{"id":45311,"date":"2025-07-01T05:24:37","date_gmt":"2025-07-01T05:24:37","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=45311"},"modified":"2025-07-01T05:24:39","modified_gmt":"2025-07-01T05:24:39","slug":"kuta-software-infinite-algebra-1-using-trigonometry-to-find-lengths-find-the-missing-side","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/kuta-software-infinite-algebra-1-using-trigonometry-to-find-lengths-find-the-missing-side\/","title":{"rendered":"Kuta Software &#8211; Infinite Algebra 1 Using Trigonometry To Find Lengths Find the missing side."},"content":{"rendered":"\n<p>Kuta Software &#8211; Infinite Algebra 1 Using Trigonometry To Find Lengths Find the missing side. Round to the nearest tenth. 1) 27\u00b0 10 2)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve for the missing side in a right triangle using trigonometry, we typically use one of the following trigonometric functions:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Sine (sin)<\/strong>: sin\u2061(\u03b8)=oppositehypotenuse\\sin(\\theta) = \\frac{\\text{opposite}}{\\text{hypotenuse}}sin(\u03b8)=hypotenuseopposite\u200b<\/li>\n\n\n\n<li><strong>Cosine (cos)<\/strong>: cos\u2061(\u03b8)=adjacenthypotenuse\\cos(\\theta) = \\frac{\\text{adjacent}}{\\text{hypotenuse}}cos(\u03b8)=hypotenuseadjacent\u200b<\/li>\n\n\n\n<li><strong>Tangent (tan)<\/strong>: tan\u2061(\u03b8)=oppositeadjacent\\tan(\\theta) = \\frac{\\text{opposite}}{\\text{adjacent}}tan(\u03b8)=adjacentopposite\u200b<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">For question 1:<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Angle \u03b8=27\u2218\\theta = 27^\\circ\u03b8=27\u2218<\/li>\n\n\n\n<li>One side is given as 10 (you didn&#8217;t specify whether it\u2019s adjacent, opposite, or hypotenuse, so I\u2019ll provide an explanation assuming the missing side is the opposite or adjacent).<\/li>\n<\/ul>\n\n\n\n<p>Assume the given side is the <strong>adjacent<\/strong> side, and we need to find the <strong>opposite<\/strong> side. We can use the <strong>tangent<\/strong> function:tan\u2061(\u03b8)=oppositeadjacent\\tan(\\theta) = \\frac{\\text{opposite}}{\\text{adjacent}}tan(\u03b8)=adjacentopposite\u200b<\/p>\n\n\n\n<p>Substitute the known values:tan\u2061(27\u2218)=opposite10\\tan(27^\\circ) = \\frac{\\text{opposite}}{10}tan(27\u2218)=10opposite\u200b<\/p>\n\n\n\n<p>Now, solve for the opposite side:opposite=10\u22c5tan\u2061(27\u2218)\\text{opposite} = 10 \\cdot \\tan(27^\\circ)opposite=10\u22c5tan(27\u2218)opposite\u224810\u22c50.5095\\text{opposite} \\approx 10 \\cdot 0.5095opposite\u224810\u22c50.5095opposite\u22485.1\\text{opposite} \\approx 5.1opposite\u22485.1<\/p>\n\n\n\n<p>Thus, the missing side (opposite) is approximately <strong>5.1<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">For question 2:<\/h3>\n\n\n\n<p>If you have a specific angle or side for question 2, feel free to provide it, and I can calculate the missing side accordingly!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-24.jpeg\" alt=\"\" class=\"wp-image-45312\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-24.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-24-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-24-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Kuta Software &#8211; Infinite Algebra 1 Using Trigonometry To Find Lengths Find the missing side. Round to the nearest tenth. 1) 27\u00b0 10 2) The Correct Answer and Explanation is: To solve for the missing side in a right triangle using trigonometry, we typically use one of the following trigonometric functions: For question 1: Given: [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-45311","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45311","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=45311"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45311\/revisions"}],"predecessor-version":[{"id":45313,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45311\/revisions\/45313"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=45311"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=45311"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=45311"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}