Page 64 - mathsvol1ch1to3ans

64

Solution:
a cos A + b cos B + c cos C = 2R sin A cos A + 2R sin B cos B + 2R sin C cos C Using Sine Rule

= R (sin 2A + sin 2B + sin 2C)

= R (2 sin(A + B) cos(A − B) + sin 2C)

= R (2 sin C cos(A − B) + 2 sin C cos C)

= 2R sin C (cos(A − B) − cos(A + B))
= 2R sin C (2 sin A sin B)
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= 2R sin A sin C sin B

= a sin C sin B

B − C
◦
6. In a 4ABC, ∠A = 60 . Prove that b + c = 2a cos .
2
Solution:
b + c (sin B + sin C)
= R
2a 2R sin(A)

B + C B − C
2 sin cos
2 2
=
A A
4 sin cos
2
2
A B − C
cos cos
2 2
=
A A
2 sin cos
2 2
B − C
cos
= 2
2 sin(30) ◦
B − C
b + c = 2a cos
2
7. In a 4ABC, prove the following

A A
(i) a sin + B = (b + c) sin
2 2
Solution:
B + C B − C A B − C
2 sin cos 2 cos cos
b + c k sin B + k sin C sin B + sin C 2 2 2 2
= = = =
a k sin A sin A A A A A
2 sin cos 2 sin cos
2 2 2 2

A B − C B + C A π A
(b + c) sin = a cos = a cos B − = a cos B + − = a sin B +
2 2 2 2 2 2
A
(ii) a(cos B + cos C) = 2(b + c) sin 2
2
Solution: