Page 65 - mathsvol1ch1to3ans

65

a k sin A
=
b + c k sin B + k sin C
A A
2 sin cos
= 2 2
B + C B − C
2 sin cos
2 2
A A
2 sin cos
= 2 2
A B − C
2 cos cos
2 2
A
sin
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= 2
B − C
cos
2
A B + C
sin cos
= 2 2
B + C B − C
cos cos
2 2
A B + C
2 sin cos
= 2 2
B + C B − C
2 cos cos
2 2
A B + C
2 sin cos
= 2 2
cos B + cos C
A
2 sin 2
= 2
cos B + cos C

A
Hence a(cos B + cos C) = 2(b + c) sin 2
2
2
a − c 2 sin(A − C)
(iii) =
b 2 sin(A + C)
Solution:
2
a − c 2 (a + c)(a − c)
=
b sin(A − C) b sin(A − C)
2
2
R 2 sin A − sin C
=
R sin(A + C) sin(A − C)
2
2
sin A − sin C
= R
2
2
sin A − sin C
= R
2
a − c 2 sin(A − C)
Hence = Rb sin(A − C) =
b 2 sin(A + C)
a sin(B − C) b sin(C − A) c sin(A − B)
(iv) = =
2
2
2
b − c 2 c − a 2 a − b 2
Solution: