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Question: Answered & Verified by Expert
In a $\triangle A B C, \frac{a}{b^2-c^2}+\frac{c}{b^2-a^2}=0$, then $B$ is equal to
MathematicsProperties of TrianglesAP EAMCETAP EAMCET 2001
Options:
  • A $\frac{\pi}{2}$
  • B $\frac{\pi}{4}$
  • C $\frac{2 \pi}{3}$
  • D $\frac{\pi}{3}$
Solution:
1349 Upvotes Verified Answer
The correct answer is: $\frac{\pi}{3}$
We have,
$\begin{aligned} & \frac{a}{b^2-c^2}+\frac{c}{b^2-a^2}=0 \\ & \Rightarrow \quad \frac{2 R \sin A}{4 R^2\left(\sin ^2 B-\sin ^2 C\right)} \\ & \quad+\frac{2 R \sin C}{4 R^2\left(\sin ^2 B-\sin ^2 A\right)}=0\end{aligned}$
$\begin{aligned} & \Rightarrow \quad \frac{1}{2 R} \cdot \frac{\sin A}{\sin (B+C) \sin (B-C)} \\ & +\frac{1}{2 R} \cdot \frac{\sin C}{\sin (B+A) \sin (B-A)}=0 \\ & \Rightarrow \quad \frac{1}{2 R} \cdot \frac{\sin A}{\sin A-\sin (B-C)} \\ & +\frac{1}{2 R} \cdot \frac{\sin C}{\sin C \sin (B-A)}=0 \\ & \rightarrow \quad \frac{1}{2 R \sin (R-C)}+\frac{1}{2 R \sin (B-A)}=0 \\ & \Rightarrow \quad \sin (B-C)+\sin (B-A)=0 \\ & \Rightarrow \quad 2 \sin \frac{2 B-A-C}{2} \cos \frac{-C+A}{2}=0 \\ & \Rightarrow \quad \sin \frac{2 B-A-C}{2}=0 \\ & \Rightarrow \quad 2 l l=A+C \\ & 2 R=180^n-B \\ & 3 b=180^{\circ} \\ & B=\frac{\pi}{3} \\ & \end{aligned}$

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