Question

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

• A. $b^2-a^2=a^2+c^2$
• B. $a^2,b^2,c^2$ are in $A.P.$
• 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: $\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}\cdots \left(i\right)$
As $A,B,C$ are angles of triangle
$A+B+C=\pi$
$A=\pi -(B+C)$
So, $\sin A=\sin (B+C)\cdots \left(ii\right)$
$\sin B=\sin (A+C)\cdots \left(iii\right)$
From $\left(i\right),\left(ii\right),\left(iii\right)$
$\Rightarrow \frac{\sin (B+C)}{\sin (A+C)}=\frac{\sin (A-C)}{\sin (C-B)}$
$\Rightarrow \sin (B+C)\sin (C-B)=\sin (A-C)\sin (A+C)$
$\Rightarrow {\sin }^{2}C-{\sin }^{2}B={\sin }^{2}A-{\sin }^{2}C$
$\Rightarrow 2{\sin }^{2}C={\sin }^{2}A+{\sin }^{2}B$
By sine rule,
$2c^2=a^2+b^2$
$\Rightarrow b^2,c^2,a^2$ are in $A.P.$