Page 63 - mathsvol1ch1to3ans

63

2π c
sin − A = sin B
3 b
√
!
2
π
= √ sin
3 3
√ √
!
2 3
= √
3 2
1
= √
2
π

= sin
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4
2π π
− A =
3 4
8π − 5π π
A = =
12 4
sin A
3. In a 4ABC, if cos C = , show that the triangle is isosceles.
2 sin B
Solution:

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

sin A + sin(B − C) = sin A

sin(B − C) = 0

B − C = 0

B = C
sin B c − a cos B
4. In a 4ABC, prove that = .
sin C b − a cos C
Solution:
c − a cos B 2R sin C − 2R sin A cos B
=
b − a cos C 2R sin B − 2R sin A cos C
2 sin C − sin(A + B) − sin(A − B)
=
2 sin B − sin(A + C) − sin(A − C)
2 sin C − sin C − sin(A − B)
=
2 sin B − sin B − sin(A − C)
sin C − sin(A − B)
=
sin B − sin(A − C)
sin(A + B) − sin(A − B)
=
sin(A + C) − sin(A − C)
2 sin B cos A
=
2 sin C cos A
sin B
=
sin C
5. In a 4ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.