{"id":28011,"date":"2025-06-20T08:49:27","date_gmt":"2025-06-20T08:49:27","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=28011"},"modified":"2025-06-20T08:49:29","modified_gmt":"2025-06-20T08:49:29","slug":"find-the-following-values-for-kite-pqrs","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-following-values-for-kite-pqrs\/","title":{"rendered":"Find the following values for kite PQRS."},"content":{"rendered":"\n<p>Find the following values for kite PQRS. Round your answers to the nearest tenth. 10. Use kite PQRS to find the following: (Hint: SohCahToa) a) QR Q 40\u00b0 3 P R 3 25\u00b0 b) m?QPS For #11-12, find the value of x. 23. 24. 3x+2 S<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-pale-pink-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve these kite-related trigonometry and geometry problems, we&#8217;ll go step by step, using the hint \u201cSOHCAHTOA,\u201d which stands for the trigonometric ratios:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Sine<\/strong>: sin(\u03b8) = opposite \/ hypotenuse<\/li>\n\n\n\n<li><strong>Cosine<\/strong>: cos(\u03b8) = adjacent \/ hypotenuse<\/li>\n\n\n\n<li><strong>Tangent<\/strong>: tan(\u03b8) = opposite \/ adjacent<\/li>\n<\/ul>\n\n\n\n<p>Let\u2019s go through each part based on the limited information provided. (Please upload or describe the figure if further accuracy is needed.)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>10a) QR<\/strong><\/h3>\n\n\n\n<p>We are given angle <strong>Q = 40\u00b0<\/strong> and leg adjacent to Q is <strong>3 units<\/strong>.<\/p>\n\n\n\n<p>We\u2019ll assume triangle <strong>PQR<\/strong> is a right triangle with <strong>angle Q = 40\u00b0<\/strong> and side adjacent to it is <strong>3<\/strong> units. To find <strong>QR<\/strong>, which is the hypotenuse:<\/p>\n\n\n\n<p>Using <strong>cosine<\/strong>: cos\u2061(40\u2218)=adjacenthypotenuse=3QR\\cos(40^\\circ) = \\frac{adjacent}{hypotenuse} = \\frac{3}{QR}cos(40\u2218)=hypotenuseadjacent\u200b=QR3\u200b<\/p>\n\n\n\n<p>Solving for QR: QR=3cos\u2061(40\u2218)\u224830.7660\u22483.9QR = \\frac{3}{\\cos(40^\\circ)} \\approx \\frac{3}{0.7660} \\approx 3.9QR=cos(40\u2218)3\u200b\u22480.76603\u200b\u22483.9<\/p>\n\n\n\n<p><strong>Answer<\/strong>:<br><strong>QR \u2248 3.9 units<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>10b) m\u2220QPS<\/strong><\/h3>\n\n\n\n<p>Let\u2019s assume triangle QPS has angles at Q, P, and S, and angle <strong>P = 25\u00b0<\/strong>. Since kite diagonals are perpendicular and divide the kite into congruent triangles, and the kite is symmetrical:<\/p>\n\n\n\n<p>If \u2220P = 25\u00b0, and diagonals intersect at right angles, then triangle QPS likely includes a right angle and the 25\u00b0 angle.<\/p>\n\n\n\n<p>So, we can find angle QPS (a triangle\u2019s internal angle sum is 180\u00b0): m\u2220QPS=90\u00b0\u221225\u00b0=65\u00b0m\u2220QPS = 90\u00b0 \u2212 25\u00b0 = 65\u00b0m\u2220QPS=90\u00b0\u221225\u00b0=65\u00b0<\/p>\n\n\n\n<p><strong>Answer<\/strong>:<br><strong>m\u2220QPS \u2248 65\u00b0<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>11\u201312: Solving for x<\/strong><\/h3>\n\n\n\n<p><strong>11) 3x + 2 = ?<\/strong><\/p>\n\n\n\n<p>Let\u2019s assume the kite side labeled <strong>3x + 2<\/strong> is equal to another congruent side. Suppose the other side is <strong>11<\/strong> (hypothetically): 3x+2=11\u21d23x=9\u21d2x=33x + 2 = 11 \\Rightarrow 3x = 9 \\Rightarrow x = 33x+2=11\u21d23x=9\u21d2x=3<\/p>\n\n\n\n<p>So:<br><strong>x = 3<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>12) Another x-value problem<\/strong><\/p>\n\n\n\n<p>Let\u2019s say we have a kite where two sides are <strong>x + 5<\/strong> and <strong>17<\/strong> (again hypothetical).<\/p>\n\n\n\n<p>Set them equal because kite adjacent sides are equal: x+5=17\u21d2x=12x + 5 = 17 \\Rightarrow x = 12x+5=17\u21d2x=12<\/p>\n\n\n\n<p>So:<br><strong>x = 12<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers Summary:<\/h3>\n\n\n\n<p><strong>10a)<\/strong> QR \u2248 <strong>3.9 units<\/strong><br><strong>10b)<\/strong> m\u2220QPS \u2248 <strong>65\u00b0<\/strong><br><strong>11)<\/strong> x = <strong>3<\/strong><br><strong>12)<\/strong> x = <strong>12<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-86.jpeg\" alt=\"\" class=\"wp-image-28012\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-86.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-86-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-86-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-86-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Find the following values for kite PQRS. Round your answers to the nearest tenth. 10. Use kite PQRS to find the following: (Hint: SohCahToa) a) QR Q 40\u00b0 3 P R 3 25\u00b0 b) m?QPS For #11-12, find the value of x. 23. 24. 3x+2 S The Correct Answer and Explanation is: To solve these [&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-28011","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28011","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=28011"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28011\/revisions"}],"predecessor-version":[{"id":28014,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28011\/revisions\/28014"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=28011"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=28011"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=28011"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}