{"id":170704,"date":"2024-06-03T16:55:34","date_gmt":"2024-06-03T11:25:34","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170704"},"modified":"2024-06-03T16:56:35","modified_gmt":"2024-06-03T11:26:35","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-3-ex-3-9","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-3-ex-3-9\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9"},"content":{"rendered":"

Access to comprehensive Class 11 ISC Maths Solutions<\/a> Chapter 3 Trigonometry Ex 3.9 encourages independent learning.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9<\/h2>\n

Very short answer\/objective questions (1 to 4) :<\/span><\/p>\n

Question 1.
\nIn a \u2206ABC, if a = 4, b = 5 and c = 6, find cos C.
\nSolution:
\nGiven a = 4, b = 5 and c = 6 using cosine formula, we have
\ncos C = \\(\\frac{a^2+b^2-c^2}{2 a b}\\)
\n= \\(\\frac{4^2+5^2-6^2}{2 \\times 4 \\times 5}\\)
\n= \\(\\frac{16+25-36}{40}\\)
\n= \\(\\frac{5}{40}=\\frac{1}{8}\\)<\/p>\n

Question 2.
\nIn a \u2206ABC, if a = 7, c = 8 and \u2220B = 45\u00b0, find area of \u2206ABC.
\nSolution:
\nGiven a = 7 ; c = 8 ; \u2220B = 45\u00b0
\n\u2234 area of \u2206ABC = \\(\\frac{1}{2}\\) \u00d7 a \u00d7 c
\n= \\(\\frac{1}{2}\\) \u00d7 7 \u00d7 8 \u00d7 sin 45\u00b0
\n= \\(\\frac{28}{\\sqrt{2}}\\)
\n= 14\u221a2 sq. units<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 3.
\nIn a \u2206ABC, if a = 4, b = 6 and c = 8, find the value of 4 cos B + 3 cos C.
\nSolution:
\nGiven a = 4, b = 6, c = 8
\n\u2234 cos B = \\(\\frac{a^2+c^2-b^2}{2 a c}\\)
\n= \\(\\frac{4^2+8^2-6^2}{2 \\times 4 \\times 8}\\)
\n= \\(\\frac{80-36}{64}\\)
\n= \\(\\frac{44}{64}=\\frac{11}{16}\\)
\nand cos C = \\(\\frac{a^2+b^2-c^2}{2 a b}\\)
\n= \\(\\frac{4^2+6^2-8^2}{2 \\times 4 \\times 6}\\)
\n= \\(-\\frac{12}{48}=-\\frac{1}{4}\\)
\nThus 4 cos B + 3 cos C = \\(4 \\times \\frac{11}{16}+3\\left(-\\frac{1}{4}\\right)\\)
\n= \\(\\frac{11-3}{4}\\)
\n= 2<\/p>\n

Question 4.
\nIn a \u2206ABC, if a = 3, b = 5 and c = 6, then verify that c = a cos B + b cos A.
\nSolution:
\nGiven a = 3, b = 5 and c = 6
\nR.H.S = a cos B + b cos A
\n= \\(a\\left(\\frac{a^2+c^2-b^2}{2 a c}\\right)+b\\left(\\frac{b^2+c^2-a^2}{2 b c}\\right)\\)
\n= \\(\\frac{1}{2 c}\\) [a2<\/sup> + c2<\/sup> – b2<\/sup> + b2<\/sup> + c2<\/sup> – a2<\/sup>]
\n= \\(\\frac{2 c^2}{2 c}\\)
\n= c = L.H.S<\/p>\n

\"ML<\/p>\n

Short and long answer questions (5 to 25) :<\/span><\/p>\n

Question 5.
\n2(b ccos A + ca cos B + ab cos C) = a2<\/sup> + b2<\/sup> + c2<\/sup>.
\nSolution:
\nLH.S. = 2 (bc cos A + ca cos B + ah cos C)
\n= 2 \\(\\left[b c\\left(\\frac{b^2+c^2-a^2}{2 b c}\\right)+c a\\left(\\frac{c^2+a^2-b^2}{2 c a}\\right)+a b\\left(\\frac{a^2+b^2-c^2}{2 a b}\\right)\\right]\\)
\n[using cosine’s formula]
\n= \\(\\frac{2}{2}\\) [b2<\/sup> + c2<\/sup> – a2<\/sup> – c2<\/sup> + a2<\/sup> – b2<\/sup> + a2<\/sup> + b2<\/sup> – c2<\/sup>]
\n= a2<\/sup> + b2<\/sup> + c2<\/sup>
\n= R.H.S.<\/p>\n

Question 6.
\n\\(\\frac{\\cos \\mathrm{A}}{a}+\\frac{\\cos \\mathrm{B}}{b}+\\frac{\\cos \\mathrm{C}}{c}=\\frac{a^2+b^2+c^2}{2 a b c}\\)
\nSolution:
\nL.H.S.<\/p>\n

\"ML<\/p>\n

Question 7.
\n(a + b) sin \\(\\frac{C}{2}\\) = c cos \\(\\frac{A-B}{2}\\)
\nSolution:
\nIn any \u2206ABC, using sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\) = k (say)
\n\u21d2 a = k sin A ;
\nb = k sin B
\nand c = k sin C<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 8.
\n(a – b) cos \\(\\frac{C}{2}\\) = c sin \\(\\frac{A-B}{2}\\)
\nSolution:
\nIn any \u2206ABC, using sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\) = k (say)
\n\u21d2 a = k sin A ;
\nb = k sin B
\nand c = k sin C<\/p>\n

\"ML<\/p>\n

Question 9.
\n(c – b)2<\/sup> cos2<\/sup> \\(\\frac{A}{2}\\) + (c + b)2<\/sup> sin2<\/sup> \\(\\frac{A}{2}\\) = a2<\/sup>.
\nSolution:
\nL.H.S. = (c – b)2<\/sup> cos2<\/sup> \\(\\frac{A}{2}\\) + (c + b)2<\/sup> sin2<\/sup> \\(\\frac{A}{2}\\)
\n= (b2<\/sup> + c2<\/sup>) \\(\\left(\\cos ^2 \\frac{\\mathrm{A}}{2}+\\sin ^2 \\frac{\\mathrm{A}}{2}\\right)\\) – 2 bc \\(\\left(\\cos ^2 \\frac{\\mathrm{A}}{2}-\\sin ^2 \\frac{\\mathrm{A}}{2}\\right)\\)
\n= (b2<\/sup> + c2<\/sup>) \u00d7 1 – 2bc cos A
\n= (b2<\/sup> + c2<\/sup>) – 2bc \\(\\left(\\frac{b^2+c^2-a^2}{2 b c}\\right)\\)
\n= b2<\/sup> + c2<\/sup> – b2<\/sup> – c2<\/sup> + a2<\/sup>
\n= a2<\/sup>
\n= R.H.S.<\/p>\n

Question 10.
\n\\(\\frac{a^2 \\sin (B-C)}{\\sin B+\\sin C}+\\frac{b^2 \\sin (C-A)}{\\sin C+\\sin A}+\\frac{c^2 \\sin (A-B)}{\\sin A+\\sin B}\\) = 0
\nSolution:
\nL.H.S. = \\(\\frac{a^2 \\sin (B-C)}{\\sin B+\\sin C}+\\frac{b^2 \\sin (C-A)}{\\sin C+\\sin A}+\\frac{c^2 \\sin (A-B)}{\\sin A+\\sin B}\\)<\/p>\n

\"ML<\/p>\n

= k2<\/sup> sin A (sin B – sin C) + k2<\/sup> sin B (sin C – sin A) + k2<\/sup> sin C (sin A – sin B)
\n= k2<\/sup> [sin A sin B – sin A sin C + sin B sin C – sin A sin B + sin C sin A – sin B sin C]
\n= k2<\/sup> \u00d7 0
\n= 0 = R.H.S.<\/p>\n

\"ML<\/p>\n

Question 11.
\n\\(\\frac{a^2-b^2}{\\cos A+\\cos B}+\\frac{b^2-c^2}{\\cos B+\\cos C}+\\frac{c^2-a^2}{\\cos C+\\cos A}\\) = 0.
\nSolution:
\nL.H.S. = \\(\\frac{a^2-b^2}{\\cos A+\\cos B}+\\frac{b^2-c^2}{\\cos B+\\cos C}+\\frac{c^2-a^2}{\\cos C+\\cos A}\\)
\nIn any \u2206 ABC, bby sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\) = k (say)
\n\u21d2 a = k sin A ;
\nb = k sin B
\nand c = k sin C
\n\u2234 L.H.S. = \\(\\frac{k^2\\left[\\sin ^2 \\mathrm{~A}-\\sin ^2 \\mathrm{~B}\\right]}{\\cos \\mathrm{A}+\\cos \\mathrm{B}}+\\frac{k^2\\left[\\sin ^2 \\mathrm{~B}-\\sin ^2 \\mathrm{C}\\right]}{\\cos \\mathrm{B}+\\cos \\mathrm{C}}+\\frac{k^2\\left[\\sin ^2 \\mathrm{C}-\\sin ^2 \\mathrm{~A}\\right]}{\\cos \\mathrm{C}+\\cos \\mathrm{A}}\\)
\n= \\(\\frac{k^2\\left[\\cos ^2 \\mathrm{~B}-\\cos ^2 \\mathrm{~A}\\right]}{\\cos \\mathrm{B}+\\cos \\mathrm{A}}+\\frac{k^2\\left[\\cos ^2 \\mathrm{C}-\\cos ^2 \\mathrm{~B}\\right]}{\\cos \\mathrm{C}+\\cos \\mathrm{B}}+\\frac{k^2\\left[\\cos ^2 \\mathrm{~A}-\\cos ^2 \\mathrm{C}\\right]}{\\cos \\mathrm{A}+\\cos \\mathrm{C}}\\)
\n[\u2235 sin2<\/sup> \u03b8 + cos2<\/sup> \u03b8 = 1
\nsin2<\/sup> \u03b8 = 1 – cos2<\/sup> \u03b8]
\n= k2<\/sup> [cos B – cos A + cos C – cos B + cos A – cos C]
\n= k2<\/sup> \u00d7 0
\n= 0 = R.H.S.<\/p>\n

Question 12.
\nb2<\/sup> sin 2C + c2<\/sup> sin 2B = 2bc sin A
\nSolution:
\nL.H.S. = b2<\/sup> sin 2C + c2<\/sup> sin 2B
\n= 2b2<\/sup> sin C cos C + 2c2<\/sup> sin B cos B
\nBy sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\) = k
\n\u21d2 a = k sin A ;
\nb = k sin B
\nand c = k sin C
\n\u2234 L.H.S. = 2b2<\/sup> \u00d7 \\(\\frac{c}{k}\\left(\\frac{b^2+a^2-c^2}{2 a b}\\right)\\) + 2c2<\/sup> \u00d7 \\(\\frac{b}{k} \\frac{\\left(c^2+a^2-b^2\\right)}{2 a c}\\)
\n= \\(\\frac{b c}{a k}\\) (b2<\/sup> + a2<\/sup> – c2<\/sup>) + \\(\\frac{b c}{a k}\\) (c2<\/sup> + a2<\/sup> – b2<\/sup>)
\n= \\(\\frac{b c}{a k}\\) [b2<\/sup> + a2<\/sup> – c2<\/sup> + c2<\/sup> + a2<\/sup> – b2<\/sup>]
\n= \\(\\frac{b c}{a k}\\) \u00d7 2a2<\/sup>
\n= \\(\\frac{2 a b c}{k}\\)
\n= 2bc \\(\\left(\\frac{a}{k}\\right)\\)
\n= 2bc sin A [using sine formula]
\n= R.H.S.<\/p>\n

Question 13.
\n\\(\\frac{b^2-c^2}{a} \\cos \\mathrm{A}+\\frac{c^2-a^2}{b} \\cos \\mathrm{B}+\\frac{a^2-b^2}{c} \\cos \\mathrm{C}\\) = 0
\nSolution:
\nL.H.S. = \\(\\frac{b^2-c^2}{a} \\cos \\mathrm{A}+\\frac{c^2-a^2}{b} \\cos \\mathrm{B}+\\frac{a^2-b^2}{c} \\cos \\mathrm{C}\\)
\n= \\(\\frac{b^2-c^2}{a}\\left(\\frac{b^2+c^2-a^2}{2 b c}\\right)+\\frac{c^2-a^2}{b}\\left(\\frac{c^2+a^2-b^2}{2 a c}\\right)+\\frac{a^2-b^2}{c}\\left(\\frac{a^2+b^2-c^2}{2 a b}\\right)\\)
\n[using cosine’s formula]
\n= \\(\\frac{1}{2 a b c}\\) [(b2<\/sup> – c2<\/sup>) (b2<\/sup> + c2<\/sup> – a2<\/sup>) + (c2<\/sup> – a2<\/sup>) (c2<\/sup> + a2<\/sup> – b2<\/sup>) + (a2<\/sup> – b2<\/sup>) (a2<\/sup> + b2<\/sup> – c2<\/sup>)]
\n= \\(\\frac{1}{2 a b c}\\) [b4<\/sup> – c4<\/sup> – a2<\/sup> (b2<\/sup> – c2<\/sup>) + c4<\/sup> – a4<\/sup> – b2<\/sup> (c2<\/sup> – a2<\/sup>) + a4<\/sup> – b4<\/sup> – c2<\/sup> (a2<\/sup> – b2<\/sup>)]
\n= \\(\\frac{1}{2 a b c}\\) [- a2<\/sup>b2<\/sup> + a2<\/sup>c2<\/sup> – b2<\/sup>c2<\/sup> + a2<\/sup>b2<\/sup> – c2<\/sup>a2<\/sup> + b2<\/sup>c2<\/sup>]
\n= \\(\\frac{1}{2 a b c}\\) \u00d7 0
\n= 0 = R.H.S.<\/p>\n

\"ML<\/p>\n

Question 14.
\nb2<\/sup> cos 2A – a2<\/sup> cos 2B = b2<\/sup> – a2<\/sup>.
\nSolution:
\nUsing sine formula
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\) = k
\n\u21d2 a = k sin A ;
\nb = k sin B
\nand c = k sin C
\nL.H.S. = b2<\/sup> cos 2A – a2<\/sup> cos 2B
\n= b2<\/sup> (1 – 2 sin2<\/sup> A) – a2<\/sup> (1 – 2 sin2<\/sup> B)
\n= b2<\/sup> – a2<\/sup> – 2b2<\/sup> \\(\\left(\\frac{a}{k}\\right)^2\\) + 2a2<\/sup> \\(\\left(\\frac{b}{k}\\right)^2\\)
\n= b2<\/sup> – a2<\/sup> – \\(\\frac{2 b^2 a^2}{k^2}+\\frac{2 a^2 b^2}{k^2}\\)
\n= b2<\/sup> – a2<\/sup>
\n= R.H.S.<\/p>\n

Question 15.
\n(c2<\/sup> – a2<\/sup> + b2<\/sup>) tan A = (a2<\/sup> – b2<\/sup> + c2<\/sup>) tan B = (b2<\/sup> – c2<\/sup> + a2<\/sup>) tan C.
\nSolution:
\nUsing sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\) = k
\n\u21d2 a = k sin A ;
\nb = k sin B
\nand c = k sin C
\n\u21d2 a = k sin A ;
\nb = k sin B
\nand c = k sin C
\nNow (c2<\/sup> – a2<\/sup> + b2<\/sup>) tan A<\/p>\n

\"ML<\/p>\n

Thus,
\n(c2<\/sup> – a2<\/sup> + b2<\/sup>) tan A = (a2<\/sup> – b2<\/sup> + c2<\/sup>) tan B = (b2<\/sup> – c2<\/sup> + a2<\/sup>) tan C
\n[using (1), (2) and (3)].<\/p>\n

Question 16.
\n2 [a sin2<\/sup> \\(\\frac{C}{2}\\) + c sin2<\/sup> \\(\\frac{A}{2}\\)] = c + a – b
\nSolution:
\nLH.S. = \\(\\left[a \\sin ^2 \\frac{\\mathrm{C}}{2}+c \\sin ^2 \\frac{\\mathrm{A}}{2}\\right]\\)
\n= 2 \\(\\left[a\\left(\\frac{1-\\cos \\mathrm{C}}{2}\\right)+c\\left(\\frac{1-\\cos \\mathrm{A}}{2}\\right)\\right]\\)
\n= a + c – a cos C – c cos A
\n= a + c – b [using projection formula)
\n= R.H.S.<\/p>\n

\"ML<\/p>\n

Question 17.
\nIf in a triangle ABC, sin 2A + sin 2B = sin 2C, prove that either A = 90\u00b0 or B = 90\u00b0.
\nSolution:
\nGiven sin 2A + sin 2B = sin 2C
\n\u21d2 2 sin (A + B) cos (A – B) = 2 sin C cos C
\n\u21d2 sin [\u03c0 – C] cos (A – B) = sin C cos C
\n[\u2235 A + B + C = \u03c0
\n\u21d2 A + B = \u03c0 – C]
\n\u21d2 sin C [cos (A – B) – cos C] = 0
\n\u21d2 sin C [cos (A – B) – cos (\u03c0 – \\(\\overline{A+B}\\))] = 0
\n\u21d2 sin C [cos(A – B) cos (A + B)] = 0
\n\u21d2 sin C [2 cos A cos B] = 0
\neither sin C = 0 or cos A = 0 or cos B = 0
\nC = 0 or A = 90\u00b0 or B = 90\u00b0
\nsince A, B and C are the angles of \u2206ABC
\n\u2234 neither of the angles A, B and C is 0.
\nHence either A = 90\u00b0 or B = 90\u00b0.<\/p>\n

Question 18.
\nIf in a triangle ABC, sin2<\/sup> A + sin2<\/sup> B = sin2<\/sup> C. prove that \u00bfABC is right angled.
\nSolution:
\nGiven sin2<\/sup> A + sin2<\/sup> B = sin2<\/sup> C ………………..(1)
\nusing sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\) = k
\n\u21d2 a = k sin A ;
\nb = k sin B
\nand c = k sin C
\nThus from (1) ; we have
\n\\(\\frac{a^2}{k^2}+\\frac{b^2}{k^2}=\\frac{c^2}{k^2}\\)
\n\u21d2 a2<\/sup> + b2<\/sup> = c2<\/sup>
\nSo the triangles ABC is right angled \u2206 at C.
\nsince pythagoras theorem holds.<\/p>\n

\"ML<\/p>\n

Question 19.
\nIf in a triangle ABC, \\(\\frac{\\cos \\mathrm{A}}{a}=\\frac{\\cos \\mathrm{B}}{b}\\), show that the triangle is isosceles.
\nSolution:
\nGiven, \\(\\frac{\\cos \\mathrm{A}}{a}=\\frac{\\cos \\mathrm{B}}{b}\\)
\n\u21d2 b cos A = a cos B ………………..(1)
\nIn any ABC, using sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\) = k
\n\u2234 from (1) ;
\nk sin B cos A = k sin A cos B
\n\u21d2 2 sin B cosA = 2 sin A cos B
\n\u21d2 sin (A + B) – sin (A – B) = sin (A + B) + sin (A – B)
\n\u21d2 2 sin (A – B) = 0
\n\u21d2 A – B = 0
\n\u21d2 A = B
\nHence the given triangle is isosceless.<\/p>\n

\"ML<\/p>\n

Question 20.
\nIn triangle ABC, if a = 18, b = 24 and c = 30 then find
\n(i) cos A, cos B, cos C
\n(ii) sin A sin B, sin C.
\nSolution:
\n(i) Given a = 18, b = 24 ; c = 30
\nUsing cosine formula, we have<\/p>\n

\"ML<\/p>\n

(ii) Using sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\)
\n\u21d2 \\(\\frac{18}{\\sin A}=\\frac{24}{\\sin B}=\\frac{30}{\\sin \\frac{\\pi}{2}}\\)
\n\u21d2 sin A = \\(=\\frac{18}{30}=\\frac{3}{5}\\) ;
\nsin B = \\(\\frac{24}{30}=\\frac{4}{5}\\)
\nand sin C = 1<\/p>\n

Question 21.
\nIf a = 7 cm, b = 5 cm and c = 3 cm, prove that the triangle ha 14ij angle.
\nSolution:
\nGiven a = 7 cm ;
\nb = 5 cm
\nand c = 3 cm
\n\u2234 cos A = \\(\\frac{b^2+c^2-a^2}{2 b c}\\)
\n= \\(\\frac{5^2+3^2-7^2}{2 \\times 5 \\times 3}\\)
\n= \\(\\frac{34-49}{30}\\)
\n= \\(-\\frac{15}{30}=-\\frac{1}{2}\\)
\ncos A = – cos \\(\\frac{\\pi}{3}\\)
\n= cos (\u03c0 – \\(\\frac{\\pi}{3}\\))
\n= cos \\(\\frac{2 \\pi}{3}\\)
\n\u21d2 A = \\(\\frac{2 \\pi}{3}\\)
\nClearly one of the angle of given \u2206 be 120\u00b0 > 90\u00b0.
\nHence one of the angle ofi be obtuse angle.<\/p>\n

\"ML<\/p>\n

Question 22.
\nIf the angles of a triangle are in the ratio 1 : 2 : 3, prove that the corresponding sides of the triangle are in ratio 1 : \u221a3 : 2.
\nSolution:
\nLet the angles of triangle be x\u00b0, 2x\u00b0 and 3x\u00b0
\nsince, x\u00b0 + 2x\u00b0 + 3x\u00b0 = 180\u00b0
\n\u21d2 6x\u00b0 = 180\u00b0
\n\u21d2 x = 330\u00b0
\nHence the angles of a triangle are 30\u00b0, 60\u00b0 and 90\u00b0.
\nUsing sine formula, we have
\n\\(\\frac{a}{\\sin \\mathrm{A}}=\\frac{b}{\\sin \\mathrm{B}}=\\frac{c}{\\sin \\mathrm{C}}\\)
\n\u21d2 \\(\\frac{a}{\\sin 30^{\\circ}}=\\frac{b}{\\sin 60^{\\circ}}=\\frac{c}{\\sin 90^{\\circ}}\\)
\n\u21d2 \\(\\frac{a}{\\frac{1}{2}}=\\frac{b}{\\frac{\\sqrt{3}}{2}}=\\frac{c}{1}\\)
\n\u21d2 \\(\\frac{a}{1}=\\frac{\\dot{b}}{\\sqrt{3}}=\\frac{c}{2}\\)
\nHence a : b : c :: 1 : \u221a3 : 2
\nThus, the corresponding sides of the triangle are in the ratio 1 : \u221a3 : 2.<\/p>\n

Question 23.
\nIn a ABC, a = 1, b = \u221a3 and C = \\(\\frac{\\pi}{6}\\). Find the other two angles and the third side.
\nSolution:
\nGiven a = 1 ;
\nb = \u221a3
\nand C = \\(\\frac{\\pi}{6}\\) or 30\u00b0
\nIn only \u2206ABC, using sine formula, we have<\/p>\n

\"ML<\/p>\n

\u21d2 3 = 4 – c2<\/sup>
\n\u21d2 c2<\/sup> = 1
\n\u21d2 c = 1 [\u2235 c > 0]
\n\u2234 from (1) ;
\n\\(\\frac{1}{\\sin A}=\\frac{\\sqrt{3}}{\\sin B}=\\frac{1}{\\frac{1}{2}}\\)
\n\u21d2 sin A = \\(\\frac{1}{2}\\)
\n\u21d2 A = \\(\\frac{\\pi}{6}\\)
\nSince A. B and C are the angles of \u2206ABC.
\n\u2234 A + B + C = 180\u00b0
\n\u21d2 B = 180\u00b0 – 30\u00b0 – 30\u00b0 = 120\u00b0<\/p>\n

Question 24.
\nTwo boats leave a place at the same time. One travels 56 km in the direction N 40\u00b0 E, while the other travels 48 km in the direction S 80\u00b0 E. What is the distance between the boats?
\nSolution:
\nClearly \u2220POQ = 180\u00b0 – 40\u00b0 – 80\u00b0 = 60\u00b0
\nusing cosine\u2019s formula, we have
\ncos 60\u00b0 = \\(\\frac{\\mathrm{OP}^2+\\mathrm{OQ}^2-\\mathrm{PQ}^2}{2 \\mathrm{OP} \\cdot \\mathrm{OQ}}\\)
\n\u21d2 \\(\\frac{1}{2}=\\frac{56^2+48^2-\\mathrm{PQ}^2}{2 \\times 56 \\times 48}\\)<\/p>\n

\"ML<\/p>\n

\u21d2 56 \u00d7 48 = 562<\/sup> + 482<\/sup> – PQ2<\/sup>
\n\u21d2 PQ2<\/sup> = 3136 + 2304 – 2688 = 2752
\n\u21d2 PQ = \\(\\sqrt{2752}\\) cm = 52.5 km<\/p>\n

\"ML<\/p>\n

Question 25.
\nTwo trees A and B are on the same side of the river. From a point C in the river the distance of the trees A and B is 250 m and 300 m respectively. 1f the angle C is 45, find the distance between the trees. Use \u221a2 = 1.44.
\nSolution:
\nUsing cosine formula, we have
\ncos C = \\(\\frac{a^2+b^2-c^2}{2 a b}\\)
\n= \\(\\frac{\\mathrm{CA}^2+\\mathrm{CB}^2-\\mathrm{AB}^2}{2 \\mathrm{CA} \\cdot \\mathrm{CB}}\\)
\n\u21d2 cos 45\u00b0 = \\(\\frac{(250)^2+(300)^2-\\mathrm{AB}^2}{2 \\times 250 \\times 300}\\)<\/p>\n

\"ML<\/p>\n

\u21d2 \\(\\frac{1}{\\sqrt{2}}\\) \u00d7 2 \u00d7 250 \u00d7 300 = 62500 + 90000 – AB2<\/sup>
\n\u21d2 AB2<\/sup> = 152500 – 75000 \u221a2
\n\u21d2 AB2<\/sup> = 152500 – 106050 = 46450
\n[\u2235 \u221a2 = 1.414]
\n\u21d2 AB = \\(\\sqrt{46450}\\) = 215.5 m
\nHence, the required distance between trees be 215.5 metre.<\/p>\n","protected":false},"excerpt":{"rendered":"

Access to comprehensive Class 11 ISC Maths Solutions Chapter 3 Trigonometry Ex 3.9 encourages independent learning. ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 Very short answer\/objective questions (1 to 4) : Question 1. In a \u2206ABC, if a = 4, b = 5 and c = 6, find cos …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170704"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170704"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170704\/revisions"}],"predecessor-version":[{"id":170716,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170704\/revisions\/170716"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170704"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170704"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170704"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}