If $\frac{\sin A}{\sin C}=\frac{\sin(A-B)}{\sin(B-C)}$, then the sides of ΔABC are in:
Answer & explanation
Correct answer: option 4
$\sin A . \sin (B - C) = \sin C \sin(A - B)$
$⇒ \sin(B + C) . \sin(B -C) = \sin(A - B) \sin(A + B) ⇒ \sin^2 B - \sin^2 C = \sin^2 A - \sin^2 B$
$⇒ a^2, b^2 c^2$ are in A.P. (using sin rule)