{"id":63549,"date":"2021-08-24T10:22:00","date_gmt":"2021-08-24T04:52:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63549"},"modified":"2021-08-26T10:27:45","modified_gmt":"2021-08-26T04:57:45","slug":"rd-sharma-solutions-class-11-maths-chapter-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 4 &#8211; Measurement of Angles (Updated For 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-119069\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-4.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 4 \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span style=\"font-weight: 400;\"><strong>RD Sharma Solutions Class 11 Maths Chapter 4:<\/strong>\u00a0<\/span><span style=\"font-weight: 400;\">If you are new to this chapter and found this topic tricky to understand such\u00a0<\/span><span style=\"font-weight: 400;\">as angles and radians measure, their conversions, and inter-conversions then this article belongs to you. In this article, we are going to talk about the <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a> Chapter 4 Measurement of Angles.<\/span><\/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-69e7533f9574a\" 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 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69e7533f9574a\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#download-rd-sharma-solutions-class-11-maths-chapter-4-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 4 PDF\">Download RD Sharma Solutions Class 11 Maths Chapter 4 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#exercise-wise-rd-sharma-solutions-class-11-maths-chapter-4\" title=\"Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 4\">Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 4<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#rd-sharma-class-11-maths-chapter-4-exercise-41-solutions\" title=\"RD Sharma Class 11 Maths Chapter 4 Exercise 4.1 Solutions\">RD Sharma Class 11 Maths Chapter 4 Exercise 4.1 Solutions<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#important-topics-from-rd-sharma-solutions-class-11-maths-chapter-4\" title=\"Important Topics from RD Sharma Solutions Class 11 Maths Chapter 4\">Important Topics from RD Sharma Solutions Class 11 Maths Chapter 4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#rd-sharma-solutions-class-11-maths-chapter-4-%e2%80%93-measurement-of-angles\" title=\"RD Sharma Solutions Class 11 Maths Chapter 4 &#8211; Measurement of Angles\">RD Sharma Solutions Class 11 Maths Chapter 4 &#8211; Measurement of Angles<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#access-rd-sharma-solutions-class-11-maths-chapter-4\" title=\"Access RD Sharma Solutions Class 11 Maths Chapter 4\">Access RD Sharma Solutions Class 11 Maths Chapter 4<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#1-find-the-degree-measure-corresponding-to-the-following-radian-measures-use-%cf%80-227-i-9%cf%805-ii-5%cf%806-iii-18%cf%805-c-iv-3-c-v-11c-vi-1c\" title=\"1. Find the degree measure corresponding to the following radian measures (Use \u03c0 = 22\/7) (i) 9\u03c0\/5\u00a0(ii) -5\u03c0\/6\u00a0(iii) (18\u03c0\/5)\u00a0c\u00a0(iv) (-3)\u00a0c\u00a0(v) 11c\u00a0(vi) 1c\">1. Find the degree measure corresponding to the following radian measures (Use \u03c0 = 22\/7) (i) 9\u03c0\/5\u00a0(ii) -5\u03c0\/6\u00a0(iii) (18\u03c0\/5)\u00a0c\u00a0(iv) (-3)\u00a0c\u00a0(v) 11c\u00a0(vi) 1c<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#2-find-the-radian-measure-corresponding-to-the-following-degree-measures-i-300o-ii-35o-iii-56o-iv135o-v-300o-vi-7o-30%e2%80%b2-vii-125o-30%e2%80%99-viii-47o-30%e2%80%b2\" title=\"2. Find the radian measure corresponding to the following degree measures: (i) 300o\u00a0(ii) 35o\u00a0(iii) -56o\u00a0(iv)135o\u00a0(v) -300o (vi) 7o\u00a030\u2032 (vii) 125o\u00a030\u2019 (viii) -47o\u00a030\u2032\">2. Find the radian measure corresponding to the following degree measures: (i) 300o\u00a0(ii) 35o\u00a0(iii) -56o\u00a0(iv)135o\u00a0(v) -300o (vi) 7o\u00a030\u2032 (vii) 125o\u00a030\u2019 (viii) -47o\u00a030\u2032<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#3-the-difference-between-the-two-acute-angles-of-a-right-angled-triangle-is-2%cf%805-radians-express-the-angles-in-degrees\" title=\"3. The difference between the two acute angles of a right-angled triangle is 2\u03c0\/5 radians. Express the angles in degrees.\">3. The difference between the two acute angles of a right-angled triangle is 2\u03c0\/5 radians. Express the angles in degrees.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#4-one-angle-of-a-triangle-is-23x-grades-and-another-is-32x-degrees-while-the-third-is-%cf%80x75-radians-express-all-the-angles-in-degrees\" title=\"4. One angle of a triangle is 2\/3x\u00a0grades,\u00a0and another is 3\/2x\u00a0degrees while the third is \u03c0x\/75\u00a0radians. Express all the angles in degrees.\">4. One angle of a triangle is 2\/3x\u00a0grades,\u00a0and another is 3\/2x\u00a0degrees while the third is \u03c0x\/75\u00a0radians. Express all the angles in degrees.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#5-find-the-magnitude-in-radians-and-degrees-of-the-interior-angle-of-a-regular-i-pentagon-ii-octagon-iii-heptagon-iv-duodecagon\" title=\"5. Find the magnitude, in radians and degrees, of the interior angle of a regular: (i) Pentagon (ii) Octagon (iii) Heptagon (iv) Duodecagon.\">5. Find the magnitude, in radians and degrees, of the interior angle of a regular: (i) Pentagon (ii) Octagon (iii) Heptagon (iv) Duodecagon.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#6-the-angles-of-a-quadrilateral-are-in-ap-and-the-greatest-angle-is-120o-express-the-angles-in-radians\" title=\"6. The angles of a quadrilateral are in A.P., and the greatest angle is 120o. Express the angles in radians.\">6. The angles of a quadrilateral are in A.P., and the greatest angle is 120o. Express the angles in radians.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#7-the-angles-of-a-triangle-are-in-ap-and-the-number-of-degrees-in-the-least-angle-is-to-the-number-of-degrees-in-the-mean-angle-as-1-120-find-the-angle-in-radians\" title=\"7. The angles of a triangle are in A.P., and the number of degrees in the least angle is to the number of degrees in the mean angle as 1:120. Find the angle in radians.\">7. The angles of a triangle are in A.P., and the number of degrees in the least angle is to the number of degrees in the mean angle as 1:120. Find the angle in radians.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#8-the-angle-in-one-regular-polygon-is-to-that-in-another-as-3-2-and-the-number-of-sides-in-first-is-twice-that-in-the-second-determine-the-number-of-sides-of-two-polygons\" title=\"8. The angle in one regular polygon is to that in another as 3:2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.\">8. The angle in one regular polygon is to that in another as 3:2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#9-the-angles-of-a-triangle-are-in-ap-such-that-the-greatest-is-5-times-the-least-find-the-angles-in-radians\" title=\"9. The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.\">9. The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#10-the-number-of-sides-of-two-regular-polygons-is-5-4-and-the-difference-between-their-angles-is-9o-find-the-number-of-sides-of-the-polygons\" title=\"10. The number of sides of two regular polygons is 5:4 and the difference between their angles is 9o. Find the number of sides of the polygons.\">10. The number of sides of two regular polygons is 5:4 and the difference between their angles is 9o. Find the number of sides of the polygons.<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#key-take-away-from-class-11-maths-rd-sharma-solutions\" title=\"Key Take Away from Class 11 Maths RD Sharma Solutions\">Key Take Away from Class 11 Maths RD Sharma Solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#access-other-important-chapters-of-rd-sharma-solutions-class-11-maths-chapter-4\" title=\"Access Other Important Chapters of RD Sharma Solutions Class 11 Maths Chapter 4\">Access Other Important Chapters of RD Sharma Solutions Class 11 Maths Chapter 4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#faqs-on-rd-sharma-solutions-class-11-maths-chapter-4\" title=\"FAQs on RD Sharma Solutions Class 11 Maths Chapter 4.\u00a0\">FAQs on RD Sharma Solutions Class 11 Maths Chapter 4.\u00a0<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-20\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#how-to-solve-questions-from-rd-sharma-solutions-class-11-maths-chapter-4\" title=\"How to solve questions from RD Sharma Solutions Class 11 Maths Chapter 4?\">How to solve questions from RD Sharma Solutions Class 11 Maths Chapter 4?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-21\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-4\" title=\"From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 4?\">From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 4?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-22\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/#can-i-access-the-rd-sharma-solutions-maths-without-the-internet\" title=\"Can I access the RD Sharma Solutions Maths without the internet? \">Can I access the RD Sharma Solutions Maths without the internet? <\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-4-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 4 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">The RD Sharma solutions are available in hard copies in the market. But in case you don\u2019t want to buy the harder version you can download the pdf of the RD Sharma Solutions booklet from the link given below.<\/span><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-4-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 4<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-4-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-11-maths-chapter-4\"><\/span>Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4-exercise-4-1\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma Solutions Class11 Exercise-4.1<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-11-maths-chapter-4-exercise-41-solutions\"><\/span><span style=\"background-color: initial;\">RD Sharma Class 11 Maths Chapter 4 Exercise 4.1 Solutions<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">\u00a0The exercise-4.1 starts with a basic introduction of angles, radians, and the way to convert angles into radians. <\/span><span style=\"font-weight: 400;\">In this exercise, you will encounter questions like these; <\/span><span style=\"font-weight: 400;\">Convert 40 degrees 60 minutes into radians.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">RD Sharma is the one-stop <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\">RD Sharma Solution<\/a> to every problem we encounter in this chapter. Please go through the entire concept carefully as the concepts are going to help you in the chapters related to trigonometry.<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-topics-from-rd-sharma-solutions-class-11-maths-chapter-4\"><\/span>Important Topics from RD Sharma Solutions Class 11 Maths Chapter 4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">The chapter has only a few topics important being listed below:<\/span><\/p>\n<ul>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Radians<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Conversion of degree into radians<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Solving mixed questions based on degree minutes and seconds into radians.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-11-maths-chapter-4-%e2%80%93-measurement-of-angles\"><\/span>RD Sharma Solutions Class 11 Maths Chapter 4 &#8211; Measurement of Angles<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">RD Sharma is the ultimate solution to every problem you encounter in class 11 mathematics. These are the many advantages of using RD Sharma Solutions:<\/span><\/p>\n<ul>\n<li><span style=\"font-weight: 400;\">Once you are done with the chapter, you can solve any questions that pop up in the examinations<\/span><\/li>\n<li><span style=\"font-weight: 400;\">Consider this as a key to unlock your concepts when feeling stuck.<\/span><\/li>\n<li><span style=\"font-weight: 400;\">You will get a deeper understanding of concepts by the step-by-step solutions given in the book.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-4\"><\/span>Access <strong>RD Sharma Solutions Class 11 Maths Chapter 4<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"1-find-the-degree-measure-corresponding-to-the-following-radian-measures-use-%cf%80-227-i-9%cf%805-ii-5%cf%806-iii-18%cf%805-c-iv-3-c-v-11c-vi-1c\"><\/span>1. Find the degree measure corresponding to the following radian measures (Use \u03c0 = 22\/7)<br \/>(i) 9\u03c0\/5\u00a0(ii) -5\u03c0\/6\u00a0(iii) (18\u03c0\/5)<sup>\u00a0c<\/sup>\u00a0(iv) (-3)<sup>\u00a0c<\/sup>\u00a0(v) 11<sup>c<\/sup>\u00a0(vi) 1<sup>c<\/sup><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that \u03c0 rad = 180\u00b0\u00a0\u21d2\u00a01 rad = 180\u00b0\/ \u03c0<\/p>\n<p><strong>(i)<\/strong>\u00a0\u00a09\u03c0\/5<\/p>\n<p>[(180\/\u03c0) \u00d7 (9\u03c0\/5)]<sup>\u00a0o<\/sup><\/p>\n<p>\u00a0<\/p>\n<p>Substituting the value of \u03c0 = 22\/7<\/p>\n<p>[180\/22 \u00d7 7 \u00d7 9 \u00d7 22\/(7\u00d75)]<\/p>\n<p>(36 \u00d7 9) \u00b0<\/p>\n<p>324\u00b0<\/p>\n<p>\u2234 Degree measure of 9\u03c0\/5\u00a0is 324\u00b0<\/p>\n<p><strong>(ii)<\/strong>\u00a0-5\u03c0\/6<strong>\u00a0<\/strong><\/p>\n<p>[(180\/\u03c0) \u00d7 (-5\u03c0\/6)]<sup>\u00a0o<\/sup><\/p>\n<p>\u00a0<\/p>\n<p>Substituting the value of \u03c0 = 22\/7<\/p>\n<p>[180\/22 \u00d7 7 \u00d7 -5 \u00d7 22\/(7\u00d76) ]<\/p>\n<p>(30 \u00d7 -5) \u00b0<\/p>\n<p>\u2013 (150) \u00b0<\/p>\n<p>\u2234 Degree measure of -5\u03c0\/6\u00a0is -150\u00b0<\/p>\n<p><strong>(iii)<\/strong>\u00a0(18\u03c0\/5)<\/p>\n<p>[(180\/\u03c0) \u00d7 (18\u03c0\/5)]<sup>\u00a0o<\/sup><\/p>\n<p>\u00a0<\/p>\n<p>Substituting the value of \u03c0 = 22\/7<\/p>\n<p>[180\/22 \u00d7 7 \u00d7 18 \u00d7 22\/(7\u00d75)]<\/p>\n<p>(36 \u00d7 18) \u00b0<\/p>\n<p>648\u00b0<\/p>\n<p>\u2234 Degree measure of 18\u03c0\/5\u00a0is 648\u00b0<\/p>\n<p><strong>(iv)\u00a0<\/strong>(-3)<sup>\u00a0c<\/sup><strong>\u00a0<\/strong><\/p>\n<p>[(180\/\u03c0) \u00d7 (-3)]<sup>\u00a0o<\/sup><\/p>\n<p>\u00a0<\/p>\n<p>Substituting the value of \u03c0 = 22\/7<\/p>\n<p>[180\/22 \u00d7 7 \u00d7 -3]<sup>\u00a0o<\/sup><\/p>\n<p>\u00a0<\/p>\n<p>(-3780\/22)<sup>\u00a0o<\/sup><\/p>\n<p>(-171 18\/22)<sup>\u00a0o<\/sup><\/p>\n<p>(-171<sup>\u00a0o\u00a0<\/sup>(18\/22 \u00d7 60)\u2019)<\/p>\n<p>(-171<sup>o<\/sup>\u00a0(49 1\/11)\u2019)<\/p>\n<p>(-171<sup>o<\/sup>\u00a049\u2032 (1\/11 \u00d7 60)\u2019)<\/p>\n<p>\u2013 (171\u00b0 49\u2032 5.45\u201d)<\/p>\n<p>\u2248 \u2013 (171\u00b0 49\u2032 5\u201d)<\/p>\n<p>\u2234 Degree measure of (-3)<sup>\u00a0c<\/sup><strong>\u00a0<\/strong>is -171\u00b0 49\u2032 5\u201d<\/p>\n<p><strong>(v)<\/strong>\u00a011<sup>c<\/sup><\/p>\n<p>(180\/ \u03c0 \u00d7 11)<sup>\u00a0o<\/sup><\/p>\n<p>Substituting the value of \u03c0 = 22\/7<\/p>\n<p>(180\/22 \u00d7 7 \u00d7 11)<sup>\u00a0o<\/sup><\/p>\n<p>(90 \u00d7 7) \u00b0<\/p>\n<p>630\u00b0<\/p>\n<p>\u2234 Degree measure of 11<sup>c<\/sup>\u00a0is 630\u00b0<\/p>\n<p><strong>(vi)\u00a0<\/strong>1<sup>c<\/sup><\/p>\n<p>(180\/ \u03c0 \u00d7 1)<sup>\u00a0o<\/sup><\/p>\n<p>Substituting the value of \u03c0 = 22\/7<\/p>\n<p>(180\/22 \u00d7 7 \u00d7 1)<sup>\u00a0o<\/sup><\/p>\n<p>(1260\/22)\u00a0<sup>o<\/sup><\/p>\n<p>(57 3\/11)\u00a0<sup>o<\/sup><\/p>\n<p>(57<sup>o<\/sup>\u00a0(3\/11 \u00d7 60)\u2019)<\/p>\n<p>(57<sup>o<\/sup>\u00a0(16 4\/11)\u2019)<\/p>\n<p>(57<sup>o<\/sup>\u00a016\u2032 (4\/11 \u00d7 60)\u2019)<\/p>\n<p>(57<sup>o<\/sup>\u00a016\u2032 21.81\u201d)<\/p>\n<p>\u2248 (57<sup>o<\/sup>\u00a016\u2032 21\u201d)<\/p>\n<p>\u2234 Degree measure of 1<sup>c<\/sup>\u00a0is 57<sup>o<\/sup>\u00a016\u2032 21\u201d<\/p>\n<h3><span class=\"ez-toc-section\" id=\"2-find-the-radian-measure-corresponding-to-the-following-degree-measures-i-300o-ii-35o-iii-56o-iv135o-v-300o-vi-7o-30%e2%80%b2-vii-125o-30%e2%80%99-viii-47o-30%e2%80%b2\"><\/span>2. Find the radian measure corresponding to the following degree measures:<br \/>(i) 300<sup>o<\/sup>\u00a0(ii) 35<sup>o<\/sup>\u00a0(iii) -56<sup>o<\/sup>\u00a0(iv)135<sup>o<\/sup>\u00a0(v) -300<sup>o<\/sup><br \/>(vi) 7<sup>o<\/sup>\u00a030\u2032 (vii) 125<sup>o<\/sup>\u00a030\u2019 (viii) -47<sup>o<\/sup>\u00a030\u2032<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that 180\u00b0 = \u03c0 rad\u00a0\u21d2\u00a01\u00b0 = \u03c0\/ 180 rad<\/p>\n<p><strong>(i)<\/strong>\u00a0300\u00b0<\/p>\n<p>(300 \u00d7 \u03c0\/180) rad<\/p>\n<p>5\u03c0\/3<\/p>\n<p>\u2234 Radian measure of 300<sup>o<\/sup>\u00a0is 5\u03c0\/3<\/p>\n<p><strong>(ii)<\/strong>\u00a035\u00b0<\/p>\n<p>(35 \u00d7 \u03c0\/180) rad<\/p>\n<p>7\u03c0\/36<\/p>\n<p>\u2234 Radian measure of 35<sup>o<\/sup>\u00a0is 7\u03c0\/36<\/p>\n<p><strong>(iii)<\/strong>\u00a0-56\u00b0<\/p>\n<p>(-56 \u00d7 \u03c0\/180) rad<\/p>\n<p>-14\u03c0\/45<\/p>\n<p>\u2234 Radian measure of -56\u00b0 is -14\u03c0\/45<\/p>\n<p><strong>(iv)<\/strong>\u00a0135\u00b0<\/p>\n<p>(135 \u00d7 \u03c0\/180) rad<\/p>\n<p>3\u03c0\/4<\/p>\n<p>\u2234 Radian measure of 135\u00b0 is 3\u03c0\/4<\/p>\n<p><strong>(v)<\/strong>\u00a0-300\u00b0<\/p>\n<p>(-300 \u00d7 \u03c0\/180) rad<\/p>\n<p>-5\u03c0\/3<\/p>\n<p>\u2234 Radian measure of -300\u00b0 is -5\u03c0\/3<\/p>\n<p><strong>(vi)<\/strong>\u00a07\u00b0 30\u2032<\/p>\n<p>We know that, 30\u2032 = (1\/2) \u00b0<\/p>\n<p>7\u00b0 30\u2032 = (7 1\/2) \u00b0<\/p>\n<p>= (15\/2)<sup>\u00a0o<\/sup><\/p>\n<p>= (15\/2 \u00d7 \u03c0\/180) rad<\/p>\n<p>= \u03c0\/24<\/p>\n<p>\u2234 Radian measure of 7\u00b0 30\u2032 is \u03c0\/24<\/p>\n<p><strong>(vii)<\/strong>\u00a0125\u00b0 30\u2032<\/p>\n<p>We know that, 30\u2032 = (1\/2) \u00b0<\/p>\n<p>125\u00b0 30\u2019 = (125 1\/2) \u00b0<\/p>\n<p>= (251\/2)<sup>\u00a0o<\/sup><\/p>\n<p>= (251\/2 \u00d7 \u03c0\/180) rad<\/p>\n<p>= 251\u03c0\/360<\/p>\n<p>\u2234 Radian measure of 125\u00b0 30\u2032 is 251\u03c0\/360<\/p>\n<p><strong>(viii)<\/strong>\u00a0-47\u00b0 30\u2032<\/p>\n<p>We know that, 30\u2032 = (1\/2) \u00b0<\/p>\n<p>-47\u00b0 30\u2019 = \u2013 (47 1\/2) \u00b0<\/p>\n<p>= \u2013 (95\/2)<sup>\u00a0o<\/sup><\/p>\n<p>= \u2013 (95\/2 \u00d7 \u03c0\/180) rad<\/p>\n<p>= \u2013 19\u03c0\/72<\/p>\n<p>\u2234 Radian measure of -47\u00b0 30\u2032 is \u2013 19\u03c0\/72<\/p>\n<h3><span class=\"ez-toc-section\" id=\"3-the-difference-between-the-two-acute-angles-of-a-right-angled-triangle-is-2%cf%805-radians-express-the-angles-in-degrees\"><\/span>3. The difference between the two acute angles of a right-angled triangle is 2\u03c0\/5 radians. Express the angles in degrees.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given the difference between the two acute angles of a right-angled triangle is 2\u03c0\/5 radians.<\/p>\n<p>We know that \u03c0 rad = 180\u00b0\u00a0\u21d2\u00a01 rad = 180\u00b0\/ \u03c0<\/p>\n<p>Given:<\/p>\n<p>2\u03c0\/5<\/p>\n<p>(2\u03c0\/5 \u00d7 180\/ \u03c0)<sup>\u00a0o<\/sup><\/p>\n<p>Substituting the value of \u03c0 = 22\/7<\/p>\n<p>(2\u00d722\/(7\u00d75) \u00d7 180\/22 \u00d7 7)<\/p>\n<p>(2\/5 \u00d7 180) \u00b0<\/p>\n<p>72\u00b0<\/p>\n<p>Let one acute angle be x\u00b0 and the other acute angle be 90\u00b0 \u2013 x\u00b0.<\/p>\n<p>Then,<\/p>\n<p>x\u00b0 \u2013 (90\u00b0 \u2013 x\u00b0) = 72\u00b0<\/p>\n<p>2x\u00b0 \u2013 90\u00b0 = 72\u00b0<\/p>\n<p>2x\u00b0 = 72\u00b0 + 90\u00b0<\/p>\n<p>2x\u00b0 = 162\u00b0<\/p>\n<p>x\u00b0 = 162\u00b0\/ 2<\/p>\n<p>x\u00b0 = 81\u00b0 and<\/p>\n<p>90\u00b0 \u2013 x\u00b0 = 90\u00b0 \u2013 81\u00b0<\/p>\n<p>= 9\u00b0<\/p>\n<p>\u2234 The angles are 81<sup>o<\/sup>\u00a0and 9<sup>o<\/sup><\/p>\n<h3><span class=\"ez-toc-section\" id=\"4-one-angle-of-a-triangle-is-23x-grades-and-another-is-32x-degrees-while-the-third-is-%cf%80x75-radians-express-all-the-angles-in-degrees\"><\/span>4. One angle of a triangle is 2\/3x\u00a0grades,\u00a0and another is 3\/2x\u00a0degrees while the third is \u03c0x\/75\u00a0radians. Express all the angles in degrees.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>One angle of a triangle is 2x\/3 grades and another is 3x\/2 degree while the third is \u03c0x\/75 radians.<\/p>\n<p>We know that, 1 grad = (9\/10)<sup>\u00a0o<\/sup><\/p>\n<p>2\/3x grad = (9\/10) (2\/3x)<sup>\u00a0o<\/sup><\/p>\n<p>= 3\/5x<sup>o<\/sup><\/p>\n<p>We know that, \u03c0 rad = 180\u00b0\u00a0\u21d2\u00a01 rad = 180\u00b0\/ \u03c0<\/p>\n<p>Given: \u03c0x\/75<\/p>\n<p>(\u03c0x\/75 \u00d7 180\/\u03c0)<sup>\u00a0o<\/sup><\/p>\n<p>(12\/5x)<sup>\u00a0o<\/sup><\/p>\n<p>We know that, the sum of the angles of a triangle is 180\u00b0.<\/p>\n<p>3\/5x<sup>o<\/sup>\u00a0+ 3\/2x<sup>o<\/sup>\u00a0+ 12\/5x<sup>o<\/sup>\u00a0= 180<sup>o<\/sup><\/p>\n<p>(6+15+24)\/10x<sup>o<\/sup>\u00a0= 180<sup>o<\/sup><\/p>\n<p>Upon cross-multiplication we get,<\/p>\n<p>45x<sup>o<\/sup>\u00a0= 180<sup>o<\/sup>\u00a0\u00d7 10<sup>o<\/sup><\/p>\n<p>= 1800<sup>o<\/sup><\/p>\n<p>x<sup>o<\/sup>\u00a0= 1800<sup>o<\/sup>\/45<sup>o<\/sup><\/p>\n<p>= 40<sup>o<\/sup><\/p>\n<p>\u2234\u00a0The angles of the triangle are:<\/p>\n<p>3\/5x<sup>o<\/sup>\u00a0= 3\/5 \u00d7 40<sup>o<\/sup>\u00a0= 24<sup>o<\/sup><\/p>\n<p>3\/2x<sup>o<\/sup>\u00a0= 3\/2 \u00d7 40<sup>o<\/sup>\u00a0= 60<sup>o<\/sup><\/p>\n<p>12\/5 x<sup>o<\/sup>\u00a0= 12\/5 \u00d7 40<sup>o<\/sup>\u00a0= 96<sup>o<\/sup><\/p>\n<h3><span class=\"ez-toc-section\" id=\"5-find-the-magnitude-in-radians-and-degrees-of-the-interior-angle-of-a-regular-i-pentagon-ii-octagon-iii-heptagon-iv-duodecagon\"><\/span>5. Find the magnitude, in radians and degrees, of the interior angle of a regular:<br \/>(i) Pentagon (ii) Octagon (iii) Heptagon (iv) Duodecagon.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that the sum of the interior angles of a polygon = (n \u2013 2) \u03c0<\/p>\n<p>And each angle of polygon = sum of interior angles of polygon \/ number of sides<\/p>\n<p>Now, let us calculate the magnitude of<\/p>\n<p><strong>(i)<\/strong>\u00a0Pentagon<\/p>\n<p>Number of sides in pentagon = 5<\/p>\n<p>Sum of interior angles of pentagon = (5 \u2013 2) \u03c0 = 3\u03c0<\/p>\n<p>\u2234\u00a0Each angle of pentagon = 3\u03c0\/5 \u00d7 180<sup>o<\/sup>\/ \u03c0 = 108<sup>o<\/sup><\/p>\n<p><strong>(ii)<\/strong>\u00a0Octagon<\/p>\n<p>Number of sides in octagon = 8<\/p>\n<p>Sum of interior angles of octagon = (8 \u2013 2) \u03c0 = 6\u03c0<\/p>\n<p>\u2234\u00a0Each angle of octagon = 6\u03c0\/8 \u00d7 180<sup>o<\/sup>\/ \u03c0 = 135<sup>o<\/sup><\/p>\n<p><strong>(iii)<\/strong>\u00a0Heptagon<\/p>\n<p>Number of sides in heptagon = 7<\/p>\n<p>Sum of interior angles of heptagon = (7 \u2013 2) \u03c0 = 5\u03c0<\/p>\n<p>\u2234\u00a0Each angle of heptagon = 5\u03c0\/7 \u00d7 180<sup>o<\/sup>\/ \u03c0 = 900<sup>o<\/sup>\/7 = 128<sup>o<\/sup>\u00a034\u2032 17\u201d<\/p>\n<p><strong>(iv)<\/strong>\u00a0Duodecagon<\/p>\n<p>Number of sides in duodecagon = 12<\/p>\n<p>Sum of interior angles of duodecagon = (12 \u2013 2) \u03c0 = 10\u03c0<\/p>\n<p>\u2234\u00a0Each angle of duodecagon = 10\u03c0\/12 \u00d7 180<sup>o<\/sup>\/ \u03c0 = 150<sup>o<\/sup><\/p>\n<h3><span class=\"ez-toc-section\" id=\"6-the-angles-of-a-quadrilateral-are-in-ap-and-the-greatest-angle-is-120o-express-the-angles-in-radians\"><\/span>6. The angles of a quadrilateral are in A.P., and the greatest angle is 120<sup>o<\/sup>. Express the angles in radians.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the angles of quadrilateral be (a \u2013 3d) \u00b0, (a \u2013 d) \u00b0, (a + d) \u00b0 and (a + 3d) \u00b0.<\/p>\n<p>We know that, the sum of angles of a quadrilateral is 360\u00b0.<\/p>\n<p>a \u2013 3d + a \u2013 d + a + d + a + 3d = 360\u00b0<\/p>\n<p>4a = 360\u00b0<\/p>\n<p>a = 360\/4<\/p>\n<p>= 90\u00b0<\/p>\n<p>Given:<\/p>\n<p>The greatest angle = 120\u00b0<\/p>\n<p>a + 3d = 120\u00b0<\/p>\n<p>90\u00b0 + 3d = 120\u00b0<\/p>\n<p>3d = 120\u00b0 \u2013 90\u00b0<\/p>\n<p>3d = 30\u00b0<\/p>\n<p>d = 30\u00b0\/3<\/p>\n<p>= 10<sup>o<\/sup><\/p>\n<p>\u2234 The angles are:<\/p>\n<p>(a \u2013 3d) \u00b0 = 90\u00b0 \u2013 30\u00b0 = 60\u00b0<\/p>\n<p>(a \u2013 d) \u00b0 = 90\u00b0 \u2013 10\u00b0 = 80\u00b0<\/p>\n<p>(a + d) \u00b0 = 90\u00b0 + 10\u00b0 = 100\u00b0<\/p>\n<p>(a + 3d) \u00b0 = 120\u00b0<\/p>\n<p>Angles of quadrilateral in radians:<\/p>\n<p>(60 \u00d7 \u03c0\/180) rad = \u03c0\/3<\/p>\n<p>(80 \u00d7 \u03c0\/180) rad = 4\u03c0\/9<\/p>\n<p>(100 \u00d7 \u03c0\/180) rad = 5\u03c0\/9<\/p>\n<p>(120 \u00d7 \u03c0\/180) rad = 2\u03c0\/3<\/p>\n<h3><span class=\"ez-toc-section\" id=\"7-the-angles-of-a-triangle-are-in-ap-and-the-number-of-degrees-in-the-least-angle-is-to-the-number-of-degrees-in-the-mean-angle-as-1-120-find-the-angle-in-radians\"><\/span>7. The angles of a triangle are in A.P., and the number of degrees in the least angle is to the number of degrees in the mean angle as 1:120. Find the angle in radians.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the angles of the triangle be (a \u2013 d) \u00b0, a\u00b0 and (a + d) \u00b0.<\/p>\n<p>We know that, the sum of the angles of a triangle is 180\u00b0.<\/p>\n<p>a \u2013 d + a + a + d = 180\u00b0<\/p>\n<p>3a = 180\u00b0<\/p>\n<p>a = 60\u00b0<\/p>\n<p>Given:<\/p>\n<p>Number of degrees in the least angle \/ Number of degrees in the mean angle = 1\/120<\/p>\n<p>(a-d)\/a = 1\/120<\/p>\n<p>(60-d)\/60 = 1\/120<\/p>\n<p>(60-d)\/1 = 1\/2<\/p>\n<p>120-2d = 1<\/p>\n<p>2d = 119<\/p>\n<p>d = 119\/2<\/p>\n<p>= 59.5<\/p>\n<p>\u2234 The angles are:<\/p>\n<p>(a \u2013 d) \u00b0 = 60\u00b0 \u2013 59.5\u00b0 = 0.5\u00b0<\/p>\n<p>a\u00b0 = 60\u00b0<\/p>\n<p>(a + d) \u00b0 = 60\u00b0 + 59.5\u00b0 = 119.5\u00b0<\/p>\n<p>Angles of triangle in radians:<\/p>\n<p>(0.5 \u00d7 \u03c0\/180) rad = \u03c0\/360<\/p>\n<p>(60 \u00d7 \u03c0\/180) rad = \u03c0\/3<\/p>\n<p>(119.5 \u00d7 \u03c0\/180) rad = 239\u03c0\/360<\/p>\n<h3><span class=\"ez-toc-section\" id=\"8-the-angle-in-one-regular-polygon-is-to-that-in-another-as-3-2-and-the-number-of-sides-in-first-is-twice-that-in-the-second-determine-the-number-of-sides-of-two-polygons\"><\/span>8. The angle in one regular polygon is to that in another as 3:2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the number of sides in the first polygon be 2x and<\/p>\n<p>The number of sides in the second polygon be x.<\/p>\n<p>We know that, angle of an n-sided regular polygon = [(n-2)\/n] \u03c0 radian<\/p>\n<p>The angle of the first polygon\u00a0= [(2x-2)\/2x] \u03c0 = [(x-1)\/x] \u03c0 radian<\/p>\n<p>The angle of the second polygon = [(x-2)\/x] \u03c0 radian<\/p>\n<p>Thus,<\/p>\n<p>[(x-1)\/x] \u03c0 \/ [(x-2)\/x] \u03c0 = 3\/2<\/p>\n<p>\u00a0<\/p>\n<p>(x-1)\/(x-2) = 3\/2<\/p>\n<p>Upon cross-multiplication we get,<\/p>\n<p>2x \u2013 2 = 3x \u2013 6<\/p>\n<p>3x-2x = 6-2<\/p>\n<p>x = 4<\/p>\n<p>\u2234 Number of sides in the first polygon = 2x = 2(4) = 8<\/p>\n<p>Number of sides in the second polygon = x = 4<\/p>\n<h3><span class=\"ez-toc-section\" id=\"9-the-angles-of-a-triangle-are-in-ap-such-that-the-greatest-is-5-times-the-least-find-the-angles-in-radians\"><\/span>9. The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the angles of the triangle be (a \u2013 d)<sup>\u00a0o<\/sup>, a<sup>o<\/sup>\u00a0and (a + d)<sup>\u00a0o<\/sup>.<\/p>\n<p>We know that, the sum of angles of triangle is 180\u00b0.<\/p>\n<p>a \u2013 d + a + a + d = 180\u00b0<\/p>\n<p>3a = 180\u00b0<\/p>\n<p>a = 180\u00b0\/3<\/p>\n<p>= 60<sup>o<\/sup><\/p>\n<p>Given:<\/p>\n<p>Greatest angle = 5 \u00d7 least angle<\/p>\n<p>Upon cross-multiplication,<\/p>\n<p>Greatest angle \/ least angle = 5<\/p>\n<p>(a+d)\/(a-d) = 5<\/p>\n<p>(60+d)\/(60-d) = 5<\/p>\n<p>By cross-multiplying we get,<\/p>\n<p>60 + d = 300 \u2013 5d<\/p>\n<p>6d = 240<\/p>\n<p>d = 240\/6<\/p>\n<p>= 40<\/p>\n<p>Hence, angles are:<\/p>\n<p>(a \u2013 d) \u00b0 = 60\u00b0 \u2013 40\u00b0 = 20\u00b0<\/p>\n<p>a\u00b0 = 60\u00b0<\/p>\n<p>(a + d) \u00b0 = 60\u00b0 + 40\u00b0 = 100\u00b0<\/p>\n<p>\u2234\u00a0Angles of triangle in radians:<\/p>\n<p>(20 \u00d7 \u03c0\/180) rad = \u03c0\/9<\/p>\n<p>(60 \u00d7 \u03c0\/180) rad = \u03c0\/3<\/p>\n<p>(100 \u00d7 \u03c0\/180) rad = 5\u03c0\/9<\/p>\n<h3><span class=\"ez-toc-section\" id=\"10-the-number-of-sides-of-two-regular-polygons-is-5-4-and-the-difference-between-their-angles-is-9o-find-the-number-of-sides-of-the-polygons\"><\/span>10. The number of sides of two regular polygons is 5:4 and the difference between their angles is 9<sup>o<\/sup>. Find the number of sides of the polygons.<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the number of sides in the first polygon be 5x and<\/p>\n<p>The number of sides in the second polygon be 4x.<\/p>\n<p>We know that, angle of an n-sided regular polygon = [(n-2)\/n] \u03c0 radian<\/p>\n<p>The angle of the first polygon = [(5x-2)\/5x] 180<sup>o<\/sup><\/p>\n<p>The angle of the second polygon = [(4x-1)\/4x] 180<sup>o<\/sup><\/p>\n<p>Thus,<\/p>\n<p>[(5x-2)\/5x] 180<sup>o<\/sup>\u00a0\u2013 [(4x-1)\/4x] 180<sup>o<\/sup>\u00a0= 9<\/p>\n<p>\u00a0<\/p>\n<p>180<sup>o<\/sup>\u00a0[(4(5x-2) \u2013 5(4x-2))\/20x] = 9<\/p>\n<p>Upon cross-multiplication we get,<\/p>\n<p>(20x \u2013 8 \u2013 20x + 10)\/20x = 9\/180<\/p>\n<p>2\/20x = 1\/20<\/p>\n<p>2\/x = 1<\/p>\n<p>x = 2<\/p>\n<p>\u2234Number of sides in the first polygon = 5x = 5(2) = 10<\/p>\n<p>Number of sides in the second polygon = 4x = 4(2) = 8<\/p>\n<h2><span class=\"ez-toc-section\" id=\"key-take-away-from-class-11-maths-rd-sharma-solutions\"><\/span>Key Take Away from Class 11 Maths RD Sharma Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Understanding is the key.<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Do not learn answers or the questions for the ease of solving them.<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Analyze every problem you did wrong.<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> \u00a0 \u00a0 \u00a0 <\/span><span style=\"font-weight: 400;\">Do not leave any concept in between. Go through again because everything you leave will create a void for further levels.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"access-other-important-chapters-of-rd-sharma-solutions-class-11-maths-chapter-4\"><\/span>Access Other Important Chapters of RD Sharma Solutions Class 11 Maths Chapter 4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 1 &#8211; Sets<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 2 &#8211; Relations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 3 &#8211; Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 5 &#8211; Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 6 &#8211; Graphs of Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 7 &#8211; Trigonometric Ratios of Compound Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 8 &#8211; Transformation Formulae<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 9 &#8211; Trigonometric Ratios of Multiple and Sub Multiple Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 10 &#8211; Sine and Cosine Formulae and Their Applications<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 11 &#8211; Trigonometric Equations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 12 &#8211; Mathematical Induction<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 13 &#8211; Complex Numbers<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 14 &#8211; Quadratic Equations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 15 &#8211; Linear Inequations<\/a><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">Lastly, all the best for your <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener noreferrer\">CBSE<\/a> class11 examinations!<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-4\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 4.\u00a0<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-1629781084849\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-to-solve-questions-from-rd-sharma-solutions-class-11-maths-chapter-4\"><\/span>How to solve questions from RD Sharma Solutions Class 11 Maths Chapter 4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can follow the key points mentioned in the above blog. <\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629781176363\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-4\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download link from the above blog. <\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629781228464\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-maths-without-the-internet\"><\/span>Can I access the RD Sharma Solutions Maths without the internet? <span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have to download the PDF online, you can access it offline whenever required. <\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 4:\u00a0If you are new to this chapter and found this topic tricky to understand such\u00a0as angles and radians measure, their conversions, and inter-conversions then this article belongs to you. In this article, we are going to talk about the RD Sharma Solutions Class 11 Maths Chapter 4 Measurement &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 4 &#8211; Measurement of Angles (Updated For 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-4\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 4 &#8211; Measurement of Angles (Updated For 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":119069,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63549"}],"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\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=63549"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63549\/revisions"}],"predecessor-version":[{"id":119695,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63549\/revisions\/119695"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/119069"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63549"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63549"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63549"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}