{"id":126018,"date":"2023-09-05T05:13:00","date_gmt":"2023-09-04T23:43:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126018"},"modified":"2023-12-27T11:55:36","modified_gmt":"2023-12-27T06:25:36","slug":"rd-sharma-class-10-solutions-chapter-5-trigonometric-ratios","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-trigonometric-ratios\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126020\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Trigonometric-Ratios.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 5\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Trigonometric-Ratios.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-5-Trigonometric-Ratios-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios:&nbsp;<\/strong>Students can find RD Sharma Solutions for Class 10 Maths Chapter 5 here for quick access. The RD Sharma Solutions, prepared by our experts at Kopykitab in accordance with the latest CBSE guidelines, is one such effective weapon. Here you will get a pdf of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> Chapter 5 Trigonometric Ratios.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d025939849e\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 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src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-5.pdf\", \"#example1\");<\/script><\/p>\n<p style=\"text-align: center;\"><a style=\"display: inline-block; width: auto; padding: 18px; cursor: pointer; font-weight: bold; border-radius: 40px; text-decoration: none; color: #fff; background: #ff4500; -webkit-box-shadow: 0px 2px 6px 0px rgba(0, 0, 0, 0.25); -moz-box-shadow: 0px 2px 6px 0px rgba(0, 0, 0, 0.25); box-shadow: 0px 2px 6px 0px rgba(0, 0, 0, 0.25);\" href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-5.pdf\" target=\"_blank\" rel=\"noopener noreferrer\">RD Sharma Class 10 Solutions Chapter 5 PDF- Direct Link<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-the-rd-sharma-solutions-for-class-10-maths-chapter-5-%e2%80%93-trigonometric-ratios\"><\/span>Access the RD Sharma Solutions For Class 10 Maths Chapter 5 \u2013 Trigonometric Ratios<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-5.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 5 Trigonometric Ratios ex 5.1 - 1\" width=\"226\" height=\"202\"><\/strong><\/p>\n<p><strong>(i)&nbsp;sin A = 2\/3<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,<\/p>\n<p>sin A = 2\/3 \u2026\u2026..\u2026.. (1)<\/p>\n<p>As we know, by sin definition,<\/p>\n<p>sin A&nbsp;=&nbsp;&nbsp;Perpendicular\/ Hypotenuse&nbsp;= 2\/3&nbsp;\u2026.(2)<\/p>\n<p>By comparing eq. (1) and (2), we&nbsp;have<\/p>\n<p>Opposite side = 2 and Hypotenuse = 3<\/p>\n<p>Now, on using Pythagoras theorem in \u0394&nbsp;ABC<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2 +<\/sup>&nbsp;BC<sup>2<\/sup><\/p>\n<p>Putting the values of perpendicular side (BC) and hypotenuse (AC) and for the base side as (AB), we get<\/p>\n<p>\u21d2 3<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ 2<sup>2<\/sup><\/p>\n<p>AB<sup>2&nbsp;<\/sup>= 3<sup>2<\/sup>&nbsp;\u2013 2<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 9 \u2013 4<\/p>\n<p>AB<sup>2 =&nbsp;<\/sup>5<\/p>\n<p>AB = \u221a5<\/p>\n<p>Hence, Base =&nbsp;\u221a5<\/p>\n<p>By definition,<\/p>\n<p>cos A = Base\/Hypotenuse<\/p>\n<p>\u21d2 cos A = \u221a5\/3<\/p>\n<p>Since, cosec A =&nbsp;1\/sin A = Hypotenuse\/Perpendicular<\/p>\n<p>\u21d2 cosec A = 3\/2<\/p>\n<p>And, sec A =&nbsp;Hypotenuse\/Base<\/p>\n<p>\u21d2 sec A =&nbsp;3\/\u221a5<\/p>\n<p>And, tan A =&nbsp;Perpendicular\/Base<\/p>\n<p>\u21d2 tan A =&nbsp;&nbsp;2\/\u221a5<\/p>\n<p>And, cot A =&nbsp;1\/ tan A = Base\/Perpendicular<\/p>\n<p>\u21d2 cot A =&nbsp;\u221a5\/2<\/p>\n<p><strong>(ii) cos A = 4\/5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,<\/p>\n<p>cos A = 4\/5&nbsp;\u2026\u2026.\u2026. (1)<\/p>\n<p>As we know, by cos definition,<\/p>\n<p>cos A = Base\/Hypotenuse&nbsp;\u2026. (2)<\/p>\n<p>By comparing eq. (1) and (2), we&nbsp;get<\/p>\n<p>Base = 4 and Hypotenuse = 5<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2&nbsp;<\/sup>+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and hypotenuse (AC) and for the perpendicular (BC), we get<\/p>\n<p>5<sup>2<\/sup>&nbsp;= 4<sup>2&nbsp;<\/sup>+ BC<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 5<sup>2<\/sup>&nbsp;\u2013 4<sup>2<\/sup><\/p>\n<p>BC<sup>2&nbsp;<\/sup>= 25 \u2013 16<\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 9<\/p>\n<p>BC= 3<\/p>\n<p>Hence, Perpendicular = 3<\/p>\n<p>By definition,<\/p>\n<p>sin A&nbsp;=&nbsp;Perpendicular\/Hypotenuse<\/p>\n<p>\u21d2 sin A = 3\/5<\/p>\n<p>Then, cosec A =&nbsp;1\/sin A<\/p>\n<p>\u21d2 cosec A= 1\/ (3\/5) = 5\/3 = Hypotenuse\/Perpendicular<\/p>\n<p>And, sec A = 1\/cos A<\/p>\n<p>\u21d2 sec A =Hypotenuse\/Base<\/p>\n<p>sec A =&nbsp;5\/4<\/p>\n<p>And, tan A =&nbsp;Perpendicular\/Base<\/p>\n<p>\u21d2 tan A = 3\/4<\/p>\n<p>Next, cot A =&nbsp;1\/tan A = Base\/Perpendicular<\/p>\n<p>\u2234 cot A =&nbsp;4\/3<\/p>\n<p><strong>(iii) tan \u03b8 = 11\/1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, tan&nbsp;\u03b8 = 11\u2026..\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>tan&nbsp;\u03b8 = Perpendicular\/ Base\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;get;<\/p>\n<p>Base = 1 and&nbsp;Perpendicular = 5<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and perpendicular (BC) to get hypotenuse(AC), we get<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 1<sup>2<\/sup>&nbsp;+ 11<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 1 + 121<\/p>\n<p>AC<sup>2<\/sup>= 122<\/p>\n<p>AC= \u221a122<\/p>\n<p>Hence, hypotenuse = \u221a122<\/p>\n<p>By definition,<\/p>\n<p>sin = Perpendicular\/Hypotenuse<\/p>\n<p>\u21d2 sin&nbsp;\u03b8 = 11\/\u221a122<\/p>\n<p>And, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>\u21d2 cosec&nbsp;\u03b8&nbsp;= \u221a122\/11<\/p>\n<p>Next, cos&nbsp;\u03b8 = Base\/ Hypotenuse<\/p>\n<p>\u21d2 cos&nbsp;\u03b8 = 1\/\u221a122<\/p>\n<p>And, sec&nbsp;\u03b8 = 1\/cos&nbsp;\u03b8<\/p>\n<p>\u21d2 sec&nbsp;\u03b8 = \u221a122\/1 =&nbsp;\u221a122<\/p>\n<p>And, cot&nbsp;\u03b8 &nbsp;= 1\/tan&nbsp;\u03b8<\/p>\n<p>\u2234&nbsp;cot&nbsp;\u03b8 = 1\/11<\/p>\n<p><strong>(iv) sin \u03b8 = 11\/15<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, &nbsp;sin \u03b8 = 11\/15&nbsp;\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>sin \u03b8 = Perpendicular\/ Hypotenuse&nbsp;\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we get,<\/p>\n<p>Perpendicular = 11 and Hypotenuse= 15<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base (AB), we have<\/p>\n<p>15<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+11<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 15<sup>2&nbsp;<\/sup>\u2013 11<sup>2<\/sup><\/p>\n<p>AB<sup>2&nbsp;<\/sup>= 225 \u2013 121<\/p>\n<p>AB<sup>2&nbsp;<\/sup>= 104<\/p>\n<p>AB =&nbsp;\u221a104<\/p>\n<p>AB=&nbsp;\u221a (2\u00d72\u00d72\u00d713)<\/p>\n<p>AB= 2\u221a(2\u00d713)<\/p>\n<p>AB= 2\u221a26<\/p>\n<p>Hence, Base = 2\u221a26<\/p>\n<p>By definition,<\/p>\n<p>cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos\u03b8 = 2\u221a26\/ 15<\/p>\n<p>And, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>\u2234 cosec&nbsp;\u03b8&nbsp;= 15\/11<\/p>\n<p>And, sec\u03b8&nbsp;= Hypotenuse\/Base<\/p>\n<p>\u2234 sec\u03b8 =15\/ 2\u221a26<\/p>\n<p>And,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan\u03b8 =11\/ 2\u221a26<\/p>\n<p>And,&nbsp; cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234&nbsp;cot\u03b8 =2\u221a26\/ 11<\/p>\n<p><strong>&nbsp;(v) tan \u03b1 = 5\/12<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, &nbsp;tan&nbsp;\u03b1 = 5\/12&nbsp;\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>tan \u03b1 = Perpendicular\/Base\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;get<\/p>\n<p>Base = 12 and Perpendicular side = 5<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of the base (AB) and the perpendicular (BC) to get hypotenuse (AC), we have<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 12<sup>2<\/sup>&nbsp;+ 5<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 144 + 25<\/p>\n<p>AC<sup>2<\/sup>= 169<\/p>\n<p>AC = 13 [After taking sq root on both sides]<\/p>\n<p>Hence, Hypotenuse = 13<\/p>\n<p>By definition,<\/p>\n<p>sin&nbsp;\u03b1&nbsp;&nbsp;= Perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin&nbsp;\u03b1 = 5\/13<\/p>\n<p>And, cosec&nbsp;\u03b1&nbsp;&nbsp;= Hypotenuse\/Perpendicular<\/p>\n<p>\u2234 cosec&nbsp;\u03b1&nbsp;= 13\/5<\/p>\n<p>And,&nbsp; cos&nbsp;\u03b1 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos&nbsp;\u03b1 = 12\/13<\/p>\n<p>And,&nbsp; sec&nbsp;\u03b1 =1\/cos&nbsp;\u03b1<\/p>\n<p>\u2234 sec&nbsp;\u03b1 = 13\/12<\/p>\n<p>And, tan&nbsp;\u03b1 = sin&nbsp;\u03b1\/cos&nbsp;\u03b1<\/p>\n<p>\u2234 tan&nbsp;\u03b1=5\/12<\/p>\n<p>Since, cot&nbsp;\u03b1 = 1\/tan&nbsp;\u03b1<\/p>\n<p>\u2234&nbsp;cot&nbsp;\u03b1 =12\/5<\/p>\n<p>&nbsp;<\/p>\n<p><strong>&nbsp;(vi) sin \u03b8 = \u221a3\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,&nbsp;sin&nbsp;\u03b8 =&nbsp;\u221a3\/2&nbsp;\u2026\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>sin&nbsp;\u03b8 = Perpendicular\/ Hypotenuse\u2026.(2)<\/p>\n<p>On Comparing eq. (1) and (2), we get<\/p>\n<p>Perpendicular =&nbsp;\u221a3 and&nbsp;Hypotenuse = 2<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of perpendicular (BC) and hypotenuse (AC) and get the base (AB), we get<\/p>\n<p>2<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ (\u221a3)<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 2<sup>2<\/sup>&nbsp;\u2013 (\u221a3)<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 4 \u2013 3<\/p>\n<p>AB<sup>2<\/sup>&nbsp;= 1<\/p>\n<p>AB = 1<\/p>\n<p>Thus, Base = 1<\/p>\n<p>By definition,<\/p>\n<p>cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos \u03b8 = 1\/2<\/p>\n<p>And, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>Or cosec \u03b8= Hypotenuse\/Perpendicular<\/p>\n<p>\u2234 cosec&nbsp;\u03b8&nbsp;=2\/\u221a3<\/p>\n<p>And,&nbsp; sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234&nbsp;sec \u03b8 = 2\/1<\/p>\n<p>And,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan&nbsp;\u03b8 = \u221a3\/1<\/p>\n<p>And,&nbsp; cot&nbsp;\u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234&nbsp;cot&nbsp;\u03b8 = 1\/\u221a3<\/p>\n<p><strong>(vii) cos \u03b8 = 7\/25<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,&nbsp;cos \u03b8 = 7\/25&nbsp;\u2026\u2026\u2026.. (1)<\/p>\n<p>By definition,<\/p>\n<p>cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;get;<\/p>\n<p>Base = 7 and Hypotenuse = 25<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC),<\/p>\n<p>25<sup>2<\/sup>&nbsp;= 7<sup>2&nbsp;<\/sup>+BC<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 25<sup>2<\/sup>&nbsp;\u2013 7<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 625 \u2013 49<\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 576<\/p>\n<p>BC=&nbsp;\u221a576<\/p>\n<p>BC= 24<\/p>\n<p>Hence, Perpendicular side = 24<\/p>\n<p>By definition,<\/p>\n<p>sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234 sin \u03b8&nbsp;= 24\/25<\/p>\n<p>Since, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>Also, cosec \u03b8= Hypotenuse\/Perpendicular<\/p>\n<p>\u2234 cosec&nbsp;\u03b8&nbsp;= 25\/24<\/p>\n<p>Since,&nbsp; sec&nbsp;\u03b8 = 1\/cosec&nbsp;\u03b8<\/p>\n<p>Also,&nbsp; sec \u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234 sec&nbsp;\u03b8 = 25\/7<\/p>\n<p>Since,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan&nbsp;\u03b8 = 24\/7<\/p>\n<p>Now,&nbsp; cot = 1\/tan&nbsp;\u03b8<\/p>\n<p>So,&nbsp; cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234 cot&nbsp;\u03b8 = 7\/24<\/p>\n<p><strong>(viii) tan \u03b8 = 8\/15<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have,&nbsp;tan&nbsp;\u03b8 = 8\/15&nbsp;\u2026\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>tan \u03b8 = Perpendicular\/Base&nbsp;\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we get<\/p>\n<p>Base = 15 and Perpendicular = 8<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= 15<sup>2<\/sup>&nbsp;+ 8<sup>2<\/sup><\/p>\n<p>AC<sup>2&nbsp;<\/sup>= 225 + 64<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 289<\/p>\n<p>AC =&nbsp;\u221a289<\/p>\n<p>AC = 17<\/p>\n<p>Hence, Hypotenuse = 17<\/p>\n<p>By definition,<\/p>\n<p>Since,&nbsp; sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin \u03b8 = 8\/17<\/p>\n<p>Since, cosec&nbsp;\u03b8&nbsp;= 1\/sin \u03b8<\/p>\n<p>Also, cosec \u03b8 = Hypotenuse\/Perpendicular<\/p>\n<p>\u2234 cosec \u03b8 = 17\/8<\/p>\n<p>Since,&nbsp; cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos \u03b8 = 15\/17<\/p>\n<p>Since,&nbsp; sec \u03b8 = 1\/cos \u03b8<\/p>\n<p>Also,&nbsp; sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234 sec \u03b8 = 17\/15<\/p>\n<p>Since, cot \u03b8 = 1\/tan \u03b8<\/p>\n<p>Also, &nbsp;cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234 cot \u03b8 = 15\/8<\/p>\n<p><strong>(ix) cot \u03b8 = 12\/5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, cot&nbsp;\u03b8 = 12\/5 \u2026\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>cot \u03b8 = 1\/tan \u03b8<\/p>\n<p>cot \u03b8 = Base\/Perpendicular&nbsp;\u2026\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;have<\/p>\n<p>Base = 12 and Perpendicular side = 5<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and perpendicular (BC) to get the hypotenuse (AC),<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 12<sup>2<\/sup>&nbsp;+ 5<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>= 144 + 25<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= 169<\/p>\n<p>AC =&nbsp;\u221a169<\/p>\n<p>AC = 13<\/p>\n<p>Hence, Hypotenuse = 13<\/p>\n<p>By definition,<\/p>\n<p>Since,&nbsp; sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin \u03b8= 5\/13<\/p>\n<p>Since, cosec&nbsp;\u03b8&nbsp;= 1\/sin \u03b8<\/p>\n<p>Also, cosec \u03b8= Hypotenuse\/Perpendicular<\/p>\n<p>\u2234 cosec&nbsp;\u03b8&nbsp;= 13\/5<\/p>\n<p>Since,&nbsp; cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos \u03b8 = 12\/13<\/p>\n<p>Since,&nbsp; sec \u03b8 = 1\/cos\u03b8<\/p>\n<p>Also,&nbsp; sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234 sec \u03b8 = 13\/12<\/p>\n<p>Since,&nbsp; tan\u03b8 = 1\/cot \u03b8<\/p>\n<p>Also,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234 tan \u03b8 = 5\/12<\/p>\n<p><strong>(x) &nbsp;sec \u03b8 = 13\/5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, sec \u03b8 = 13\/5\u2026\u2026.\u2026 (1)<\/p>\n<p>By definition,<\/p>\n<p>sec&nbsp;\u03b8 = Hypotenuse\/Base\u2026\u2026\u2026\u2026. (2)<\/p>\n<p>On Comparing eq. (1) and (2), we&nbsp;get<\/p>\n<p>Base = 5 and&nbsp;Hypotenuse = 13<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>And putting the value of the base side (AB) and hypotenuse (AC) to get the perpendicular side (BC),<\/p>\n<p>13<sup>2<\/sup>&nbsp;= 5<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 13<sup>2&nbsp;<\/sup>\u2013 5<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>=169 \u2013 25<\/p>\n<p>BC<sup>2<\/sup>= 144<\/p>\n<p>BC=&nbsp;\u221a144<\/p>\n<p>BC = 12<\/p>\n<p>Hence, Perpendicular = 12<\/p>\n<p>By definition,<\/p>\n<p>Since, &nbsp;sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234 sin \u03b8= 12\/13<\/p>\n<p>Since, cosec&nbsp;\u03b8= 1\/ sin \u03b8<\/p>\n<p>Also, cosec \u03b8= Hypotenuse\/Perpendicular<\/p>\n<p>\u2234&nbsp;cosec&nbsp;\u03b8&nbsp;= 13\/12<\/p>\n<p>Since cos \u03b8= 1\/sec \u03b8<\/p>\n<p>Also,&nbsp; cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234 cos \u03b8 = 5\/13<\/p>\n<p>Since,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan \u03b8 = 12\/5<\/p>\n<p>Since,&nbsp; cot \u03b8 = 1\/tan \u03b8<\/p>\n<p>Also,&nbsp; cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234 cot \u03b8 = 5\/12<\/p>\n<p><strong>(xi) &nbsp;cosec \u03b8 = \u221a10<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have, cosec&nbsp;\u03b8&nbsp;= \u221a10\/1&nbsp; &nbsp;\u2026\u2026..\u2026 (1)<\/p>\n<p>By definition,<\/p>\n<p>cosec \u03b8 = Hypotenuse\/ Perpendicular \u2026\u2026.\u2026.(2)<\/p>\n<p>And, cosec\u03b8 = 1\/sin \u03b8<\/p>\n<p>On comparing eq.(1) and(2), we get<\/p>\n<p>Perpendicular side = 1 and&nbsp;Hypotenuse =&nbsp;\u221a10<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC,<\/p>\n<p>AC<sup>2&nbsp;<\/sup>= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of perpendicular (BC) and hypotenuse (AC) to get the base side (AB),<\/p>\n<p>(\u221a10)<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ 1<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>= (\u221a10)<sup>2<\/sup>&nbsp;\u2013 1<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>= 10 \u2013 1<\/p>\n<p>AB =&nbsp;\u221a9<\/p>\n<p>AB = 3<\/p>\n<p>So, Base side = 3<\/p>\n<p>By definition,<\/p>\n<p>Since,&nbsp; sin \u03b8 = Perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin \u03b8 = 1\/\u221a10<\/p>\n<p>Since,&nbsp; cos&nbsp;\u03b8 = Base\/Hypotenuse<\/p>\n<p>\u2234&nbsp;cos \u03b8 = 3\/\u221a10<\/p>\n<p>Since,&nbsp; sec \u03b8 = 1\/cos \u03b8<\/p>\n<p>Also, sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234&nbsp;sec \u03b8 = \u221a10\/3<\/p>\n<p>Since,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan \u03b8 = 1\/3<\/p>\n<p>Since,&nbsp; cot \u03b8 = 1\/tan \u03b8<\/p>\n<p>\u2234 cot \u03b8 = 3\/1<\/p>\n<p><strong>(xii)&nbsp; cos \u03b8 =12\/15<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We have;&nbsp;cos&nbsp;\u03b8 = 12\/15&nbsp;\u2026\u2026\u2026. (1)<\/p>\n<p>By definition,<\/p>\n<p>cos&nbsp;\u03b8 = Base\/Hypotenuse\u2026\u2026\u2026 (2)<\/p>\n<p>By comparing eq. (1) and (2), we&nbsp;get;<\/p>\n<p>Base =12 and Hypotenuse = 15<\/p>\n<p>Now, using Pythagoras\u2019 theorem in \u0394 ABC, we get<\/p>\n<p>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>+ BC<sup>2<\/sup><\/p>\n<p>Putting the value of base (AB) and hypotenuse (AC) to get the perpendicular (BC),<\/p>\n<p>15<sup>2<\/sup>&nbsp;= 12<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 15<sup>2<\/sup>&nbsp;\u2013 12<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>&nbsp;= 225 \u2013 144<\/p>\n<p>BC&nbsp;<sup>2<\/sup>= 81<\/p>\n<p>BC =&nbsp;\u221a81<\/p>\n<p>BC = 9<\/p>\n<p>So, Perpendicular = 9<\/p>\n<p>By definition,<\/p>\n<p>Since,&nbsp; sin \u03b8 = perpendicular\/Hypotenuse<\/p>\n<p>\u2234&nbsp;sin&nbsp;\u03b8 = 9\/15 = 3\/5<\/p>\n<p>Since, cosec&nbsp;\u03b8&nbsp;= 1\/sin&nbsp;\u03b8<\/p>\n<p>Also, cosec \u03b8 = Hypotenuse\/Perpendicular<\/p>\n<p>\u2234&nbsp;cosec&nbsp;\u03b8= 15\/9 = 5\/3<\/p>\n<p>Since,&nbsp; sec \u03b8 = 1\/cos \u03b8<\/p>\n<p>Also,&nbsp; sec&nbsp;\u03b8 = Hypotenuse\/Base<\/p>\n<p>\u2234 sec&nbsp;\u03b8 = 15\/12 = 5\/4<\/p>\n<p>Since,&nbsp; tan \u03b8 = Perpendicular\/Base<\/p>\n<p>\u2234&nbsp;tan&nbsp;\u03b8 = 9\/12 = 3\/4<\/p>\n<p>Since,&nbsp; cot&nbsp;\u03b8 = 1\/tan&nbsp;\u03b8<\/p>\n<p>Also,&nbsp; cot \u03b8 = Base\/Perpendicular<\/p>\n<p>\u2234 cot&nbsp;\u03b8 = 12\/9 = 4\/3<\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-class-10-maths-chapter-5-%e2%80%93-trigonometric-ratios\"><\/span>Exercise-Wise RD Sharma Class 10 Maths Chapter 5 \u2013 Trigonometric Ratios<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table>\n<tbody>\n<tr>\n<td style=\"text-align: center;\"><strong>RD Sharma Class 10 Solutions Chapter 5 Exercises<\/strong><\/td>\n<\/tr>\n<tr>\n<td><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-1\/\">RD Sharma Solutions for Class 10 Chapter 5 Exercise 5.1<\/a><\/td>\n<\/tr>\n<tr>\n<td><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-2\/\">RD Sharma Solutions for Class 10 Chapter 5 Exercise 5.2<\/a><\/td>\n<\/tr>\n<tr>\n<td><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-exercise-5-3\/\"><span style=\"font-size: inherit; background-color: initial;\">RD Sharma Solutions for Class 10 Chapter 5 Exercise 5.3<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-vsaqs\/\">RD Sharma Solutions for Class 10 Chapter 5 VSAQs<\/a><span style=\"font-size: inherit; background-color: initial;\"><br><\/span><\/td>\n<\/tr>\n<tr>\n<td><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-mcqs\/\">RD Sharma Solutions for Class 10 Chapter 5 MCQs<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Download RD Sharma&nbsp;<a href=\"http:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 10 Maths Chapter 5 Solutions to boost your preparations for the exam. If you have any queries, feel free to ask us in the comment section.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"frequently-asked-questions-on-rd-sharma-class-10-maths-solutions-chapter-5\"><\/span>Frequently Asked Questions on RD Sharma Class 10 Maths Solutions Chapter 5<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631187476003\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-solutions-class-10-maths-chapter-5-free-pdf\"><\/span>Where can I get RD Sharma Solutions Class 10 Maths Chapter 5 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma for Class 10 Maths Solutions Chapter 5 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631187499531\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-most-important-topics-covered-in-chapter-5-of-rd-sharma-solutions-for-class-10-maths\"><\/span>What are the most important topics covered in Chapter 5 of RD Sharma Solutions for Class 10 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Trigonometry is the science of measuring triangles, and it is covered in the RD Sharma Solutions for Class 10 Maths Chapter 5 books. In addition, the trigonometric ratios and their relationships will be covered in this chapter. Also discussed are the trigonometric ratios of some specific angles.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631187640088\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"in-rd-sharma-solutions-for-class-10-maths-chapter-5-how-many-exercises-are-there\"><\/span>In RD Sharma Solutions for Class 10 Maths Chapter 5, how many exercises are there?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>In RD Sharma Solutions for Class 10 Maths Chapter 5, there are three exercises. Based on the notions of trigonometric ratios, the first exercise comprises 18 questions with sub-questions, the second exercise contains 20 questions with sub-questions, and the third exercise contains 8 questions with sub-questions.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1681973535493\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-rd-sharma-class-10-solutions-chapter-5-trigonometric-ratios-pdf\"><\/span>How much does it cost to download the RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download NCERT Solutions for RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios PDF for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1681973664921\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-class-10-solutions-chapter-5-trigonometric-ratios-pdf-offline\"><\/span>Can I access the RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios PDF online, you can access it offline whenever you want.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios:&nbsp;Students can find RD Sharma Solutions for Class 10 Maths Chapter 5 here for quick access. The RD Sharma Solutions, prepared by our experts at Kopykitab in accordance with the latest CBSE guidelines, is one such effective weapon. Here you will get a pdf of RD Sharma &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-5-trigonometric-ratios\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 5- Trigonometric Ratios (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126020,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126018"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126018"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126018\/revisions"}],"predecessor-version":[{"id":527271,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126018\/revisions\/527271"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126020"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126018"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126018"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126018"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}