Question

In $\Delta ABC$, if $\sin A:\sin C=\sin (A-B):\sin (B-C)$, then $a^2,b^2$ and $c^2$ are in

• A. A.P.
• B. G.P
• C. H.P.
• D. None of these

Answer 1

The correct option is A A.P.

Given:

$\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$
$\frac{\sin (\pi -(B+C)}{\sin (\pi -(A+B)}=\frac{\sin (A-B)}{\sin (B-C)}$
$\frac{\sin (B+C)}{\sin (A+B)}=\frac{\sin (A-B)}{\sin (B-C)}$
$2\sin (B+C)\sin (B-C)=2\sin (A+B)\sin (A-B)$
$\cos 2C-\cos 2B=\cos 2B-\cos 2A$
$2\cos 2B=\cos 2A+\cos 2C$
$2(1-2{\sin }^{2}B)=2-2{\sin }^{2}A-2{\sin }^{2}C$
$1-2{\sin }^{2}B=1-{\sin }^{2}A-{\sin }^{2}C$
$2{\sin }^{2}B={\sin }^{2}A+{\sin }^{2}C$
Hence,
${\sin }^{2}A,{\sin }^{2}B,{\sin }^{2}C$ are in A.P.
Hence, by sine rule $a^2,b^2,c^2$ will be in A.P.

Hence, option A.