Question

Let $\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}$, where $A,B,C$ are angles of a triangle $ABC$. If the lengths of the sides opposite these angles are $a,b,c$ respectively, then

• A. $a^2,b^2,c^2$ are in A.P.
• B. $b^2-a^2=a^2+c^2$
• C. $b^2,c^2,a^2$ are in A.P.
• D. $c^2,a^2,b^2$ are in A.P.

Answer 1

The correct option is C $b^2,c^2,a^2$ are in A.P.

Given: $\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}$
$\Rightarrow \frac{\sin (\pi -(C+B\left)\right)}{\sin (\pi -(A+C\left)\right)}=\frac{\sin (A-C)}{\sin (C-B)}$
$\Rightarrow \sin (C+B)\sin (C-B)=\sin (A+C)\sin (A-C)$
$\Rightarrow {\sin }^{2}C-{\sin }^{2}B={\sin }^{2}A-{\sin }^{2}C$
$\Rightarrow 2{\sin }^{2}C={\sin }^{2}A+{\sin }^{2}B$
By sin rule, we have
$a^2+b^2=2c^2$
Hence, $b^2,c^2,a^2$ are in A.P.