{"id":6071,"date":"2023-09-09T07:40:03","date_gmt":"2023-09-09T07:40:03","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6071"},"modified":"2023-09-09T07:40:03","modified_gmt":"2023-09-09T07:40:03","slug":"ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 8 (Introduction to Trigonometry)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 8 Introduction to Trigonometry\u00a0 <\/strong>Exercise 8.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 8 Introduction to Trigonometry<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-8-introduction-to-trigonometry-ex-8-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 8.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 8.3<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Evaluate :<br \/>\n<\/strong><strong>(i) sin 18\u00b0\/cos 72\u00b0<br \/>\n(ii) tan 26\u00b0\/cot 64\u00b0<br \/>\n(iii)\u00a0\u00a0cos 48\u00b0 \u2013 sin 42\u00b0<br \/>\n(iv)\u00a0\u00a0cosec 31\u00b0 \u2013 sec 59\u00b0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) sin 18\u00b0\/cos 72\u00b0<br \/>\n<\/strong>= sin (90\u00b0 \u2013 18\u00b0) \/cos 72\u00b0<br \/>\n= cos 72\u00b0 \/cos 72\u00b0<br \/>\n= 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) tan 26\u00b0\/cot 64\u00b0<br \/>\n<\/strong>= tan (90\u00b0 \u2013 26\u00b0)\/cot 64\u00b0<br \/>\n= cot 64\u00b0\/cot 64\u00b0<br \/>\n= 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii)\u00a0cos 48\u00b0 \u2013 sin 42\u00b0<br \/>\n<\/strong>= cos (90\u00b0 \u2013 42\u00b0) \u2013 sin 42\u00b0<br \/>\n= sin 42\u00b0 \u2013 sin 42\u00b0<br \/>\n= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) cosec 31\u00b0 \u2013 sec 59\u00b0<br \/>\n<\/strong>= cosec (90\u00b0 \u2013 59\u00b0) \u2013 sec 59\u00b0<br \/>\n= sec 59\u00b0 \u2013 sec 59\u00b0<br \/>\n= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. \u00a0Show that:<br \/>\n<\/strong><strong>(i) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0 = 1<br \/>\n(ii) cos 38\u00b0 cos 52\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0 = 0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0<br \/>\n<\/strong>= tan (90\u00b0 \u2013 42\u00b0) tan (90\u00b0 \u2013 67\u00b0) tan 42\u00b0 tan 67\u00b0<br \/>\n= cot 42\u00b0 cot 67\u00b0 tan 42\u00b0 tan 67\u00b0<br \/>\n= (cot 42\u00b0 tan 42\u00b0) (cot 67\u00b0 tan 67\u00b0)<br \/>\n(Since, tan A\u00b0 cot A\u00b0 = 1)<br \/>\n= 1 \u00d7 1<br \/>\n= 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) cos 38\u00b0 cos 52\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0<br \/>\n<\/strong>= cos (90\u00b0 \u2013 52\u00b0) cos (90\u00b0-38\u00b0) \u2013 sin 38\u00b0 sin 52\u00b0<br \/>\n= sin 52\u00b0 sin 38\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0<br \/>\n= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. If tan 2A = cot (A \u2013 18\u00b0), where 2A is an acute angle, find the value of A<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong><br \/>\ntan 2A = cot (A &#8211; 18\u00b0)<br \/>\nWe know that tan 2A = cot (90\u00b0 \u2013 2A)<br \/>\n\u21d2 cot (90\u00b0 \u2013 2A) = cot (A -18\u00b0)<br \/>\n\u21d2 90\u00b0 \u2013 2A = A &#8211; 18\u00b0<br \/>\n\u21d2\u00a0108\u00b0 = 3A<br \/>\n\u21d2 A = 108\u00b0\/3<br \/>\n\u21d2 A = 36\u00b0<br \/>\nTherefore, the value of A = 36\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4.\u00a0\u00a0If tan A = cot B, prove that A + B = 90\u00b0.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong><br \/>\ntan A = cot B<br \/>\nWe know that cot B = tan (90\u00b0 \u2013 B)<br \/>\nTo prove A + B = 90\u00b0,<br \/>\ntan A = tan (90\u00b0 \u2013 B)<br \/>\n\u21d2 A = 90\u00b0 \u2013 B<br \/>\n\u21d2 A + B = 90\u00b0<br \/>\nHence Proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. If sec 4A = cosec (A \u2013 20\u00b0), where 4A is an acute angle, find the value of A.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong><br \/>\nsec 4A = cosec (A \u2013 20\u00b0)<br \/>\nWe know that sec 4A = cosec (90\u00b0 \u2013 4A)<br \/>\n\u21d2 cosec (90\u00b0 \u2013 4A) = cosec (A \u2013 20\u00b0)<br \/>\n\u21d2 90\u00b0 \u2013 4A= A &#8211; 20\u00b0<br \/>\n\u21d2 110\u00b0 = 5A<br \/>\n\u21d2 A = 110\u00b0\/ 5<br \/>\n\u21d2 A = 22\u00b0<br \/>\nTherefore, the value of A = 22\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. If A, B and C are interior angles of a triangle ABC, then show that<br \/>\n<\/strong><strong><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{sin\\left&amp;space;(\\frac{B+C}{2}&amp;space;\\right&amp;space;)&amp;space;=&amp;space;cos&amp;space;\\frac{A}{2}}\" alt=\"\\mathbf{sin\\left (\\frac{B+C}{2} \\right ) = cos \\frac{A}{2}}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong> \u00a0In a triangle, sum of all the interior angles <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">A + B\u00a0+ C = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 B\u00a0+ C = 180\u00b0 &#8211; A<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (B + C)\/2 = (180\u00b0 &#8211; A)\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 (B + C)\/2 = (90\u00b0 &#8211; A\/2)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 sin (B + C)\/2 = sin (90\u00b0 &#8211; A\/2) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 sin (B + C)\/2 = cos A\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7.\u00a0Express sin 67\u00b0 + cos 75\u00b0 in terms of trigonometric ratios of angles between 0\u00b0 and 45\u00b0.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong><br \/>\nsin 67\u00b0 + cos 75\u00b0<br \/>\n= sin (90\u00b0 \u2013 23\u00b0) + cos (90\u00b0 \u2013 15\u00b0)<br \/>\n= cos 23\u00b0 + sin 15\u00b0<br \/>\n<\/span><\/p>\n<p><span style=\"color: #0000ff; font-family: Georgia, Palatino;\"><b><i><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\">Go Back To Chapters<\/a><\/i><\/b><\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 8 (Introduction to Trigonometry)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 8 Introduction to Trigonometry\u00a0 Exercise 8.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 8 Introduction to Trigonometry NCERT Class 10 Maths Solution [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1091,1095,1044,1049,1048],"class_list":["post-6071","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-8-introduction-to-trigonometry-solutions","tag-ncert-class-10-mathematics-exercise-8-3-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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