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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

```

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