? The triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving <strong>non-right triangles<\/strong>.\r\n\r\n&nbsp;\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165008\/CNX_Precalc_Figure_08_01_0012.jpg\" alt=\"A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted line perpendicular to the ground side connecting the airplane vertex with the ground - an altitude line.\" width=\"487\" height=\"134\" \/>\r\n<h2>Using the Law of Sines to Solve Oblique Triangles<\/h2>\r\nIn any triangle, we can draw an <strong>altitude<\/strong>, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.\r\n\r\nAny triangle that is not a right triangle is an <strong>oblique triangle<\/strong>. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<ol>\r\n \t<li><strong>ASA (angle-side-angle)<\/strong> We know the measurements of two angles and the included side.<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165011\/CNX_Precalc_Figure_08_01_0022.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma's values are known, as is the side opposite beta, between alpha and gamma.\" width=\"487\" height=\"141\" \/><\/li>\r\n \t<li><strong>AAS (angle-angle-side)<\/strong> We know the measurements of two angles and a side that is not between the known angles.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165013\/CNX_Precalc_Figure_08_01_0032.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma are known, as is the side opposite alpha, between beta and gamma.\" width=\"487\" height=\"141\" \/><\/li>\r\n \t<li><strong>SSA (side-side-angle)<\/strong> We know the measurements of two sides and an angle that is not between the known sides.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165016\/CNX_Precalc_Figure_08_01_0042.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha is the only angle known. Two sides are known. The first is opposite alpha, between beta and gamma, and the second is opposite gamma, between alpha and beta.\" width=\"487\" height=\"141\" \/><\/li>\r\n<\/ol>\r\n<\/section>Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165018\/CNX_Precalc_Figure_08_01_0052.jpg\" alt=\"An oblique triangle consisting of sides a, b, and c, and angles alpha, beta, and gamma. Side c is opposide angle gamma and is the horizontal base of the triangle. Side b is opposite angle beta, and side a is opposite angle alpha. There is a dotted perpendicular line - an altitude - from the gamma angle to the horizontal base c.\" width=\"487\" height=\"143\" \/>\r\n\r\nUsing the right triangle relationships, we know that [latex]\\sin \\alpha =\\frac{h}{b}[\/latex] and [latex]\\sin \\beta =\\frac{h}{a}[\/latex]. Solving both equations for [latex]h[\/latex] gives two different expressions for [latex]h[\/latex].\r\n<div style=\"text-align: center;\">[latex]h=b\\sin \\alpha \\text{ and }h=a\\sin \\beta [\/latex]<\/div>\r\nWe then set the expressions equal to each other.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}b\\sin \\alpha &amp;=a\\sin \\beta \\\\ \\left(\\frac{1}{ab}\\right)\\left(b\\sin \\alpha \\right)&amp;=\\left(a\\sin \\beta \\right)\\left(\\frac{1}{ab}\\right) &amp;&amp; \\text{Multiply both sides by}\\frac{1}{ab}. \\\\ \\frac{\\sin \\alpha }{a}&amp;=\\frac{\\sin \\beta }{b} \\end{align}[\/latex]<\/div>\r\nSimilarly, we can compare the other ratios.\r\n<div style=\"text-align: center;\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\gamma }{c}\\text{ and }\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/div>\r\nCollectively, these relationships are called the <strong>Law of Sines<\/strong>.\r\n<div style=\"text-align: center;\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\lambda }{c}[\/latex]<\/div>\r\nNote the standard way of labeling triangles: angle [latex]\\alpha [\/latex] (alpha) is opposite side [latex]a[\/latex]; angle [latex]\\beta [\/latex] (beta) is opposite side [latex]b[\/latex]; and angle [latex]\\gamma [\/latex] (gamma) is opposite side [latex]c[\/latex].\r\n\r\nWhile calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>Law of Sines<\/h3>\r\nGiven a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The <strong>Law of Sines<\/strong> is based on proportions and is presented symbolically two ways.\r\n<p style=\"text-align: center;\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{a}{\\sin \\alpha }=\\frac{b}{\\sin \\beta }=\\frac{c}{\\sin \\gamma }[\/latex]<\/p>\r\nTo solve an oblique triangle, use any pair of applicable ratios.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the triangle to the nearest tenth.<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165022\/CNX_Precalc_Figure_08_01_0072.jpg\" alt=\"An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.\" width=\"487\" height=\"200\" \/>[reveal-answer q=\"604625\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"604625\"]The three angles must add up to 180 degrees. From this, we can determine that\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\beta =180^\\circ -50^\\circ -30^\\circ =100^\\circ \\end{align}[\/latex]<\/p>\r\nTo find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle [latex]\\alpha =50^\\circ [\/latex] and its corresponding side [latex]a=10[\/latex]. We can use the following proportion from the Law of Sines to find the length of [latex]c[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\frac{\\sin \\left(30^\\circ \\right)}{c} \\\\ &amp;c\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\sin \\left(30^\\circ \\right) &amp;&amp; \\text{Multiply both sides by }c. \\\\ &amp;c=\\sin \\left(30^\\circ \\right)\\frac{10}{\\sin \\left(50^\\circ \\right)} &amp;&amp; \\text{Multiply by the reciprocal to isolate }c. \\\\ &amp;c\\approx 6.5 \\end{align}[\/latex]<\/p>\r\nSimilarly, to solve for [latex]b[\/latex], we set up another proportion.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\frac{\\sin \\left(100^\\circ \\right)}{b} \\\\ &amp;b\\sin \\left(50^\\circ \\right)=10\\sin \\left(100^\\circ \\right) &amp;&amp; \\text{Multiply both sides by }b. \\\\ &amp;b=\\frac{10\\sin \\left(100^\\circ \\right)}{\\sin \\left(50^\\circ \\right)} &amp;&amp; \\text{Multiply by the reciprocal to isolate }b. \\\\ &amp;b\\approx 12.9\\end{align}[\/latex]<\/p>\r\nTherefore, the complete set of angles and sides is\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} \\alpha =50^\\circ,\\beta =100^\\circ,\\gamma =30^\\circ \\\\ a=10,b\\approx 12.9,c\\approx 6.5 \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Solve the triangle to the nearest tenth.<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165024\/CNX_Precalc_Figure_08_01_0082.jpg\" alt=\"An oblique triangle with standard labels. Angle alpha is 98 degrees, angle gamma is 43 degrees, and side b is of length 22. Side b is the horizontal base.\" width=\"487\" height=\"247\" \/>[reveal-answer q=\"933392\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"933392\"][latex]\\begin{gathered}\\alpha ={98}^{\\circ },\\beta ={39}^{\\circ },\\gamma ={43}^{\\circ } \\\\ a=34.6, b=22, c=23.8\\end{gathered}[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]149230[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the Law of Sines to solve oblique triangles.<\/li>\n<li>Find the area of an oblique triangle using the sine function.<\/li>\n<li>Solve applied problems using the Law of Sines.<\/li>\n<\/ul>\n<\/section>\n<p>Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the <strong>angle of elevation<\/strong> measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? The triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving <strong>non-right triangles<\/strong>.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165008\/CNX_Precalc_Figure_08_01_0012.jpg\" alt=\"A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted line perpendicular to the ground side connecting the airplane vertex with the ground - an altitude line.\" width=\"487\" height=\"134\" \/><\/p>\n<h2>Using the Law of Sines to Solve Oblique Triangles<\/h2>\n<p>In any triangle, we can draw an <strong>altitude<\/strong>, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.<\/p>\n<p>Any triangle that is not a right triangle is an <strong>oblique triangle<\/strong>. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<ol>\n<li><strong>ASA (angle-side-angle)<\/strong> We know the measurements of two angles and the included side.<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165011\/CNX_Precalc_Figure_08_01_0022.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma's values are known, as is the side opposite beta, between alpha and gamma.\" width=\"487\" height=\"141\" \/><\/li>\n<li><strong>AAS (angle-angle-side)<\/strong> We know the measurements of two angles and a side that is not between the known angles.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165013\/CNX_Precalc_Figure_08_01_0032.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma are known, as is the side opposite alpha, between beta and gamma.\" width=\"487\" height=\"141\" \/><\/li>\n<li><strong>SSA (side-side-angle)<\/strong> We know the measurements of two sides and an angle that is not between the known sides.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165016\/CNX_Precalc_Figure_08_01_0042.jpg\" alt=\"An oblique triangle consisting of angles alpha, beta, and gamma. Alpha is the only angle known. Two sides are known. The first is opposite alpha, between beta and gamma, and the second is opposite gamma, between alpha and beta.\" width=\"487\" height=\"141\" \/><\/li>\n<\/ol>\n<\/section>\n<p>Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165018\/CNX_Precalc_Figure_08_01_0052.jpg\" alt=\"An oblique triangle consisting of sides a, b, and c, and angles alpha, beta, and gamma. Side c is opposide angle gamma and is the horizontal base of the triangle. Side b is opposite angle beta, and side a is opposite angle alpha. There is a dotted perpendicular line - an altitude - from the gamma angle to the horizontal base c.\" width=\"487\" height=\"143\" \/><\/p>\n<p>Using the right triangle relationships, we know that [latex]\\sin \\alpha =\\frac{h}{b}[\/latex] and [latex]\\sin \\beta =\\frac{h}{a}[\/latex]. Solving both equations for [latex]h[\/latex] gives two different expressions for [latex]h[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]h=b\\sin \\alpha \\text{ and }h=a\\sin \\beta[\/latex]<\/div>\n<p>We then set the expressions equal to each other.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}b\\sin \\alpha &=a\\sin \\beta \\\\ \\left(\\frac{1}{ab}\\right)\\left(b\\sin \\alpha \\right)&=\\left(a\\sin \\beta \\right)\\left(\\frac{1}{ab}\\right) && \\text{Multiply both sides by}\\frac{1}{ab}. \\\\ \\frac{\\sin \\alpha }{a}&=\\frac{\\sin \\beta }{b} \\end{align}[\/latex]<\/div>\n<p>Similarly, we can compare the other ratios.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\gamma }{c}\\text{ and }\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/div>\n<p>Collectively, these relationships are called the <strong>Law of Sines<\/strong>.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\lambda }{c}[\/latex]<\/div>\n<p>Note the standard way of labeling triangles: angle [latex]\\alpha[\/latex] (alpha) is opposite side [latex]a[\/latex]; angle [latex]\\beta[\/latex] (beta) is opposite side [latex]b[\/latex]; and angle [latex]\\gamma[\/latex] (gamma) is opposite side [latex]c[\/latex].<\/p>\n<p>While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>Law of Sines<\/h3>\n<p>Given a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The <strong>Law of Sines<\/strong> is based on proportions and is presented symbolically two ways.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b}=\\frac{\\sin \\gamma }{c}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{a}{\\sin \\alpha }=\\frac{b}{\\sin \\beta }=\\frac{c}{\\sin \\gamma }[\/latex]<\/p>\n<p>To solve an oblique triangle, use any pair of applicable ratios.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the triangle to the nearest tenth.<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165022\/CNX_Precalc_Figure_08_01_0072.jpg\" alt=\"An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.\" width=\"487\" height=\"200\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q604625\">Show Solution<\/button><\/p>\n<div id=\"q604625\" class=\"hidden-answer\" style=\"display: none\">The three angles must add up to 180 degrees. From this, we can determine that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\beta =180^\\circ -50^\\circ -30^\\circ =100^\\circ \\end{align}[\/latex]<\/p>\n<p>To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle [latex]\\alpha =50^\\circ[\/latex] and its corresponding side [latex]a=10[\/latex]. We can use the following proportion from the Law of Sines to find the length of [latex]c[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\frac{\\sin \\left(30^\\circ \\right)}{c} \\\\ &c\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\sin \\left(30^\\circ \\right) && \\text{Multiply both sides by }c. \\\\ &c=\\sin \\left(30^\\circ \\right)\\frac{10}{\\sin \\left(50^\\circ \\right)} && \\text{Multiply by the reciprocal to isolate }c. \\\\ &c\\approx 6.5 \\end{align}[\/latex]<\/p>\n<p>Similarly, to solve for [latex]b[\/latex], we set up another proportion.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &\\frac{\\sin \\left(50^\\circ \\right)}{10}=\\frac{\\sin \\left(100^\\circ \\right)}{b} \\\\ &b\\sin \\left(50^\\circ \\right)=10\\sin \\left(100^\\circ \\right) && \\text{Multiply both sides by }b. \\\\ &b=\\frac{10\\sin \\left(100^\\circ \\right)}{\\sin \\left(50^\\circ \\right)} && \\text{Multiply by the reciprocal to isolate }b. \\\\ &b\\approx 12.9\\end{align}[\/latex]<\/p>\n<p>Therefore, the complete set of angles and sides is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} \\alpha =50^\\circ,\\beta =100^\\circ,\\gamma =30^\\circ \\\\ a=10,b\\approx 12.9,c\\approx 6.5 \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Solve the triangle to the nearest tenth.<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165024\/CNX_Precalc_Figure_08_01_0082.jpg\" alt=\"An oblique triangle with standard labels. Angle alpha is 98 degrees, angle gamma is 43 degrees, and side b is of length 22. Side b is the horizontal base.\" width=\"487\" height=\"247\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q933392\">Show Solution<\/button><\/p>\n<div id=\"q933392\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{gathered}\\alpha ={98}^{\\circ },\\beta ={39}^{\\circ },\\gamma ={43}^{\\circ } \\\\ a=34.6, b=22, c=23.8\\end{gathered}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm149230\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149230&theme=lumen&iframe_resize_id=ohm149230&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":6,"menu_order":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":221,"module-header":"learn_it","content_attributions":[{"type":"cc-attribution","description":"Precalculus","author":"OpenStax College","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/222"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":11,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/222\/revisions"}],"predecessor-version":[{"id":4735,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/222\/revisions\/4735"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/221"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/222\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=222"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=222"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=222"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=222"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}