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Chapter 10: Sine and Cosine Formulae and Their Applications – Exercise 10.2 Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.1 The area of a triangle ABC is given by Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.2 The area of a triangle ABC is given by Therefore, Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.3 We have, a = 4, b = 6 and c = 8 Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.4 In any ∆ABC, we have we have, a = 18, b = 24, c = 30 Therefore, Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.5 b(c cos A – a cos C) = c2 - a2 RHS = c2 - a2 = k2 sin2C - k2 sin2 A = k2(sin2 C - sin2 A) = k2sin(C + A). sin(C - A) = k2 sin(π - B).sin(C - A) = k2 sin B.sin(C - A) = k sin B.k sin(C - A) = bk sin (C-A) = bk(sin C.cos A - sin A.con C) = b(k sin C.cos A - k sin A. cos C) = b(b cos A - a cos C) = LHS Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.6 c(a cos B - b cos A) = ac.cos B - bc cos A Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.7 2(bc cos A + ca cos B + ab cos C) = a2 + b2 + c2 LHS = 2bc cos A + 2ca cos B + 2ab cos C = b2 + c2 - a2 + a2 + c2 - b2 + a2 + b2 - c2 = a2 + b2 + c2 = RHS Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.8 For any ∆ABC, We have therefore, Also, Now, Hence proved. Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.10 In any ∆ABC, we have a = b cos C +c cos B b = c cos A + a cos C c = a cos B + b cos A Therefore, L.H.S = a(cos B + cos C - 1) + b(cos C + cos A - 1) + c(cos A + cos B - 1) = a cos B + a cos C - a + b cos C + b cos A - b + c cos A + c cos B - c = c - b cos A + a cos C - a + a - c cos B + b cos A - b + b - a cos C + c cos B - c = 0 = RHS Hence proved. Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.11 By sine rule we have K sin A = a, K sin B = b, k sin C = c a cos A + b cos B + cos C = k sin A cos A + k sin B cos B + k sin C cos C =k[sin(A+B) cos (A – B) + sin C cos C] =k [sin C cos (A – B) – sin C cos (A + B)] =k [sin C(cos (A – B) – cos(A + B))] = k sin C [2 sin A sin B] = 2 sin C (k sin A) sin B Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.12 We know that by cosine rule ⇒2bc cos A = b2 + c2 - a2 ⇒ a2 = b2+c2 - 2bc cosA Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.13 LHS, = 2bc – b2 - c2 + a2 + 2ca – a2 - c2 + b2 + 2ab – b2 - a2 + c2 = a2 + b2 + c2+ 2ab + 2bc + 2ca = (a + b + c)2 = RHS Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.14 sin3 A cos(B - C) + sin3B.cos(C - A) + sin3C.cos(A - B) = sin2A sin A cos (B - C) + sin2B.sin B cos(C - A) + sin2 C. sin C. cos (A - B) = sin2A sin (π -(B + C))cos(B - C) + sin2B. sin(π - (A + C)).cos(C - A) + sin2 C. sin (π - (A + B)). cos (A - B) = sin2 A sin(B + C) cos(B - C) + sin2B. sin(C + A). cos(C - A) + sin2 C. sin(A + B).cos(A - B) = sin2A. (sin2B + sin2C) + sin2B. (sin2C + sin2A) + sin2C.(sin2A + sin2B) = sin2A.(2sinB.cosB + 2sinCcosC) + 2sin2B.(2sin C cos C + 2 sin A cos A) + sin2C. (2 sin A cos A + 2 sin B cos B) = sin2A. (2sin B cos B + 2 sin C cos C) + sin2B. (2sin C cos C + 2 sin A cos A) + sin2C. (2sin A cos A + 2 sin B cos B) = sin2A. 2 sin B cos B + sin2A. 2 sin C cos C + sin2B. 2 sin C cos S + sin2B. 2sin A cos A + sin2C. 2 sin A cos A + sin2 C. 2sin B cos B = k2a2 2kb cos B + k2a2.2kc cos C + k2b2.2ka cos C + k2b2.2 ka cos A + k2c2.2ka cos A + k2c2.2kb cos B = k3ab(a cos B + b cos A) + k3ac (a cos C + c cos A) + k3bc(c cos B + b cos C) = k3abc + k3acb + k3bca = k33abc = 3(k sin A k sin B k sin C) = 3abc = RHS Sine and Cosine Formulae and Their Applications – Exercise 10.2 – Q.15 b + c = 12λ, c + a = 13λ, a + b = 15λ (b + c + c + a + a + b) = 12λ + 13λ + 15λ 2(a + b + c) = 40λ a + b + c = 20λ b + c = 12λ and a + b + c = 20λ ⟹ a = 8λ c + a = 13λ and a + b + c = 20λ ⟹ b = 7λ a + b = 15λ and a + b + c = 20λ ⟹ c = 5λ
The area of a triangle ABC is given by
Therefore,
We have, a = 4, b = 6 and c = 8
In any ∆ABC, we have
we have,
a = 18, b = 24, c = 30
b(c cos A – a cos C) = c2 - a2
RHS
= c2 - a2
= k2 sin2C - k2 sin2 A
= k2(sin2 C - sin2 A)
= k2sin(C + A). sin(C - A)
= k2 sin(π - B).sin(C - A)
= k2 sin B.sin(C - A)
= k sin B.k sin(C - A)
= bk sin (C-A)
= bk(sin C.cos A - sin A.con C)
= b(k sin C.cos A - k sin A. cos C)
= b(b cos A - a cos C) = LHS
c(a cos B - b cos A)
= ac.cos B - bc cos A
2(bc cos A + ca cos B + ab cos C) = a2 + b2 + c2
LHS
= 2bc cos A + 2ca cos B + 2ab cos C
= b2 + c2 - a2 + a2 + c2 - b2 + a2 + b2 - c2
= a2 + b2 + c2 = RHS
For any ∆ABC, We have
therefore,
Also,
Now,
Hence proved.
a = b cos C +c cos B
b = c cos A + a cos C
c = a cos B + b cos A
L.H.S = a(cos B + cos C - 1) + b(cos C + cos A - 1) + c(cos A + cos B - 1)
= a cos B + a cos C - a + b cos C + b cos A - b + c cos A + c cos B - c
= c - b cos A + a cos C - a + a - c cos B + b cos A - b + b - a cos C + c cos B - c
= 0
= RHS
By sine rule we have
K sin A = a, K sin B = b, k sin C = c
a cos A + b cos B + cos C = k sin A cos A + k sin B cos B + k sin C cos C
=k[sin(A+B) cos (A – B) + sin C cos C]
=k [sin C cos (A – B) – sin C cos (A + B)]
=k [sin C(cos (A – B) – cos(A + B))]
= k sin C [2 sin A sin B]
= 2 sin C (k sin A) sin B
We know that by cosine rule
⇒2bc cos A = b2 + c2 - a2
⇒ a2 = b2+c2 - 2bc cosA
LHS,
= 2bc – b2 - c2 + a2 + 2ca – a2 - c2 + b2 + 2ab – b2 - a2 + c2
= a2 + b2 + c2+ 2ab + 2bc + 2ca
= (a + b + c)2 = RHS
sin3 A cos(B - C) + sin3B.cos(C - A) + sin3C.cos(A - B)
= sin2A sin A cos (B - C) + sin2B.sin B cos(C - A) + sin2 C. sin C. cos (A - B)
= sin2A sin (π -(B + C))cos(B - C) + sin2B. sin(π - (A + C)).cos(C - A)
+ sin2 C. sin (π - (A + B)). cos (A - B)
= sin2 A sin(B + C) cos(B - C) + sin2B. sin(C + A). cos(C - A)
+ sin2 C. sin(A + B).cos(A - B)
= sin2A. (sin2B + sin2C) + sin2B. (sin2C + sin2A) + sin2C.(sin2A + sin2B)
= sin2A.(2sinB.cosB + 2sinCcosC) + 2sin2B.(2sin C cos C + 2 sin A cos A)
+ sin2C. (2 sin A cos A + 2 sin B cos B)
= sin2A. (2sin B cos B + 2 sin C cos C) + sin2B. (2sin C cos C + 2 sin A cos A)
+ sin2C. (2sin A cos A + 2 sin B cos B)
= sin2A. 2 sin B cos B + sin2A. 2 sin C cos C + sin2B. 2 sin C cos S
+ sin2B. 2sin A cos A + sin2C. 2 sin A cos A + sin2 C. 2sin B cos B
= k2a2 2kb cos B + k2a2.2kc cos C + k2b2.2ka cos C
+ k2b2.2 ka cos A + k2c2.2ka cos A + k2c2.2kb cos B
= k3ab(a cos B + b cos A) + k3ac (a cos C + c cos A) + k3bc(c cos B + b cos C)
= k3abc + k3acb + k3bca
= k33abc
= 3(k sin A k sin B k sin C)
= 3abc = RHS
b + c = 12λ, c + a = 13λ, a + b = 15λ
(b + c + c + a + a + b) = 12λ + 13λ + 15λ
2(a + b + c) = 40λ
a + b + c = 20λ
b + c = 12λ and a + b + c = 20λ ⟹ a = 8λ
c + a = 13λ and a + b + c = 20λ ⟹ b = 7λ
a + b = 15λ and a + b + c = 20λ ⟹ c = 5λ
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Chapter 10: Sine and Cosine Formulae and Their...
