{"id":4391,"date":"2013-06-29T05:52:27","date_gmt":"2013-06-29T05:52:27","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=4391"},"modified":"2026-03-26T11:53:18","modified_gmt":"2026-03-26T16:53:18","slug":"finding-tangent","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/finding-tangent\/","title":{"rendered":"Finding Tangent"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass__u5EXtPwxj8\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage__u5EXtPwxj8\" data-source-videoID=\"_u5EXtPwxj8\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Finding Tangent Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Finding Tangent\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage__u5EXtPwxj8:hover {cursor:pointer;} img#videoThumbnailImage__u5EXtPwxj8 {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/773-tangent-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage__u5EXtPwxj8\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage__u5EXtPwxj8\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Finding Tangent\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass__u5EXtPwxj8\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass__u5EXtPwxj8\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage__u5EXtPwxj8\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction X23_Function() {\n  var x = document.getElementById(\"X23\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"X23_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"X23\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Using_Tangent\" class=\"smooth-scroll\">Using Tangent<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#SOHCAHTOA\" class=\"smooth-scroll\">SOHCAHTOA<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Target_Identity\" class=\"smooth-scroll\">Target Identity<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Review\" class=\"smooth-scroll\">Review<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Tangent_Practice_Questions\" class=\"smooth-scroll\">Tangent Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video on tangent!<\/p>\n<p>The word <em>tangent<\/em> has two different meanings in math. In geometry, it is used to denote when an object is touching another object at only one point, like when a line touches a circle at only one point.<\/p>\n<p>We\u2019re not concerned with that one today. Today we\u2019re exploring tangent in trigonometry, where it is one of the \u201cbig three\u201d trigonometric functions, along with <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/sine\/\">sine<\/a> and <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/cosine\/\">cosine<\/a>. <\/p>\n<p>Like those two functions, it can be used to find the measure of a side or the measure of an angle in a right triangle. But it\u2019s also a bit different than those two. It has a very different looking graph and a trig \u201cidentity\u201d that makes it especially useful. <\/p>\n<p>There\u2019s a lot to look at, so let\u2019s get started!<\/p>\n<h2><span id=\"Using_Tangent\" class=\"m-toc-anchor\"><\/span>Using Tangent<\/h2>\n<p>\nWe can use tangent to find the length of the side of a right triangle that is adjacent to an acute angle with a known measure as long as we know the measure of the side opposite that angle. We can also use it to find the opposite side if we know the adjacent side and the angle in question. The thing we em>don\u2019t<\/em> need to know to use tangent is the length of the hypotenuse, which <em>is<\/em> needed when using sine and cosine. <\/p>\n<p>Tangent, which is commonly abbreviated to three letters as T-A-N, is the ratio of the side opposite the angle we know, or want to know, over the side adjacent to that angle. The adjacent side is the one touching the angle that is NOT the hypotenuse, which is the side opposite the right angle. We can write a simple equation for tangent like this: <\/p>\n<h2><span id=\"SOHCAHTOA\" class=\"m-toc-anchor\"><\/span>SOHCAHTOA<\/h2>\n<p>\nIf you\u2019ve studied trigonometry before you\u2019re probably familiar with SOHCAHTOA, which is a way to remember how each of the \u201cbig three\u201d trig functions are formed. The third part of it, TOA, helps us remember that <strong>T<\/strong>angent is equal to the length of the <strong>O<\/strong>pposite side divided by the length of the <strong>A<\/strong>djacent side.<\/p>\n<div class=\"examplesentence\">\\(\\text{tan }\u03b8=\\frac{\\text{opposite}}{\\text{adjacent}}=\\frac{\\text{o}}{\\text{a}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nLet\u2019s look at an example problem so we can put this all together.<\/p>\n<h3><span id=\"Example_Problem\" class=\"m-toc-anchor\"><\/span>Example Problem<\/h3>\n<p>\nHere, we know an angle and the side adjacent to that angle. We don\u2019t know the length of the hypotenuse or the length of the opposite side, so it\u2019s labeled with an \\(x\\). This \\(x\\) is what we want to figure out.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-115337 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.07.41-PM.png\" alt=\"right triangle with a side length 8 and an angle measure of 36.87 degrees\" width=\"378\" height=\"296\" style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.07.41-PM.png 378w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.07.41-PM-300x235.png 300w\" sizes=\"auto, (max-width: 378px) 100vw, 378px\" \/><\/p>\n<p>First, let\u2019s take our equation and input our three values:<\/p>\n<p>The tangent of the angle we know, 36.87 degrees, is equal to the length of the opposite side (which we\u2019re trying to find) over the length of the adjacent side, which is 8. From here we can find the tangent of 36.87 degrees on a calculator. We type in 36.87 and hit the TAN key to find that it is equal to 0.75000279, which we can round to 0.750. <\/p>\n<p>So now our equation looks like this:<\/p>\n<div class=\"examplesentence\">\\(0.750=\\frac{x}{8}\\)<\/div>\n<p>\n&nbsp;<br \/>\nMultiplying both sides by 8 results in 6.000 equals x. So the measure of the opposite side is six! <\/p>\n<p>Now that we know this, we could find the rest of the measures of this triangle. We could find the hypotenuse by using the Pythagorean theorem or by using sine or cosine, and we could find the third angle by adding the two we know, 36.87 degrees and 90 degrees and subtracting from 180 degrees.<\/p>\n<p>Now, tangent functions, just like sine and cosine functions, are something that we can graph. But you\u2019ll notice that the tangent function looks different from the cosine and sine functions. Those two look like horizontal waves, like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-115334 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.08.52-PM.png\" alt=\"sine (green) and cosine (red) functions graphed\" width=\"653\" height=\"225\" style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.08.52-PM.png 653w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.08.52-PM-300x103.png 300w\" sizes=\"auto, (max-width: 653px) 100vw, 653px\" \/><\/p>\n<p>The sine wave is shown in green and the cosine wave is in red. They look the same but are out of phase with each other. The tangent wave, however, does not look like these at all. <\/p>\n<p>Here\u2019s what the tangent function looks like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-115331 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.09.13-PM.png\" alt=\"tangent function graphed\" width=\"374\" height=\"390\" style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.09.13-PM.png 374w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.09.13-PM-288x300.png 288w\" sizes=\"auto, (max-width: 374px) 100vw, 374px\" \/><\/p>\n<p>Unlike the basic sine and cosine functions, which have a range of negative one to positive one, the \\(y\\)-values of the tangent function have a range of negative infinity to positive infinity! <\/p>\n<p>There\u2019s something else here that is interesting. Let\u2019s look at where the graph touches the \\(x\\)-axis. It crosses at \\(x = 0\\) and to the right it crosses again a bit after 3 and a bit more after 6. If we were to zoom in we would see that it crosses at \\(\\pi\\) and \\(2\\pi\\)! To the left of the \\(y\\)-axis it crosses at \\(x=-\\pi\\) and \\(x=-2\\pi\\). <\/p>\n<p>So what\u2019s happening in between the vertical waves? It turns out there are these invisible vertical lines called asymptotes that indicate where the graph has no points. Here\u2019s the same graph with the asymptotes unveiled as dashed blue lines:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-115328 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.10.24-PM.png\" alt=\"tangent function graphed with asymptotes\" width=\"370\" height=\"389\" style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.10.24-PM.png 370w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Screen-Shot-2022-02-25-at-1.10.24-PM-285x300.png 285w\" sizes=\"auto, (max-width: 370px) 100vw, 370px\" \/><\/p>\n<p>The asymptotes appear when our function y equals tan x is undefined. These asymptotes appear at \\(x=\\frac{1}{2}\\pi\\), at \\(x=1\\frac{1}{2}\\pi\\), and so on. This happens because on one side of the asymptote the graph is approaching positive infinity and on the other side it\u2019s approaching negative infinity, and it can\u2019t be both! <\/p>\n<h2><span id=\"Target_Identity\" class=\"m-toc-anchor\"><\/span>Target Identity<\/h2>\n<p>\nAnother way to illustrate this is to look at this tangent identity: <\/p>\n<p>So, not only is tangent equal to opposite over adjacent like we talked about before, it\u2019s also equal to sine over cosine! If we remember the graph of cosine, it has points where it has a <i>y<\/i> value of zero. <\/p>\n<p>At those points this tangent identity would be trying to divide by zero, which is undefined. Your calculator will probably display \u201cerror\u201d if you try it!<\/p>\n<p>This is how you know you\u2019ve found an asymptote. We can also use this tangent identity to solve trigonometric identity puzzles, but we\u2019ll get into those another time.<\/p>\n<hr>\n<h2><span id=\"Review\" class=\"m-toc-anchor\"><\/span>Review<\/h2>\n<p>\nOkay, before we go, let\u2019s review some of the details:<\/p>\n<p>Tangent is the ratio of the side opposite the angle we know, or want to know, over the side adjacent to that angle, which gives us the equation:<\/p>\n<p>We use tangent to find the length of the side of a right triangle that is adjacent to an acute angle with a known measure as long as we know the measure of the side opposite that angle. When we graph tangent functions, we see that the \\(y\\)-values have a range of negative infinity to positive infinity, with vertical asymptotes indicating where the graph has no points.<\/p>\n<p>I hope this review was helpful! Thanks for watching, and happy studying!<\/p>\n<ul class=\"citelist\">\n<li><a href=\"https:\/\/www.mathopenref.com\/trigfunctions.html\"target=\"_blank\">\u201cIntroduction to the 6 Trigonometry Functions &#8211; Math Open Reference.\u201d 2011. Mathopenref.com<\/a><\/li>\n<li><a href=\"https:\/\/www.mathopenref.com\/tangent.html\"target=\"_blank\">\u201cTangent &#8211; Math Open Reference.\u201d n.d.<\/a><\/li>\n<\/ul>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Tangent_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Tangent Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSay that \\(\\triangle ABC\\) is a right triangle, where \\(\\angle B\\) is 90\u00b0 and \\(\\angle C\\) is 40\u00b0. The length of side \\(AB\\) is 10 inches. What is the length of side \\(BC\\)?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">8.4 inches<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-2\">11.9 inches<\/div><div class=\"PQ\"  id=\"PQ-1-3\">13.1 inches<\/div><div class=\"PQ\"  id=\"PQ-1-4\">15.6 inches<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Since the tangent of an angle is the ratio of the length of the opposite side of the angle over the adjacent side, we can use the equation below to find the length of Side <em>BC<\/em>.<\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(\\tan(40)=\\large{\\frac{10}{BC}}\\)<br \/>\n\\(BC=\\large{\\frac{10}{\\tan(40)}}\\)<\/p>\n<p>This simplifies to \\(BC=11.9\\text{ in}\\).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Tangent-Triangle-Example-1.svg\" alt=\"A right triangle ABC with AB = 10 in, BC = 11.9 in, angle B = 90\u00b0, and angle C = 40\u00b0.\" width=\"293.04\" height=\"234.08\" class=\"aligncenter size-full wp-image-287579\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSay that \\(\\triangle PQR\\) is a right triangle where \\(\\angle R\\) is 90\u00b0 and \\(\\angle P\\) is 28\u00b0. The length of side \\(PR\\) is 5 cm. What is the length of side \\(RQ\\)?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">2.3 cm<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-2\">2.7 cm<\/div><div class=\"PQ\"  id=\"PQ-2-3\">4.4 cm<\/div><div class=\"PQ\"  id=\"PQ-2-4\">9.4 cm<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The tangent of an angle is equivalent to the ratio of the opposite side over the adjacent side of an angle. Since we have the measure of \\(\\angle P\\) and the length of side \\(PR\\), we can use the following equation to solve for the length of \\(RQ\\):<\/p>\n<p style=\"text-align: center; line-height: 45px\">\\(\\tan(28)=\\large{\\frac{RQ}{5}}\\)<br \/>\n\\(RQ=\\text{tan}(28)\\times5\\)<\/p>\n<p>Therefore, \\(RQ=2.7\\text{ cm}\\).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Tangent-Triangle-Example-2.svg\" alt=\"A right triangle PQR with PR = 5 cm, QR = 2.7 cm, angle P = 28\u00b0, and a right angle at R.\" width=\"367.84\" height=\"214.72\" class=\"aligncenter size-full wp-image-287582\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSay that \\(\\triangle LMN\\) is a right triangle, where \\(\\angle M\\) is 90\u00b0. The length of side \\(LM\\) is 7 meters, and the length of side \\(MN\\) is 10 meters. What is the measure of \\(\\angle N\\)? Round your answer to the nearest whole number.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-3-1\">35\u00b0<\/div><div class=\"PQ\"  id=\"PQ-3-2\">44\u00b0<\/div><div class=\"PQ\"  id=\"PQ-3-3\">55\u00b0<\/div><div class=\"PQ\"  id=\"PQ-3-4\">84\u00b0<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>To find the measure of an angle we use arctangent, or the inverse of tangent.<\/p>\n<p>In this case, \\(\\tan(N)=\\frac{7}{10}\\) becomes \\(N=\\arctan(\\frac{7}{10})\\), where \\(N\\) represents the measure of \\(\\angle N\\), which simplifies to 35\u00b0.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Tangent-Triangle-Example-3.svg\" alt=\"Right triangle LMN with LM = 7 m, MN = 10 m, and angle N = 35 degrees. Right angle is at M.\" width=\"280.72\" height=\"205.04\" class=\"aligncenter size-full wp-image-287585\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSay that \\(\\triangle EFG\\) is a right triangle, where \\(\\angle F\\) is 90\u00b0. Given the equations \\(\\cos(x)=\\frac{24}{25}\\) and \\(\\sin(x)=\\frac{7}{25}\\), where \\(x\\) is the measure of \\(\\angle G\\), use tangent to find the measure of \\(\\angle G\\). Round your answer to the nearest whole number.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-4-1\">16\u00b0<\/div><div class=\"PQ\"  id=\"PQ-4-2\">17\u00b0<\/div><div class=\"PQ\"  id=\"PQ-4-3\">56\u00b0<\/div><div class=\"PQ\"  id=\"PQ-4-4\">73\u00b0<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Sine we are given the cosine and sine of \\(x\\) and we know that \\(\\tan=\\frac{\\sin}{\\cos}\\), we can find the tangent of \\(\\angle G\\), which is \\(\\tan(x)=\\frac{7}{24}\\).<\/p>\n<p>Now that we have tangent, we can use arctan, or the inverse of tangent, to find the measure of \\(\\angle G\\), which is 16\u00b0:<\/p>\n<p style=\"text-align: center\">\\(x=\\arctan(\\large{\\frac{7}{24}})\\)<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Tangent-Triangle-Example-4.svg\" alt=\"A right triangle EFG with sides EF = 7 m, FG = 24 m, right angle at F, and angle EGF labeled as 16 degrees.\" width=\"161.04\" height=\"344.08\" class=\"aligncenter size-full wp-image-287588\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nIn \\(\\triangle STU\\), which is a right triangle, \\(\\angle U\\) is 90\u00b0 and \\(\\tan(x)=\\frac{15}{8}\\), where \\(x\\) is the measure of \\(\\angle T\\). What is the length of side \\(ST\\)?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">32.2 units<\/div><div class=\"PQ\"  id=\"PQ-5-2\">28.1 units<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-3\">17.0 units<\/div><div class=\"PQ\"  id=\"PQ-5-4\">12.7 units<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The tangent of an angle is the ratio of the side opposite the angle over the side adjacent to the angle. Using this information, we know the length of side \\(SU\\) is 15 units and the length of side \\(TU\\) is 8 units.<\/p>\n<p>Since we have the legs of the right triangle, we can use the Pythagorean theorem to solve for the hypotenuse, side \\(ST\\), which is 17 units.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Tangent-Triangle-Example-5.svg\" alt=\"A right triangle STU with sides ST = 17 in, SU = 15 in, and TU = 8 in. Angle x\u00b0 is at T and the right angle is at U.\" width=\"333.52\" height=\"212.08\" class=\"aligncenter size-full wp-image-287576\"  role=\"img\" \/><\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/calculus\/\">Return to Calculus Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Calculus Videos<\/p>\n","protected":false},"author":1,"featured_media":99901,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-4391","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-calculus-videos","7":"page_category-circle-video","8":"page_category-math-advertising-group","9":"page_category-video-pages-for-study-course-sidebar-ad","10":"page_type-video","11":"content_type-practice-questions","12":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4391","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=4391"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4391\/revisions"}],"predecessor-version":[{"id":261016,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4391\/revisions\/261016"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/99901"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=4391"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}