{"id":240,"date":"2017-08-23T10:28:49","date_gmt":"2017-08-23T10:28:49","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/mathematics\/?page_id=240"},"modified":"2017-09-25T16:10:58","modified_gmt":"2017-09-25T16:10:58","slug":"trigonometric-ratios","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/mathematics\/trigonometric-ratios\/","title":{"rendered":"Trigonometric Ratios"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\"><!--<a href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/math_05_08.html\" class=\"button button-primary\">\u2b05 Previous Lesson<\/a>--><br \/>\n<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/trigonometry\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/law-of-sines-law-of-cosines\">Next Lesson \u27a1<\/a><\/div>\n<p><!-- CONTENT BEGINS HERE --><\/p>\n<h1 id=\"title\">Trigonometric Ratios<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson, we will discuss how to solve trigonometry problems using angle and side relationships for special\u00a0right triangles and the basic trigonometric ratios. You will also learn two different standard measurements for\u00a0angles, degrees and radians, and how to convert between the two.<\/p>\n<h4>Previously Covered:<\/h4>\n<ul>\n<li>The <em><strong>Pythagorean\u00a0Theorem<\/strong><\/em> is used to find the length of a side of a\u00a0right triangle.<\/li>\n<li>A <strong><em>right triangle\u00a0<\/em><\/strong>is a triangle containing only one 90\u00b0 angle.<\/li>\n<li>An <strong><em>acute angle<\/em><\/strong> is an angle less than 90\u00b0.<\/li>\n<li>An <strong><em>obtuse angle<\/em><\/strong> is an angle greater than 90\u00b0.<\/li>\n<li>A <strong><em>right angle<\/em><\/strong> is a 90\u00b0 angle.<\/li>\n<li><strong><em>Sine<\/em><\/strong><em>,\u00a0<strong>cosine<\/strong><\/em>, and <strong><em>tangent\u00a0<\/em><\/strong>are some of the trigonometric functions.<\/li>\n<li>The sum of the interior angles of a triangle always equals 180\u00b0.<\/li>\n<\/ul>\n<section>\n<h3>What is Trigonometry?<\/h3>\n<p><abbr title=\"a collection of techniques that can be used to calculate the length of sides and angles in triangles\">Trigonometry<\/abbr> is a collection of techniques that can be used to calculate the\u00a0length of sides and the measure of angles in triangles. When\u00a0working with trigonometry, it may be helpful to think of yourself\u00a0as a detective. This module will teach you how to use the pieces of information given to you to determine everything about certain\u00a0types of triangles.<\/p>\n<p>Trigonometry is also sometimes referred to as pre-calculus\u00a0because many of the concepts in trigonometry serve as precursors\u00a0to the derivatives, integrals, and rates of change that you will\u00a0learn about in the upcoming calculus module.<\/p>\n<p>What are the components of a right triangle?<\/p>\n<p>In order to fully illustrate the trigonometric ratios that will\u00a0be covered later in this module, it is important that we first\u00a0understand the parts that make up a right triangle.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20001.JPG\" alt=\"Triangle labeled\" width=\"235\" height=\"109\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"text-decoration: none;\">Side <\/span><em><span style=\"text-decoration: none;\">c<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0the <\/span><abbr title=\"the longest side of a right triangle, which is opposite the right angle \"><span style=\"text-decoration: none;\">hypotenuse<\/span><\/abbr><span style=\"text-decoration: none;\">,\u00a0is the longest side of the right triangle and does not have a\u00a0right angle at either end. The other two sides of the triangle (<\/span><em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em><span style=\"text-decoration: none;\">and <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">)\u00a0are referred to as <\/span><abbr title=\" the two sides of a right triangle that form a right angle at their intersection.\"><span style=\"text-decoration: none;\">legs<\/span><\/abbr><span style=\"text-decoration: none;\">.\u00a0The sum of the angles in a right triangle, as with all triangles,\u00a0is 180\u00b0. So, every right triangle contains one 90\u00b0 angle\u00a0and two acute angles. (It is impossible for a right triangle to\u00a0contain an obtuse angle as it would force the sum of angles to be\u00a0greater than 180\u00b0.) You will find that the application of\u00a0trigonometric ratios often requires the parts of a triangle to be\u00a0referred to with respect to a given point of reference (usually an\u00a0angle). For instance, the diagram below shows how the sides of a\u00a0right triangle would be referred to using a given angle, in this\u00a0case\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p2_clip_image003.gif\" width=\"13\" height=\"19\" name=\"graphics4\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0as a point of reference. <\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20002.JPG\" alt=\"Triangle with angle theta\" width=\"277\" height=\"147\" name=\"graphics5\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p><span style=\"text-decoration: none;\">With respect to the angle\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p2_clip_image003_0000.gif\" width=\"13\" height=\"19\" name=\"graphics6\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0side <\/span><em><span style=\"text-decoration: none;\">b\u00a0<\/span><\/em><span style=\"text-decoration: none;\">is the adjacent side. It is immediately between the angle\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p2_clip_image003_0001.gif\" width=\"13\" height=\"19\" name=\"graphics7\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0and the 90\u00b0 angle. The adjacent side and the hypotenuse form\u00a0the angle\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p2_clip_image003_0002.gif\" width=\"13\" height=\"19\" name=\"graphics8\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0Side <\/span><em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em><span style=\"text-decoration: none;\">is the opposite side. It is located across the triangle from the\u00a0angle\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p2_clip_image003_0003.gif\" width=\"13\" height=\"19\" name=\"graphics9\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0The opposite side and adjacent side of a right triangle form the\u00a0right angle. <\/span><\/p>\n<h3>How can basic trigonometric ratios be used to analyze right triangles?<\/h3>\n<p><abbr title=\"Ratios that allow you to associate sides and angles in right triangles.\">Trigonometric ratios<\/abbr> are ratios that allow you to associate sides and angles\u00a0in right triangles. A common mnemonic device for remembering the\u00a0three basic trigonometric ratios is the term <em>soh cah toa.\u00a0<\/em>The \u201cS-O-H\u201d in this term stands for \u201csine,\u00a0opposite, hypotenuse,\u201d and is interpreted as, \u201cthe\u00a0sine of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image003.gif\" width=\"13\" height=\"19\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0is equal to the length of the opposite leg over the length of the\u00a0hypotenuse.\u201d This can be written as a formula as\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image006.gif\" width=\"185\" height=\"47\" name=\"graphics4\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0The \u201cC-A-H\u201d in this term stands for \u201ccosine,\u00a0adjacent, hypotenuse,\u201d and is interpreted as, \u201cthe\u00a0cosine of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image003_0000.gif\" width=\"13\" height=\"19\" name=\"graphics5\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0is equal to the length of the adjacent leg over the length of the\u00a0hypotenuse.\u201d This can be written as a formula as\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image010.gif\" width=\"188\" height=\"47\" name=\"graphics6\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0The \u201cT-O-A\u201d stands for \u201ctangent, opposite,\u00a0adjacent\u201d, and is interpreted as, \u201cthe tangent of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image003_0001.gif\" width=\"13\" height=\"19\" name=\"graphics7\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0is equal to the length of the opposite leg over the length of the\u00a0adjacent leg.\u201d This can be written as a formula as\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image014.gif\" width=\"169\" height=\"47\" name=\"graphics8\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0A practical application of these trigonometric ratios is shown in\u00a0the example below.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>On the figure below, which of the following is the length of\u00a0the hypotenuse?<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20003.JPG\" alt=\"Triangle Labeled a, c, and 20 m.\" width=\"235\" height=\"110\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<ol>\n<li>20.01 meters<\/li>\n<li>23.09 meters<\/li>\n<li>40 meters<\/li>\n<li>36.02 meters<\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The correct choice is B. The solution can be found by using\u00a0the trigonometric ratio that we just learned for finding cosine.\u00a0We know to use the cosine ratio because we are trying to solve\u00a0for the length of the hypotenuse, and we know the adjacent\u00a0measurement and the angle measurement at the intersection point\u00a0between the hypotenuse and the adjacent angle.<\/p>\n<p>We know from trigonometric ratios (and the C-A-H in soh cah\u00a0toa) that<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image017.gif\" width=\"129\" height=\"44\" name=\"graphics10\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p>Using the principle of substitution, we can say<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image020.gif\" width=\"108\" height=\"41\" name=\"graphics11\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p><span style=\"text-decoration: none;\">Multiply both sides of the\u00a0equation by <\/span><em><span style=\"text-decoration: none;\">c\u00a0<\/span><\/em><span style=\"text-decoration: none;\">(our unknown hypotenuse<\/span>).<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image023.gif\" width=\"115\" height=\"24\" name=\"graphics12\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>Divide both sides of the equation by\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image026.gif\" width=\"61\" height=\"24\" name=\"graphics13\" align=\"ABSMIDDLE\" border=\"0\" \/>to\u00a0find length of hypotenuse.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p3_clip_image029.gif\" width=\"153\" height=\"45\" name=\"graphics14\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<\/div>\n<\/section>\n<div class=\"callout\">\n<h4>Important Tidbit<\/h4>\n<p class=\"notebox_text\" align=\"CENTER\">It is important that you find a way that is easy for you to remember\u00a0these three\u00a0trigonometric ratios. If you have these memorized, trigonometry will be a great deal more fun and, as\u00a0the material\u00a0becomes more complex, knowing these basic building blocks will make the process much easier. If you are\u00a0having\u00a0trouble spelling soh cah toa, try making up a phrase that will help you remember these three basic\u00a0trigonometric\u00a0ratios.<\/p>\n<\/div>\n<p>There are actually six trigonometric ratios. The three\u00a0trigonometric ratios that we have not yet discussed are the\u00a0secant, cotangent, and cosecant. These ratios can be derived from\u00a0the three basic ratios we have already learned. We will discuss\u00a0these remaining ratios in more detail later in the lesson.<\/p>\n<h3>What are the short cuts for special right triangles?<\/h3>\n<p>The two common special right triangles that we will discuss in\u00a0this module are the <abbr title=\"A special right triangle consisting of two 45\u00b0 angles and one 90\u00b0 angle. The legs of the isosceles right triangle are of equal length. This triangle has a length of side ratio of aaa. \">isosceles\u00a0right triangle<\/abbr>, also known as the 45-45-90 triangle, and the\u00a030-60-90 triangle. These triangles have \u201cshort cuts\u201d\u00a0to help us solve for the lengths of their sides.<\/p>\n<p>The isosceles right triangle is composed of two 45\u00b0 acute\u00a0angles and one 90\u00b0 right angle. (Remember that the sum of\u00a0angles in a triangle must always equal 180 degrees.) The lengths\u00a0of the legs of an isosceles right triangle are always equal. For\u00a0instance, <span style=\"text-decoration: none;\">let the legs be\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image003.gif\" width=\"16\" height=\"15\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0and\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image006.gif\" width=\"16\" height=\"21\" name=\"graphics4\" align=\"TEXTTOP\" border=\"0\" \/>,\u00a0and the hypotenuse be <\/span><em><span style=\"text-decoration: none;\">c<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0Can you derive the hypotenuse in terms of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image003_0000.gif\" width=\"16\" height=\"15\" name=\"graphics5\" align=\"ABSMIDDLE\" border=\"0\" \/>?\u00a0Let\u2019s try it.<\/span><\/p>\n<p>Using the Pythagorean Theorem, we know that\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image010.gif\" width=\"83\" height=\"21\" name=\"graphics6\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<p>We know that an isosceles right triangle has two legs that are\u00a0of equal length (in this case,\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image003_0001.gif\" width=\"16\" height=\"15\" name=\"graphics7\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0and\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image006_0000.gif\" width=\"16\" height=\"21\" name=\"graphics8\" align=\"ABSMIDDLE\" border=\"0\" \/>),\u00a0so we can say that<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image015.gif\" width=\"89\" height=\"48\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>Now we can take the square root of both sides.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image018.gif\" width=\"60\" height=\"23\" name=\"graphics10\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p>So the answer is yes, we can derive the hypotenuse in terms of \u00a0<em><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image003_0002.gif\" width=\"16\" height=\"15\" name=\"graphics11\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/em><\/p>\n<p>An easy way of remembering the short cut to the isosceles right\u00a0triangle is to remember an isosceles triangle always has sides\u00a0with a ratio\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image022.gif\" width=\"56\" height=\"23\" name=\"graphics12\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>Using the figure below and the short cut we just learned for an\u00a0isosceles right triangle, solve for <span style=\"text-decoration: none;\">side\u00a0<\/span><em><span style=\"text-decoration: none;\">c<\/span><\/em><span style=\"text-decoration: none;\">.<\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20004.JPG\" alt=\"Triangle with sides a, c and 10 m\" width=\"158\" height=\"137\" name=\"graphics13\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<ol>\n<li>10 meters<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image025.gif\" width=\"39\" height=\"24\" name=\"graphics14\" align=\"ABSBOTTOM\" border=\"0\" \/>\u00a0meters<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image028.gif\" width=\"40\" height=\"23\" name=\"graphics15\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0meters<\/li>\n<li>20 meters<\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The correct choice is C. If you use the short cut for an\u00a0isosceles right triangle, you know that <em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= 10 meters because the legs of an isosceles triangle are equal.\u00a0You also know that the hypotenuse of an <\/span>isosceles right\u00a0triangle is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p4_clip_image031.gif\" width=\"88\" height=\"23\" name=\"graphics16\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<\/div>\n<\/section>\n<h3>Trigonometric Ratios<\/h3>\n<p><span style=\"text-decoration: none;\">Another type of special\u00a0right triangle is the 30-60-90 triangle. The 30-60-90 triangle has\u00a0a 30\u00b0 angle, a 60\u00b0 angle, and a 90\u00b0 angle, where the\u00a0hypotenuse is <\/span><em><span style=\"text-decoration: none;\">c<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0the side opposite the 60\u00b0 angle is <\/span><em><span style=\"text-decoration: none;\">a<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0and the side opposite the 30\u00b0 angle is <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0So in this case, using the trigonometric ratios:<\/span><\/p>\n<table>\n<tbody>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image003.gif\" width=\"87\" height=\"41\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image006.gif\" width=\"85\" height=\"41\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/><\/td>\n<\/tr>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image009.gif\" width=\"95\" height=\"24\" name=\"graphics5\" align=\"BOTTOM\" border=\"0\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image012.gif\" width=\"92\" height=\"24\" name=\"graphics6\" align=\"BOTTOM\" border=\"0\" \/><\/td>\n<\/tr>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image015.gif\" width=\"60\" height=\"45\" name=\"graphics7\" align=\"BOTTOM\" border=\"0\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image018.gif\" width=\"40\" height=\"41\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>An easy way of remembering the short cut to the 30-60-90\u00a0triangle is to remember that a 30-60-90 triangle always has sides\u00a0with a ratio\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image021.gif\" width=\"61\" height=\"24\" name=\"graphics9\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0You may wonder how we know that this ratio is valid for every\u00a030-60-90 triangle. By manipulating the information we have\u00a0available, we will prove that the ratio of a 30-60-90 triangle\u00a0always has sides with a ratio\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image021_0000.gif\" width=\"61\" height=\"24\" name=\"graphics10\" align=\"ABSMIDDLE\" border=\"0\" \/><\/p>\n<p>Since we have already proven that\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image018_0000.gif\" width=\"40\" height=\"41\" name=\"graphics11\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0<em><span style=\"text-decoration: none;\">c\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= <\/span><em><span style=\"text-decoration: none;\">c<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0and\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image015_0000.gif\" width=\"60\" height=\"45\" name=\"graphics12\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0we have the<\/span> ratio <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image027.gif\" width=\"73\" height=\"45\" name=\"graphics13\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0Now divide the ratio by c, and we have the ratio\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image030.gif\" width=\"63\" height=\"45\" name=\"graphics14\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0Lastly, you multiply the answer by 2 and\u2026Bingo! You have<br \/>\nthe ratio 1:2:<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image033.gif\" width=\"24\" height=\"24\" name=\"graphics15\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>Using the triangle below, what is the length of s<span style=\"text-decoration: none;\">ide\u00a0<\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">?<br \/>\n<\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20005.JPG\" alt=\"riangle with sides a, b, and 20 m\" width=\"158\" height=\"159\" name=\"graphics16\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<ol>\n<li>10 m<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image039.gif\" width=\"41\" height=\"45\" name=\"graphics17\" align=\"ABSMIDDLE\" border=\"0\" \/> m<\/li>\n<li>20 m<\/li>\n<li>18 m<\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>T<span style=\"text-decoration: none;\">he correct choice is A.\u00a0Since this is a 30-60-90 triangle, you can apply the\u00a0aforementioned\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image042.gif\" width=\"57\" height=\"24\" name=\"graphics18\" align=\"ABSMIDDLE\" border=\"0\" \/><br \/>\nratio in order to solve for the length of side <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0We know that side <\/span><em><span style=\"text-decoration: none;\">c\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= 20, and we know that side <\/span><em><span style=\"text-decoration: none;\">c\u00a0<\/span><\/em><span style=\"text-decoration: none;\">is twice as long as side <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em> (2:1\u00a0ratio). Arithmetically, we can now solve for side <em><span style=\"text-decoration: none;\">b\u00a0<\/span><\/em><span style=\"text-decoration: none;\">as <\/span><em><span style=\"text-decoration: none;\">b\u00a0<\/span><\/em><span style=\"text-decoration: none;\">=\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p5_clip_image045.gif\" width=\"55\" height=\"41\" name=\"graphics19\" align=\"ABSMIDDLE\" border=\"0\" \/>m.<br \/>\n<\/span><\/p>\n<\/div>\n<\/section>\n<p>The special right triangles we\u00a0have discussed are shown below with each side in terms of the\u00a0hypotenuse.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20006.JPG\" alt=\"Special Right Triangles\" width=\"574\" height=\"288\" name=\"graphics20\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<h3>What is a radian?<\/h3>\n<p><span style=\"text-decoration: none;\">Does the word <\/span><abbr title=\" The alternative to degrees when quantifying the measurement of an angle. It is best described as the radius of the circle spread out over the circumference of the circle. So, for every 360\u00b0, there are 2pi radians, or for every 180\u00b0 there are pi radians.\"><span style=\"text-decoration: none;\">radian<\/span><\/abbr><span style=\"text-decoration: none;\"> sound familiar? You may have heard this term used before in a math\u00a0class or by an engineer. We all know how to measure angles in\u00a0degrees, but scientists and engineers commonly measure angles in\u00a0radians. A radian is the length of the radius of the circle placed\u00a0along the circumference. For every 360\u00b0, there are 2<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image003.gif\" width=\"15\" height=\"15\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0radians, and for every 180\u00b0, there are\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image003_0000.gif\" width=\"15\" height=\"15\" name=\"graphics4\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0radians. In other words, if the length of the radius <\/span><em><span style=\"text-decoration: none;\">r\u00a0<\/span><\/em><span style=\"text-decoration: none;\">is equal to the length of the arc <\/span><em><span style=\"text-decoration: none;\">a<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0then\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image007.gif\" width=\"13\" height=\"19\" name=\"graphics5\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0is one radian. Therefore, the formulae to convert from radians to\u00a0degrees and degrees to radians are: <\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20007.JPG\" alt=\"Radians\" width=\"276\" height=\"179\" name=\"graphics6\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<ul>\n<li>Degrees = radians\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image010.gif\" width=\"47\" height=\"41\" name=\"graphics7\" align=\"ABSMIDDLE\" border=\"0\" \/><\/li>\n<li>Radians = degrees\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image013.gif\" width=\"47\" height=\"41\" name=\"graphics8\" align=\"ABSMIDDLE\" \/><\/li>\n<\/ul>\n<p>An easy way of converting from radians to degrees is to\u00a0remember that\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/s1_p6_html_m7141c7c5.gif\" width=\"59\" height=\"21\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/>.\u00a0So if you need to convert\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image019.gif\" width=\"17\" height=\"41\" name=\"graphics10\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0radians into degrees, you could just substitute the\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/s1_p6_html_m256fa1a8.gif\" width=\"72\" height=\"21\" name=\"graphics11\" align=\"BOTTOM\" border=\"0\" \/>.\u00a0So\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image019_0000.gif\" width=\"17\" height=\"41\" name=\"graphics12\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0radians =\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image024.gif\" width=\"80\" height=\"44\" name=\"graphics13\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0The unit circle below shows degrees and radians at different\u00a0points around the circle.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20008.JPG\" alt=\"circle with radians labeled\" width=\"373\" height=\"277\" name=\"graphics14\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>Which of the following shows\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image027.gif\" width=\"27\" height=\"41\" name=\"graphics15\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0radians converted into degrees?<\/p>\n<ol>\n<li>240\u00b0<\/li>\n<li>180\u00b0<\/li>\n<li>215\u00b0<\/li>\n<li>90\u00b0<\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The correct choice is A.\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image030.gif\" width=\"145\" height=\"44\" name=\"graphics16\" align=\"ABSMIDDLE\" border=\"0\" \/><\/p>\n<\/div>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p>Convert 115\u00b0 into radians.<\/p>\n<ol>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image033.gif\" width=\"25\" height=\"41\" name=\"graphics17\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image036.gif\" width=\"24\" height=\"19\" name=\"graphics18\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image039.gif\" width=\"33\" height=\"41\" name=\"graphics19\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image019_0001.gif\" width=\"17\" height=\"41\" name=\"graphics20\" align=\"BOTTOM\" border=\"0\" \/><\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<div id=\"Section1\" dir=\"LTR\"><\/div>\n<p>The correct choice is C. Use the formula radians = degrees\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image013_0000.gif\" width=\"47\" height=\"41\" name=\"graphics24\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image045.gif\" width=\"95\" height=\"41\" name=\"graphics25\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p6_clip_image048.gif\" width=\"47\" height=\"41\" name=\"graphics26\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<\/div>\n<\/section>\n<h3>Review of New Vocabulary and Concepts<\/h3>\n<ul>\n<li><strong><em>Trigonometry\u00a0<\/em><\/strong>is a collection of techniques that can be used to calculate the\u00a0length of sides and measure of angles in triangles.<\/li>\n<li>The <em><strong>hypotenuse <\/strong><\/em>is\u00a0the longest side of a right triangle. It is always opposite of\u00a0the right angle.<\/li>\n<li>The two sides of a right\u00a0triangle that intersect to form a right angle are referred to as\u00a0<strong><em>legs<\/em><\/strong>.<\/li>\n<li>There are three basic\u00a0<strong><em>trigonometric ratios <\/em><\/strong>that can help you\u00a0to solve for lengths of sides in right triangles.<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image003.gif\" width=\"127\" height=\"44\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image006.gif\" width=\"129\" height=\"44\" name=\"graphics4\" align=\"ABSMIDDLE\" border=\"0\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image009.gif\" width=\"112\" height=\"44\" name=\"graphics5\" align=\"ABSMIDDLE\" border=\"0\" \/><\/li>\n<li>The <strong><em>isosceles\u00a0<\/em><em>right triangle, <\/em><\/strong>or 45-45-90 triangle, is a special\u00a0right triangle that consists of two 45\u00b0 angles and one 90\u00b0\u00a0angle. The legs of the isosceles right triangle are equal in\u00a0length. This triangle has a length-of-side ratio of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image012.gif\" width=\"56\" height=\"23\" name=\"graphics6\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/li>\n<li>The <strong><em>30-60-90\u00a0triangle<\/em><\/strong> is a special right triangle that consists\u00a0of one 30\u00b0 angle, one 60\u00b0 angle, and a 90\u00b0 angle.\u00a0This triangle has a length-of-side ratio of\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image015.gif\" width=\"57\" height=\"24\" name=\"graphics7\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/li>\n<li>A <em><strong>radian\u00a0<\/strong><\/em>is an alternative to a degree when quantifying the measurement of\u00a0an angle. It is an arc along the circumference of a circle equal\u00a0in length to the radius of the circle. So for every 360\u00b0,\u00a0there are 2\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image018.gif\" width=\"15\" height=\"15\" name=\"graphics8\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0radians, and for every 180\u00b0, there are\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image018_0000.gif\" width=\"15\" height=\"15\" name=\"graphics9\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0radians.<\/li>\n<li><strong><em>Radians<\/em><\/strong> can be converted into <em><strong>degrees<\/strong><\/em> by<\/li>\n<\/ul>\n<p align=\"CENTER\">degrees = radians\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image022.gif\" width=\"47\" height=\"41\" name=\"graphics10\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<ul>\n<li><em><strong>Degrees<\/strong><\/em> can be converted into\u00a0<strong><em>radians<\/em><\/strong> by:<\/li>\n<\/ul>\n<p align=\"CENTER\">radians = degrees\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s1_p7_clip_image025.gif\" width=\"47\" height=\"41\" name=\"graphics11\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<div class=\"advance\"><!--<a href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/math_05_08.html\" class=\"button button-primary\">\u2b05 Previous Lesson<\/a>--><br \/>\n<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/trigonometry\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/law-of-sines-law-of-cosines\">Next Lesson \u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Workshop Index\u00a0Next Lesson \u27a1 Trigonometric Ratios Objective In this lesson, we will discuss how to solve trigonometry problems using angle and side relationships for special\u00a0right triangles and the basic trigonometric ratios. You will also learn two different standard measurements for\u00a0angles, degrees and radians, and how to convert between the two. Previously Covered: The Pythagorean\u00a0Theorem is [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-240","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/240","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/comments?post=240"}],"version-history":[{"count":17,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/240\/revisions"}],"predecessor-version":[{"id":817,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/240\/revisions\/817"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/media?parent=240"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}