{"id":6611,"date":"2026-07-14T09:32:51","date_gmt":"2026-07-14T14:32:51","guid":{"rendered":"https:\/\/www.eastcentral.edu\/learning-center\/?page_id=6611"},"modified":"2026-07-14T09:32:58","modified_gmt":"2026-07-14T14:32:58","slug":"solving-for-a-side-using-trig-ratios","status":"publish","type":"page","link":"https:\/\/www.eastcentral.edu\/learning-center\/solving-for-a-side-using-trig-ratios\/","title":{"rendered":"Solving for a Side Using Trig Ratios"},"content":{"rendered":"\n<p>*Given an angle and a given side, one can solve for a missing side using trig ratios.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Example 1: What is the length of the indicated side of the following triangle? The measure of angle A is 39\u00b0, and the length of side b is 65 cm.<\/li>\n<\/ul>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><a href=\"https:\/\/www.eastcentral.edu\/learning-center\/wp-content\/uploads\/sites\/27\/2026\/07\/right-triangle-2.png\"><img loading=\"lazy\" decoding=\"async\" width=\"421\" height=\"360\" src=\"https:\/\/www.eastcentral.edu\/learning-center\/wp-content\/uploads\/sites\/27\/2026\/07\/right-triangle-2.png\" alt=\"The image depicts a right triangle labeled with vertices A, B, and C. The triangle is filled with a blue color and oriented so that side BC is horizontal at the bottom, with point B on the left and point C on the right. The vertical side AB is indicated with the label &quot;c,&quot; and the hypotenuse AC is labeled with the letter &quot;b.&quot; The side BC, which is the base of the triangle, is marked with the letter &quot;a.&quot; An arrow points from the vertex C towards the side labeled \u201ca.\u201d Below the triangle, there is a question asking, \u201cWhat is the length of this side?\u201d in black text.\" class=\"wp-image-6612\" style=\"width:350px\" srcset=\"https:\/\/www.eastcentral.edu\/learning-center\/wp-content\/uploads\/sites\/27\/2026\/07\/right-triangle-2.png 421w, https:\/\/www.eastcentral.edu\/learning-center\/wp-content\/uploads\/sites\/27\/2026\/07\/right-triangle-2-300x257.png 300w\" sizes=\"auto, (max-width: 421px) 100vw, 421px\" \/><\/a><\/figure>\n<\/div>\n\n<p>The sides involved in this calculation are the opposite side (from angle A) and the hypotenuse. The trig ratio that relates the opposite side and hypotenuse is sine (opposite\/hypotenuse).<\/p>\n<p>Here is how the problem is set up: \u00a0\u00a0\u00a0\u00a0 sin 39 = <span style=\"display: inline-block; text-align: center; vertical-align: middle;\"> <span style=\"display: block; border-bottom: 1px solid #000; padding: 0 4px;\"> <i>opposite side length<\/i> <\/span> <span style=\"display: block;\"> <i>hypotenuse length<\/i> <\/span> <\/span> \u00a0=\u00a0 <span style=\"display: inline-block; text-align: center; vertical-align: middle;\"> <span style=\"display: block; border-bottom: 1px solid #000; padding: 0 4px;\"> a <\/span> <span style=\"display: block;\"> 65<\/span><\/span><\/p>\n<p>By consulting a trig ratio table or using a calculator, one can see that sin 39 \u0366 = 0.963795.<\/p>\n<p>Substituting that value in for sin 39 gives the following equation:<\/p>\n<p style=\"text-align: center;\">0.963795 = <span style=\"display: inline-block; text-align: center; vertical-align: middle;\"> <span style=\"display: block; border-bottom: 1px solid #000; padding: 0 4px;\"> a <\/span> <span style=\"display: block;\"> 65<\/span><\/span><\/p>\n<p>Solving for the variable a yields an answer of 63 cm.<\/p>\n<p><em><strong>Note:<\/strong> restrict the trig ratios you are using to sine, cosine, and tangent.<\/em><\/p>\n<ul>\n<li>Example 2: Using the same triangle from the previous problem, find the length of side c given that angle A is 39 \u0366 and the length of side a is 63 cm (the length of the hypotenuse is the same as the previous problem \u2013 65 cm).<\/li>\n<\/ul>\n<p>Side c is the adjacent side to angle A, so the trig ratio used will be cosine. Here is how the problem will be set up:<\/p>\n<p style=\"text-align: center;\">cos 39 = <span style=\"display: inline-block; text-align: center; vertical-align: middle;\"> <span style=\"display: block; border-bottom: 1px solid #000; padding: 0 4px;\"> <i>adjacent side length<\/i> <\/span> <span style=\"display: block;\"> <i>hypotenuse<\/i> <\/span> <\/span> \u00a0=\u00a0 <span style=\"display: inline-block; text-align: center; vertical-align: middle;\"> <span style=\"display: block; border-bottom: 1px solid #000; padding: 0 4px;\"> c <\/span> <span style=\"display: block;\"> 65<\/span><\/span><\/p>\n<p>By consulting a trig ratio table or by using a calculator to determine the cos 39 \u0366 one can see that the cos 39 \u00b0 = 0.266643<\/p>\n<p>Substituting that value in for cos 39 \u0366  gives the following equation:<\/p>\n<p style=\"text-align: center;\">0.266643 = <span style=\"display: inline-block; text-align: center; vertical-align: middle;\"> <span style=\"display: block; border-bottom: 1px solid #000; padding: 0 4px;\"> c <\/span> <span style=\"display: block;\"> 65<\/span><\/span><\/p>\n<p>Solving for the variable c yields an answer of 17 cm.<\/p>","protected":false},"excerpt":{"rendered":"<p>*Given an angle and a given side, one can solve for a missing side using trig ratios. The sides involved in this calculation are the opposite side (from angle A) and the hypotenuse. The trig ratio that relates the opposite side and hypotenuse is sine (opposite\/hypotenuse). Here is how the problem is set up: \u00a0\u00a0\u00a0\u00a0 [&hellip;]<\/p>\n","protected":false},"author":12,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_eb_attr":"","footnotes":""},"yst_prominent_words":[],"class_list":["post-6611","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/pages\/6611","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/users\/12"}],"replies":[{"embeddable":true,"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/comments?post=6611"}],"version-history":[{"count":3,"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/pages\/6611\/revisions"}],"predecessor-version":[{"id":6615,"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/pages\/6611\/revisions\/6615"}],"wp:attachment":[{"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/media?parent=6611"}],"wp:term":[{"taxonomy":"yst_prominent_words","embeddable":true,"href":"https:\/\/www.eastcentral.edu\/learning-center\/wp-json\/wp\/v2\/yst_prominent_words?post=6611"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}