{"id":36298,"date":"2025-06-25T06:33:44","date_gmt":"2025-06-25T06:33:44","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36298"},"modified":"2025-06-25T06:33:47","modified_gmt":"2025-06-25T06:33:47","slug":"ft-ladder-leaning-against-a-wall-forms-an-angle-of-%ce%b8-with-the-ground-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/ft-ladder-leaning-against-a-wall-forms-an-angle-of-%ce%b8-with-the-ground-2\/","title":{"rendered":"ft ladder leaning against a wall forms an angle of\u00a0\u03b8\u00a0with the ground"},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">ft ladder leaning against a wall forms an angle of&nbsp;\u03b8&nbsp;with the ground<\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"991\" height=\"325\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-311.png\" alt=\"\" class=\"wp-image-36308\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-311.png 991w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-311-300x98.png 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-311-768x252.png 768w\" sizes=\"auto, (max-width: 991px) 100vw, 991px\" \/><\/figure>\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>Let&#8217;s walk through each part of the trigonometric ladder problem in a structured way:<\/p>\n\n\n\n<p><strong>Part (a):<\/strong> Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A ladder of length 15 ft<\/li>\n\n\n\n<li>sec(\u03b8) = 3\/2<\/li>\n<\/ul>\n\n\n\n<p>We know that sec(\u03b8) is the reciprocal of cos(\u03b8), so: cos(\u03b8) = 2\/3<\/p>\n\n\n\n<p>From a right triangle formed by the wall, ground, and ladder, cos(\u03b8) = adjacent \/ hypotenuse = b \/ 15 So, b = 15 \u00d7 cos(\u03b8) = 15 \u00d7 (2\/3) = 10 ft<\/p>\n\n\n\n<p>Using the Pythagorean theorem: h\u00b2 + b\u00b2 = 15\u00b2 h\u00b2 + 100 = 225 h\u00b2 = 125 h = \u221a125 = 5\u221a5 ft<\/p>\n\n\n\n<p><strong>Part (b):<\/strong> We want to find values of \u03b8 such that h \/ b &lt; \u221a3. From trigonometric relationships: h = 15 sin(\u03b8) b = 15 cos(\u03b8) Then, h \/ b = (15 sin(\u03b8)) \/ (15 cos(\u03b8)) = tan(\u03b8) So, tan(\u03b8) &lt; \u221a3<\/p>\n\n\n\n<p>Taking the arctangent of both sides: \u03b8 &lt; arctan(\u221a3) = \u03c0\/3<\/p>\n\n\n\n<p>Therefore, all values of \u03b8 in the interval (0, \u03c0\/3) will satisfy the condition.<\/p>\n\n\n\n<p><strong>Part (c):<\/strong> We solve: 4 cos(\u03b8) \/ tan(\u03b8) = 3 csc(\u03b8)<\/p>\n\n\n\n<p>Left side becomes: 4 cos(\u03b8) \/ (sin(\u03b8) \/ cos(\u03b8)) = 4 cos\u00b2(\u03b8) \/ sin(\u03b8)<\/p>\n\n\n\n<p>Right side is: 3 \/ sin(\u03b8)<\/p>\n\n\n\n<p>Equating both: 4 cos\u00b2(\u03b8) \/ sin(\u03b8) = 3 \/ sin(\u03b8) =&gt; 4 cos\u00b2(\u03b8) = 3 =&gt; cos\u00b2(\u03b8) = 3\/4 =&gt; cos(\u03b8) = \u00b1\u221a3 \/ 2<\/p>\n\n\n\n<p>This gives: \u03b8 = \u03c0\/6 or \u03b8 = 5\u03c0\/6 (within [0, \u03c0])<\/p>\n\n\n\n<p>Only \u03b8 = \u03c0\/6 is valid since 0 &lt; \u03b8 &lt; \u03c0\/2 for a ladder angle.<\/p>\n\n\n\n<p>So the final answers are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(a) h = 5\u221a5 ft<\/li>\n\n\n\n<li>(b) \u03b8 &lt; \u03c0\/3<\/li>\n\n\n\n<li>(c) \u03b8 = \u03c0\/6<\/li>\n<\/ul>\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\/06\/learnexams-banner8-785.jpeg\" alt=\"\" class=\"wp-image-36326\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-785.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-785-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-785-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>ft ladder leaning against a wall forms an angle of&nbsp;\u03b8&nbsp;with the ground The Correct Answer and Explanation is: Let&#8217;s walk through each part of the trigonometric ladder problem in a structured way: Part (a): Given: We know that sec(\u03b8) is the reciprocal of cos(\u03b8), so: cos(\u03b8) = 2\/3 From a right triangle formed by the [&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-36298","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36298","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=36298"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36298\/revisions"}],"predecessor-version":[{"id":36327,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36298\/revisions\/36327"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36298"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36298"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36298"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}