Tiêu đề 6. Trigonometric Ratios Engineering Practice Sheet Solution [HSC 26].pdf ✅

w·KvYwgwZK AbycvZ  Engineering Practice Sheet Solution (HSC 26) 1 06 w·KvYwgwZK AbycvZ Trigonometric Ratios WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvewj 1| O †K›`a wewkó GKwU e„‡Ëi e ̈vmva© 10 cm Ges AB Pv‡ci •`N© ̈ 14 cm| †KvY AOB Gi gvb Ges Pvc AB I R ̈v AB Øviv Ave× ÿz`aZg †ÿ‡Îi †ÿÎdj wbY©q Ki| [BUET 17-18] mgvavb: †`Iqv Av‡Q, e„‡Ëi e ̈vmva©, r = 10 cm Ges AB Pv‡ci •`N© ̈, s = 14 cm O r  S A B Avgiv Rvwb, s = r   = s r = 14 10 = 1.4c (radian) Pvc AB I R ̈v AB Øviv Ave× †ÿ‡Îi †ÿÎdj = AOB Gi †ÿÎdj = 1 2 r 2  – 1 2 r.rsin = 1 2 r 2 ( – sin) = 1 2  (10)2  (1.4 – sin1.4c ) = 20.7275 eM© †m.wg. (Ans.) 2| hw` cossec  5 2 nq, Zvn‡j cosn  secn Gi gvb wbY©q Ki| [BUET 17-18] mgvavb: cos  sec  5 2  cos 1 cos  5 2  cos2 + 1 cos  5 2  2cos2 + 2  5cos  2cos2 – 5cos  2  0  2cos2 – 4cos– cos + 2 = 0  (2cos –1)(cos – 2)  0  cos  1 2 [∵ cos  2]  sec  2  cosn  secn      1 2 n  2 n  2 –n + 2n (Ans.) 3| mZ ̈/wg_ ̈v DËi `vI: [BUET 95-96] (i) cos2  4 = 1 – sin2  4 (ii) The equation sec = 1 2 has no solution (iii) A triangle can be drawn with sides a = 4, b = 10 and < A = 30 (iv) cos (– 390) = 3 2 mgvavb: (i) mZ ̈ (ii) mZ ̈ (iii) wg_ ̈v (iv) mZ ̈ weMZ mv‡j BUTex-G Avmv cÖkœvewj 4| f(x) = sinxtan2x dvskbwUi ch©vq wbY©q Ki| [BUTex 18-19] mgvavb: f(x) = sinxtan2x  f(2 + x) = sin (2 + x) tan2(2 + x) = sin(2 + x) tan (4 + 2x) = sinxtan2x Avevi, f(4 + x) = sin(4 + x) tan2(4 + x) = sin(4 + x) tan(8 + 2x) = sinxtan2x ............................................................  sinx tan2x Gi ch©vq = 2 MCQ weMZ mv‡j BUET-G Avmv cÖkœvewj 1. 21 cm e ̈vmva© wewkó GKwU e„ËKvi cvZ n‡Z 120 cwigvY e„ËKjv wb‡q e ̈vmva©Øq msjMœ K‡i †KvYK •Zwi Kiv n‡jv| †KvYKwUi f~wgi e ̈vmva© KZ n‡e? [BUET 13-14] 14 cm 21 cm 7 cm 49 cm DËi: 7 cm e ̈vL ̈v: 2r = s = r  r = r 2 = r 120 180 2 = 21  120 180  2 = 7 cm