Tiêu đề 6. Trigonometric Ratios CQ & MCQ Practice Sheet Solution [HSC 26].pdf ✅

w·KvYwgwZK AbycvZ  CQ & MCQ Practice Sheet Solution (HSC 26) 1 06 w·KvYwgwZK AbycvZ Trigonometric Ratios WRITTEN 1|  O B A C e ̈vmva© OC = 6.5 cm I BC = 5 cm [iv. †ev. 19] (K) tanx = 3 4 , cosx abvZ¥K n‡j, sinx Gi gvb wbY©q Ki| (L) DÏxc‡Ki ABC Gi †ÿ‡Î cÖgvY Ki †h, sin2A – sin2B + sin2C = 0 (M) DÏxc‡Ki Qvqv‡Niv As‡ki †ÿÎdj wbY©q Ki| mgvavb: (K) †`Iqv Av‡Q, tanx = 3 4 , cosx abvZ¥K A_©vr, x †KvYwU cÖ_g PZzf©v‡M Aew ̄’Z| j¤^ I f~wg h_vμ‡g 3 I 4 GKK|  AwZfzR = 3 2 + 42 = 5  sinx = 3 5 (Ans.) 4 5 3 (L) wPÎ Abymv‡i, B = 90 [ Aa©e„Ë ̄’ †KvY]  A = 90 – C evgcÿ = sin2A – sin2B + sin2C = sin2 (90 – C) – sin2 90 + sin2C = cos2C – 1 2 + sin2C = cos2C + sin2C – 1 = 1 – 1 = 0 = Wvbcÿ  sin2A – sin2B + sin2C = 0 (Proved) (M) †`Iqv Av‡Q, e„ËwUi e ̈vmva© OC = 6.5 cm  AC = 2  6.5 cm = 13 cm Ges BC = 5 cm †h‡nZz ABC mg‡KvYx wÎfzR Ges B = 1 mg‡KvY|  AC2 = BC2 + AB2  AB = AC2 – BC2 = 132 – 5 2 cm = 12 cm  ABC wÎfz‡Ri †ÿÎdj = 1 2  BC  AB = 1 2  5  12 = 30 cm2 Avevi, ABC Aa©e„‡Ëi †ÿÎdj = 1 2  3.1416  (6.5)2 cm2 = 66.366 cm2  Qvqv‡Niv As‡ki †ÿÎdj = (66.366 – 30) cm2 = 36.37 cm2 (Ans.) 2| `„k ̈Kí-1: f(x) = sinx `„k ̈Kí-2: OABC GKwU i¤^m Ges OAC e„ËKjv O †Kw›`aK e„‡Ëi Ask| C A B O 40 cm 45 [w`. †ev. 19] (K) cosecx + 16 sinx = 8 n‡j, cÖgvY Ki †h, sinn x + cosecn x = 22n + 2–2n (L) `„k ̈Kí-2 n‡Z ABCA As‡ki †ÿÎdj wbY©q Ki| (M) `„k ̈Kí-1 n‡Z y = f      2 – 2x ; –   x   Gi †jLwPÎ A1⁄4b Ki| mgvavb: (K) †`Iqv Av‡Q, sinx + 16cosecx = 8  sinx + 16 sinx = 8  sin2 x + 16 = 8sinx  sin2 x – 2.4.sinx + 42 = 0  (sinx – 4)2 = 0  sinx = 4 GLb sinn x + cosecn x = sinn x + 1 sinn x = 4n + 1 4 n = 22n + 2–2n (Proved) (L) C A B O 40 cm 45
2  Higher Math 1st Paper Chapter-6 GLv‡b, AOC = ABC = 45 OAC Gi †ÿÎdj = 1 2 OA.OCsin45 = 1 2  402  1 2 = 565.685 eM© †m.wg.  OABC i¤^‡mi †ÿÎdj = 2  OAC = 2  565.685 = 1131.37 eM© †m.wg. OAC e„ËKjvi †ÿÎdj wbY©q:  = 45 = 45   180 =  4 †iwWqvb Ges r = 40 †m.wg.  e„ËKjvi †ÿÎdj = 1 2 r 2  = 1 2 (40)2 .  4 = 200 eM© GKK  Qvqv‡Niv As‡ki †ÿÎdj = (1131.37 – 200) = 503.05 eM© GKK (cÖvq) (Ans.) (M) †`Iqv Av‡Q, f(x) = sinx  y = f     2 – 2x = sin     2 – 2x = cos2x  y = cos2x; –  ≤ x ≤  x Gi wewfbœ gv‡bi Rb ̈ y = cos2x Gi gvb `yB `kwgK ̄’vb ch©šÍ wbY©q Kwi: x 0   12   6   4   3  5 12   2  7 12 y 1 0.87 0.5 0 – 0.5 – 0.87 – 1 – 0.87 x  2 3  3 4  5 6  11 12   y – 0.5 0 0.5 0.87 1 † ̄‹j wba©viY: x Aÿ eivei ÿz`aZg e‡M©i 2 evû =  12 y Aÿ eivei ÿz`aZg e‡M©i 10 evû = 1 0 –  6 –  3 –  2 – 2 3 – 5 6 –   6  3  2 2 3 5 6  1 0.5 0 – 0.5 – 1 we›`y ̧‡jv‡K QK KvM‡R ̄’vcb Kwi Ges †hvM Kwi| Gfv‡eB y = cos2x Gi †jLwPÎ cÖ`Ë mxgvq Aw1⁄4Z nj| •ewkó ̈: (i) †jLwPÎwU g~j we›`yMvgx bq| (ii) †jLwPÎwU Awew”Qbœ Ges †XD AvK...wZi (iii) †jLwPÎ †_‡K †`Lv hvq †h, cos2x Gi m‡e©v”P gvb 1 Ges me©wb¤œ gvb – 1 3| `„k ̈Kí-1: A K‡jR †_‡K GKRb mvB‡Kj Av‡ivnx B K‡j‡R †cuŠQvj| K‡jR `yBwU c„w_exi †K‡›`a 18 †KvY Drcbœ K‡i| c„w_exi e ̈vmva© 6440 km. `„k ̈Kí-2: f(x) = sinx [h. †ev. 19] (K) cot 3x 2 dvsk‡bi ch©vq wbY©q Ki| (L) K‡jR `yBwUi ga ̈eZ©x `~iZ¡ wbY©q Ki| (M) y = f(2x) ; –   x   Gi †jLwPÎ A1⁄4b Ki| mgvavb: (K) cot 3x 2 dvsk‡bi ch©vq =  |Pj‡Ki mnM| =  3 2 = 2 3 (Ans.) (L) †K‡›`a Drcbœ †KvY,  = 18 =     18 60   180 †iwWqvb =  600 †iwWqvb e ̈vmva©, r = 6440 wK‡jvwgUvi  K‡jR `yBwUi g‡a ̈eZ©x `~iZ¡, S = r = 6440   600 = 33.72 (cÖvq)  wb‡Y©q `~iZ¡ 33.72 wK‡jvwgUvi| (Ans.) (M) GLv‡b, f(x) = sinx  y = f(2x) = sin2x; –  ≤ x ≤  x –  – 11 12 – 5 6 – 3 4 – 2 3 – 7 12 –  2 – 5 12 y 0 0.5 0.87 1 0.87 0.5 0 – 0.5 x –  3 –  4 –  6 –  12 0  12  6  4 y – 0.87 – 1 – 0.87 – 0.5 0 0.5 0.87 1 x  3 5 12  2 7 12 2 3 3 4 5 6 11 12 y 0.87 0.5 0 – 0.5 – 0.87 – 1 – 0.87 – 0.5 x  y 0 † ̄‹j wba©viY: x Aÿ eivei ÿz`aZg e‡M©i 2 evû =  12 y Aÿ eivei ÿz`aZg e‡M©i 10 evû = 1 0 –  6 –  3 –  2 – 2 3 – 5 6 –   6  3  2 2 3 5 6  1 0.5 0 – 0.5 – 1
w·KvYwgwZK AbycvZ  CQ & MCQ Practice Sheet Solution (HSC 26) 3 •ewkó ̈: (i) †jLwPÎwU Aew”Qbœ eμ‡iLv Ges AvK...wZ †XD Gi g‡Zv| (ii) †jLwPÎwU g~jwe›`yMvgx| (iii) sin2x Gi m‡e©v”P gvb 1 Ges me©wb¤œ gvb – 1 4| `„k ̈Kí-1: `„k ̈Kí-2:  = 58 O B A  f(x) = cotx [e. †ev. 19] (K) cos2A = 1 – cos4A n‡j, †`LvI †h, cot4A – cot2A = 1 (L) `„k ̈Kí-1 G cÖ`wk©Z e„‡Ëi cwiwa 31.416 †m.wg. n‡j e„ËwUi Qvqv‡Niv As‡ki †ÿÎdj wbY©q Ki| (M) `„k ̈Kí-2 n‡Z f      2 – x = y Gi †jLwPÎ AvuK, hLb –  2 < x <  2 mgvavb: (K) †`Iqv Av‡Q, cos2A = 1 – cos4A  cos4A = 1 – cos2A  cos3A = sin2A  cot4A = 1 sin2A [sin4A Øviv fvM K‡i]  cot4A = cosec2A  cot4A = 1 + cot2A [cosec2A – cot2A = 1]  cot4A – cot2A = 1 (Showed) (L) †`Iqv Av‡Q, e„‡Ëi cwiwa = 31.416 †m.wg. e„‡Ëi e ̈vmva© r n‡j, 2r = 31.416  r = 31.416 2 †m.wg. = 5 †m.wg. GLb AOB e„ËKjvi †ÿÎdj =  360  r 2 = 58 360 r 2 Ges m¤ú~Y© e„‡Ëi †ÿÎdj = r 2  e„ËwUi Qvqv‡Niv As‡ki †ÿÎdj = r 2 –  360 r 2 eM© †m.wg. = r 2     1 –  360 eM© †m.wg. = 3.1416  5 2     1 – 58 360 eM© †m.wg. = 3.1416  25     360 – 58 360 eM© †m.wg. = 3.1416  25  151 180 eM© †m.wg. = 65.886 eM© †m.wg. (Ans.) (M) †`Iqv Av‡Q, f(x) = cotx  y = f      2 – x = cot     2 – x = tanx  y = tanx; –  2 < x <  2 x Gi wewfbœ gv‡bi Rb ̈ y = tanx Gi gvb `yB `kwgK ̄’vb ch©šÍ wbY©q Kwi: x –  2 – 7 18 –  3 –  4 –  6 0  6  4 y –  – 2.75 – 1.73 – 1 – 0.58 0 0.58 1 x  3 7 18  2 y 1.73 2.75  x Aÿ eivei ÿz`aZg e‡M©i 3 evû =  6 Ges y Aÿ eivei ÿz`aZg e‡M©i 3 evû = 1 GKK a‡i †jLwPÎwU A1⁄4b Kwi| Y Y X X O  2 –  2 5| `„k ̈Kí-1: f() = 1 + cos2  cos `„k ̈Kí-2: g(x) = cosx [iv., Kz., P., e. †ev. 18] (K) sin2 – cos = 0 n‡j cÖgvY Ki †h, tan4 – sec2 = 0 (L) `„k ̈Kí-1 G f() = 5 2 n‡j †`LvI †h, cosn  + secn  = 2n + 2–n (M) –  2  x   2 e ̈ewa‡Z g(3x) Gi †jLwPÎ A1⁄4b Ki| mgvavb: (K) sin2 – cos = 0  sin2 = cos  sin4 = cos2  sin4 cos4 = 1 cos2  tan4 = sec2  tan4 – sec2 = 0 (Proved) (L) †`Iqv Av‡Q, f() = 5 2  1 + cos2  cos = 5 2  2cos2  + 2 = 5cos  2cos2  – 5cos + 2 = 0  2cos2  – 4cos – cos + 2 = 0  (cos – 2)(2cos – 1) = 0  cos – 2 = 0  cos = 2 (MÖnY‡hvM ̈ b‡n) A_ev, 2cos – 1 = 0  cos = 1 2  sec = 2 cosn  + secn  =     1 2 n + 2n = 2n + 2–n (Showed)