{"id":108,"date":"2021-07-23T11:09:49","date_gmt":"2021-07-23T05:39:49","guid":{"rendered":"http:\/\/mcq-questions.com\/?p=108"},"modified":"2022-03-02T10:23:50","modified_gmt":"2022-03-02T04:53:50","slug":"ncert-solutions-for-class-10-maths-chapter-8-hindi","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-8-hindi\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Hindi Medium)"},"content":{"rendered":"
These Solutions are part of NCERT Solutions for Class 10 Maths in Hindi Medium<\/a>. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry.<\/p>\n Chapter 8. \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u093f \u0915\u093e \u092a\u0930\u093f\u091a\u092f<\/strong><\/p>\n \u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 8.1 (\u0917\u0923\u093f\u0924)<\/span>\u00a0<\/strong><\/p>\n Ex 8.1 Class 10 \u0917\u0923\u093f\u0924 Q1. \u00a0<\/strong>D<\/strong>ABC\u00a0<\/strong>\u092e\u0947\u0902, \u091c\u093f\u0938\u0915\u093e \u0915\u094b\u0923 B \u0938\u092e\u0915\u094b\u0923 \u0939\u0948, AB = 24 cm \u0914\u0930 BC = 7 cm \u0939\u0948 | \u0928\u093f\u092e\u094d\u0928 \u0932\u093f\u0916\u093f\u0924 \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :<\/strong><\/p>\n (i) sin<\/strong>\u00a0A, cos<\/strong>\u00a0A<\/strong><\/p>\n (ii) sin<\/strong>\u00a0C<\/strong>, cos<\/strong>\u00a0C<\/strong><\/p>\n Solution:<\/strong><\/p>\n \u0938\u092e\u0915\u094b\u0923 \u0924\u094d\u0930\u093f\u092d\u0941\u091c DABC \u092e\u0947\u0902,<\/p>\n AB = 24 cm, BC = 7 cm<\/p>\n \u092a\u093e\u0907\u0925\u093e\u0917\u094b\u0930\u0938 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947,<\/p>\n <\/p>\n AC2<\/sup>\u00a0= AB2<\/sup>\u00a0+ BC2<\/sup><\/p>\n = 242<\/sup>\u00a0+ 72<\/sup><\/p>\n = 576 + 49<\/p>\n = 625<\/p>\n AC =\u00a0\u221a625\u00a0= 25 cm<\/p>\n \u0905\u092c \u0924\u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u093f\u092f \u0905\u0928\u0941\u092a\u093e\u0924 \u0932\u0947\u0928\u0947 \u092a\u0930<\/p>\n (i) sin A, cos A<\/strong><\/p>\n <\/p>\n For more information about NCERT Solutions Class 10 Maths<\/a> in English medium.<\/p>\n Ex 8.1 Class 10 \u0917\u0923\u093f\u0924\u00a0Q2. \u00a0<\/strong>\u0906\u0915\u0943\u0924\u093f 8.13 \u092e\u0947\u0902,\u00a0<\/strong>tan P \u2013 cot R\u00a0<\/strong>\u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f | \u00a0\u00a0<\/strong><\/p>\n Solution:<\/strong><\/p>\n PQ = 12 cm, PR = 13 cm<\/p>\n QR = ?<\/p>\n \u0938\u092e\u0915\u094b\u0923 \u0924\u094d\u0930\u093f\u092d\u0941\u091c DPQR \u00a0\u092e\u0947\u0902,<\/p>\n PQ = 12 cm, PR = 13 cm<\/p>\n \u092a\u093e\u0907\u0925\u093e\u0917\u094b\u0930\u0938 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947,<\/p>\n <\/p>\n PR2<\/sup>\u00a0= PQ2<\/sup>\u00a0+ QR2<\/sup><\/p>\n 132<\/sup>\u00a0= 122<\/sup>\u00a0+ QR2<\/sup><\/p>\n 169 = 144 + QR2<\/sup><\/p>\n 169 – 144 = QR2<\/sup><\/p>\n QR2<\/sup>\u00a0= 25<\/p>\n QR =\u00a0\u221a25\u00a0= 5 cm<\/p>\n \u0905\u092c \u0924\u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u093f\u092f \u0905\u0928\u0941\u092a\u093e\u0924 \u0932\u0947\u0928\u0947 \u092a\u0930<\/p>\n <\/a>\u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 8.2<\/strong><\/span><\/p>\n Ex 8.2 Class 10 \u0917\u0923\u093f\u0924 Q1. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u0947 \u092e\u093e\u0928 \u0928\u093f\u0915\u093e\u0932\u093f\u090f:<\/strong><\/p>\n (i)\u00a0<\/strong>sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0<\/strong><\/p>\n \u0939\u0932:<\/strong>\u00a0sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0<\/p>\n \u0938\u092d\u0940 \u0924\u094d\u0930\u093f\u0915\u094b\u0902\u0923\u092e\u093f\u0924\u0940\u092f \u0905\u0928\u0941\u092a\u093e\u0924\u094b\u0902 \u0915\u093e \u092e\u093e\u0928 \u0930\u0916\u0928\u0947 \u092a\u0930<\/p>\n <\/p>\n \u00a0(ii) 2 tan2<\/sup>\u00a045\u00b0 + cos2<\/sup>\u00a030\u00b0 \u2013 sin2<\/sup>\u00a060\u00b0<\/strong><\/p>\n \u0939\u0932:<\/strong>\u00a02 tan2<\/sup>\u00a045\u00b0 + cos2<\/sup>\u00a030\u00b0 \u2013 sin2<\/sup>\u00a060\u00b0<\/p>\n Ex 8.2 Class 10 \u0917\u0923\u093f\u0924\u00a0Q2. \u0938\u0939\u0940 \u0935\u093f\u0915\u0932\u094d\u092a \u091a\u0941\u0928\u093f\u090f \u0914\u0930 \u0905\u092a\u0928\u0947 \u0935\u093f\u0915\u0932\u094d\u092a \u0915\u093e \u0914\u091a\u093f\u0924\u094d\u092f \u0926\u0940\u091c\u093f\u090f :\u00a0<\/strong><\/p>\n <\/p>\n (A) sin 60\u00b0 (B) cos 60\u00b0 (C) tan 60\u00b0 (D) sin 30\u00b0<\/strong><\/p>\n (iii) sin 2A = 2 sin A \u0924\u092c \u0938\u0924\u094d\u092f \u0939\u094b\u0924\u093e \u0939\u0948, \u091c\u092c\u0915\u093f A \u092c\u0930\u093e\u092c\u0930 \u0939\u0948 :<\/strong><\/p>\n (A) 0\u00b0 \u00a0\u00a0\u00a0\u00a0(B) 30\u00b0 \u00a0\u00a0\u00a0(C) 45\u00b0 \u00a0\u00a0\u00a0\u00a0(D) 60\u00b0<\/strong><\/p>\n \u0939\u0932:<\/strong>\u00a0sin 2A = 2 sin A<\/p>\n \u21d2\u00a02 sin A cos A = 2 sin A [ \u00a0sin 2x = 2 sin x cos x]<\/p>\n \u21d2\u00a0cos A = 2 sin A – 2 sin A<\/p>\n \u21d2 cos A = 0<\/p>\n \u2234 \u00a0 \u00a0 \u00a0 \u00a0A = 0o<\/sup><\/p>\n \u0935\u093f\u0915\u0932\u094d\u092a (A) \u0938\u0924\u094d\u092f \u0939\u0948 |<\/p>\n A \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (iii) \u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930<\/p>\n A + B = 60\u00b0<\/p>\n \u21d2 45\u00b0 + B = 60\u00b0<\/p>\n \u21d2 B = 60\u00b0 – 45\u00b0<\/p>\n \u21d2 B = 15\u00b0<\/p>\n A = 45\u00b0, B = 15\u00b0<\/p>\n Ex 8.2 Class 10 \u0917\u0923\u093f\u0924\u00a0Q4. \u092c\u0924\u093e\u0907\u090f \u0915\u093f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947\u0902 \u0938\u0947 \u0915\u094c\u0928-\u0915\u094c\u0928 \u0938\u0924\u094d\u092f \u0939\u0948\u0902 \u092f\u093e \u0905\u0938\u0924\u094d\u092f \u0939\u0948\u0902 | \u0915\u093e\u0930\u0923 \u0938\u0939\u093f\u0924 \u0905\u092a\u0928\u0947 \u0909\u0924\u094d\u0924\u0930 \u0915\u0940 \u092a\u0941\u0937\u094d\u091f\u093f \u0915\u0940\u091c\u093f\u090f |<\/strong><\/p>\n (i) sin (A + B) = sin A + sin B.<\/strong><\/p>\n (ii)\u00a0<\/strong>\u03b8\u00a0\u092e\u0947\u0902 \u0935\u0943\u0926\u094d\u0927\u093f \u0939\u094b\u0928\u0947 \u0915\u0947 \u0938\u093e\u0925 sin\u00a0<\/strong>\u03b8\u00a0\u0915\u0947 \u092e\u093e\u0928 \u092e\u0947\u0902 \u092d\u0940 \u0935\u0943\u0926\u094d\u0927\u093f \u0939\u094b\u0924\u0940 \u0939\u0948 |<\/strong><\/p>\n (iii)\u00a0<\/strong>\u03b8\u00a0\u092e\u0947\u0902 \u0935\u0943\u0926\u094d\u0927\u093f \u0939\u094b\u0928\u0947 \u0915\u0947 \u0938\u093e\u0925 cos\u00a0<\/strong>\u03b8\u00a0\u0915\u0947 \u092e\u093e\u0928 \u092e\u0947\u0902 \u092d\u0940 \u0935\u0943\u0926\u094d\u0927\u093f \u0939\u094b\u0924\u0940 \u0939\u0948 |<\/strong><\/p>\n (iv)\u00a0<\/strong>\u03b8\u00a0\u0915\u0947 \u0938\u092d\u0940 \u092e\u093e\u0928\u094b\u0902 \u092a\u0930 sin\u00a0<\/strong>\u03b8\u00a0= cos\u00a0<\/strong>\u03b8<\/p>\n (v) A = 0<\/strong>\u00b0<\/strong>\u00a0\u092a\u0930 cot A \u092a\u0930\u093f\u092d\u093e\u0937\u093f\u0924 \u0928\u0939\u0940\u0902 \u0939\u0948 |<\/strong><\/p>\n \u0909\u0924\u094d\u0924\u0930:<\/strong><\/p>\n (i)<\/strong> \u0926\u093f\u092f\u093e \u0917\u092f\u093e \u0915\u0925\u0928 \u0905\u0938\u0924\u094d\u092f \u0939\u0948 |<\/p>\n (ii)<\/strong> \u0926\u093f\u092f\u093e \u0917\u092f\u093e \u0915\u0925\u0928 \u0938\u0924\u094d\u092f \u0939\u0948 |<\/p>\n (iii)<\/strong> \u0926\u093f\u092f\u093e \u0917\u092f\u093e \u0915\u0925\u0928 \u0905\u0938\u0924\u094d\u092f \u0939\u0948 |<\/p>\n (iv)<\/strong> \u0926\u093f\u092f\u093e \u0917\u092f\u093e \u0915\u0925\u0928 \u0905\u0938\u0924\u094d\u092f \u0939\u0948 |<\/p>\n (v)<\/strong> \u0926\u093f\u092f\u093e \u0917\u092f\u093e \u0915\u0925\u0928 \u0938\u0924\u094d\u092f \u0939\u0948 |<\/p>\n <\/a>\u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 8.3<\/strong><\/span><\/p>\n Ex 8.3 Class 10 \u0917\u0923\u093f\u0924 Q1. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u093e \u092e\u093e\u0928 \u0928\u093f\u0915\u093e\u0932\u093f\u090f:<\/strong><\/p>\n (iii) cos 48<\/strong>\u00b0<\/strong>\u00a0–<\/strong>\u00a0sin 42<\/strong>\u00b0<\/strong><\/p>\n \u0939\u0932:<\/strong>\u00a0cos 48\u00b0 – sin 42\u00b0<\/p>\n \u21d2\u00a0 sin(90\u00b0 – 48\u00b0) – sin 42\u00b0<\/p>\n \u21d2\u00a0 sin 42\u00b0 – sin 42\u00b0 = 0<\/p>\n (iv) cosec 31\u00b0 – sec 59\u00b0<\/strong><\/p>\n \u0939\u0932:<\/strong>\u00a0cosec 31\u00b0 – sec 59\u00b0<\/p>\n \u21d2\u00a0 sec (90\u00b0 – 31\u00b0) – sec 59\u00b0\u00a0 [ cosec q = sec (90\u00b0 – q) ]<\/p>\n \u21d2\u00a0 sec 59\u00b0 – sec 59\u00b0 = 0<\/p>\n Ex 8.3 Class 10 \u0917\u0923\u093f\u0924\u00a0Q2.\u00a0 \u0926\u093f\u0916\u093e\u0907\u090f \u0915\u093f<\/strong><\/p>\n (i) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0 = 1<\/strong><\/p>\n \u0939\u0932:\u00a0(i) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0 = 1<\/p>\n LHS = tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0<\/p>\n = cot (90\u00b0 – 48\u00b0) tan (90\u00b0 – 23\u00b0) tan 42\u00b0 tan 67\u00b0<\/p>\n = cot 42\u00b0 cot 67\u00b0 tan 42\u00b0 tan 67\u00b0<\/p>\n = (cot 42\u00b0 \u00d7 tan 42\u00b0) (cot 67\u00b0 \u00d7 tan 67\u00b0)<\/p>\n = 1 \u00d7 1\u00a0\u00a0 [ cot A\u00a0\u00d7 tan A\u00a0= 1 ]<\/p>\n = 1<\/p>\n LHS = RHS<\/p>\n (ii) cos 38\u00b0 cos 52\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0 = 0<\/strong><\/p>\n \u0939\u0932:\u00a0(ii) cos 38\u00b0 cos 52\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0 = 0<\/p>\n LHS = cos 38\u00b0 cos 52\u00b0\u00a0 sin 38\u00b0 sin 52\u00b0<\/p>\n = sin (90\u00b0 – 38\u00b0) cos 52\u00b0 \u2013 cos (90\u00b0 – 38\u00b0) sin 52\u00b0<\/p>\n = sin 52\u00b0 cos 52\u00b0 – cos 52\u00b0 sin 52\u00b0<\/p>\n = sin 52\u00b0 (cos 52\u00b0 – cos 52\u00b0)<\/p>\n = sin 52\u00b0 \u00d7 0<\/p>\n = 0<\/p>\n LHS = RHS<\/p>\n Ex 8.3 Class 10 \u0917\u0923\u093f\u0924\u00a0Q3.\u00a0 \u092f\u0926\u093f tan 2A = cot(A\u00a0<\/strong>–<\/strong>\u00a018<\/strong>\u00b0<\/strong>), \u091c\u0939\u093e\u0901 2A \u090f\u0915 \u0928\u094d\u092f\u0942\u0928\u0915\u094b\u0923 \u0939\u0948, \u0924\u094b A \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f |<\/strong><\/p>\n \u0939\u0932:\u00a0<\/strong>tan 2A = cot(A – 18\u00b0),<\/p>\n \u21d2\u00a0cot (90\u00b0 – 2A) = cot(A – 18\u00b0)<\/p>\n \u0926\u094b\u0928\u094b\u0902 \u092a\u0915\u094d\u0937\u094b\u0902 \u092e\u0947\u0902 \u0924\u0941\u0932\u0928\u093e \u0915\u0930\u0928\u0947 \u092a\u0930<\/p>\n \u21d2\u00a0 \u00a0 \u00a090\u00b0 – 2A = A – 18\u00b0<\/p>\n \u21d2\u00a0 \u00a090\u00b0 + 18\u00b0 = A + 2A<\/p>\n \u21d2 3A = 108\u00b0<\/p>\n <\/p>\n Ex 8.3 Class 10 \u0917\u0923\u093f\u0924\u00a0Q4. \u00a0\u092f\u0926\u093f tan A = cot B, \u0924\u094b \u0938\u093f\u0926\u094d\u0927 \u0915\u0940\u091c\u093f\u090f \u0915\u093f A + B = 90<\/strong>\u00b0<\/strong><\/p>\n \u0939\u0932:<\/strong>\u00a0tan A = cot B\u00a0\u00a0\u00a0 \u0926\u093f\u092f\u093e \u0939\u0948 |<\/p>\n \u21d2\u00a0 tan A = tan (90\u00b0 – B)\u00a0 \u0924\u0941\u0932\u0928\u093e \u0915\u0930\u0928\u0947 \u092a\u0930<\/p>\n \u21d2\u00a0\u00a0\u00a0\u00a0\u00a0 A = 90\u00b0 – B<\/p>\n \u21d2\u00a0 A + B = 90\u00b0 \u00a0Proved<\/p>\n Ex 8.3 Class 10 \u0917\u0923\u093f\u0924\u00a0Q5. \u092f\u0926\u093f sec 4A = cosec(A\u00a0<\/strong>–<\/strong>\u00a020<\/strong>\u00b0<\/strong>), \u091c\u0939\u093e\u0901 4A \u090f\u0915 \u0928\u094d\u092f\u0942\u0928\u0915\u094b\u0923 \u0939\u0948, \u0924\u094b A \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f |<\/strong><\/p>\n \u0939\u0932:<\/strong>\u00a0sec 4A = cosec(A – 20\u00b0)<\/p>\n \u21d2\u00a0cosec (90\u00b0 – 4A) = cosec(A – 20\u00b0)\u00a0\u00a0[ sec\u00a0<\/strong>q<\/strong>\u00a0= (90<\/strong>\u00b0<\/strong>–\u00a0<\/strong>q<\/strong>) ]<\/strong><\/p>\n \u0924\u0941\u0932\u0928\u093e \u0915\u0930\u0928\u0947 \u092a\u0930<\/p>\n \u21d2\u00a090\u00b0 – 4A = A – 20\u00b0<\/p>\n \u21d2\u00a090\u00b0 + 20\u00b0 = A + 4A<\/p>\n \u21d2\u00a0 \u00a05A = 110\u00b0<\/p>\n Ex 8.3 Class 10 \u0917\u0923\u093f\u0924\u00a0Q7. sin 67\u00b0 + cos 75\u00b0 \u0915\u094b 0\u00b0 \u0914\u0930 45\u00b0 \u0915\u0947 \u092c\u0940\u091a \u0915\u0947 \u0915\u094b\u0923\u094b\u0902 \u0915\u0947 \u0924\u094d\u0930\u093f\u0915\u094b\u0923\u092e\u093f\u0924\u093f\u092f \u0905\u0928\u0941\u092a\u093e\u0924\u094b\u0902 \u0915\u0947 \u092a\u0926\u094b\u0902 \u092e\u0947\u0902 \u0935\u094d\u092f\u0915\u094d\u0924 \u0915\u0940\u091c\u093f\u090f |<\/strong><\/p>\n
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