Question

If `(sin "A")/(sin "C") = (sin ("A - B"))/(sin ("B - C"))`, then show that a², b², c² are in A.P.

Answer 1

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.