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Question
If `(sin "A")/(sin "C") = (sin ("A - B"))/(sin ("B - C"))`, then show that a2, b2, c2 are in A.P.
Solution
By sine rule,
`"sin A"/"a" = "sin B"/"b" = "sin C"/"c" = "k"`
∴ sin A = ka, sin B = kb, sin C = kc
Now, `(sin "A")/(sin "C") = (sin ("A - B"))/(sin ("B - C"))`
∴ sin A . sin (B - C) = sin C. sin (A - B)
∴ sin [π - (B + C)]. sin (B - C)
= sin [π - (A + B)]. sin(A - B) .....[∵ A + B + C = π]
∴ sin (B + C). sin (B - C) = sin (A + B). sin (A - B)
∴ sin2B - sin2C = sin2A - sin2B
∴ 2 sin2B = sin2A + sin2C
∴ 2k2b2 = k2a2 + k2c2
∴ 2b2 = a2 + c2
Hence, a2, b2, c2 are in A.P.
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