Question

In $△ABC$, if $\frac{a}{b^2-c^2}+\frac{c}{b^2-a^2}=0$, then $B$ is equal to

• A. $\frac{\pi }{2}$
• B. $\pi /4$
• C. $2\pi /3$
• D. $\pi /3$

Answer 1

Answer: (D)

$\pi /3$

We have
$△ABC$, if $\frac{a}{b^2-c^2}+\frac{c}{b^2-a^2}=k$ (say)

$\therefore \frac{a}{b^2-c^2}+\frac{c}{b^2-a^2}=0$

$\Rightarrow \frac{k\sin A}{k^2({\sin }^{2}B-{\sin }^{2}C)}+\frac{k\sin C}{k^2({\sin }^{2}B-{\sin }^{2}A)}=0$

$\Rightarrow \frac{\sin A}{\sin (B+C)\sin (B-C)}+\frac{\sin C}{\sin (B+A)\sin (B-A)}=0$

$\Rightarrow \frac{1}{\sin (B-C)}+\frac{1}{\sin (B-A)}=0$ since $[A+B+C=\pi ]$

$\Rightarrow \sin (A-B)=\sin (B-C)$

$\Rightarrow A-B=B-C$

$\Rightarrow A+C=2B\Rightarrow B={60}^o$