Page 62 - mathsvol1ch1to3ans

62
√
5 + 1
(x) cos 2θ =
4
◦
cos 2θ = cos 36 .
π
◦
θ = 18 =
5
π
θ = 2nπ ±
5
2
(xi) 2 cos x − 7 cos x + 3 = 0 (2 cos (x) − 1) (cos (x) − 3) = 0
1
cos(x) = or cos(x) = 3
2
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π
Since cos(x) = 3 has no solution,the only solution is x = + 2πn
3
Exercise 3.9:
sin A sin(A − B)
2
2
2
1. In a 4ABC, if = , prove that a , b , c are in Arithmetic Progression.
sin C sin(B − C)
Solution:
sin A sin(B − C) = sin C sin(A − B)
sin(B + C) sin(B − C) = sin(A + B) sin(A − B)

2
2
2
2
sin B − sin C = sin A − sin B
2
2
2
2
2
2
R (b − c ) = R (a − b )
2
2b 2 = a + c 2
2
2
2
Hence a , b , c are in Arithmetic Progression.
√ √
2. The angles of a triangle ABC, are in Arithmetic Progression and if b : c = 3 : 2, find ∠A.
Solution:
Since ∠A, ∠B, ∠C are in A.P, we have 2B = A + C.
π π 2π
But A + B + C = π ⇒ 2B + B = π. That is, B = . Hence A + C = π − =
3 3 3
sin B sin C
From Sine Rule, = .
b c

2π
sin − A
sin B 3
We have =
b c