Q68 / 160 · TS EAMCET 2016 · Mathematics
In $\triangle A B C$, if $2 R+r=r_2$, then $\angle B$ is equal to
- A$\frac{\pi}{3}$
- B$\frac{\pi}{4}$
- C$\frac{\pi}{6}$
- Correct answerD$\frac{\pi}{2}$
Solution
We have,
$$
\begin{array}{lrl}
& 2 R+r & =r_2 \\
\Rightarrow & 2 R & =r_2-r \\
\Rightarrow & & \\
2 R=4 R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2}-4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\
\Rightarrow & & 2 R=4 R \sin \frac{B}{2} \cos \left(\frac{A+C}{2}\right) \\
\Rightarrow & & 1=2 \sin ^2 \frac{B}{2} \Rightarrow \sin ^2 \frac{B}{2}=\frac{1}{2}
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow & \sin \frac{B}{2} & =\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4} \\
\Rightarrow & & \frac{B}{2} & =\frac{\pi}{4} \\
& & B & =\frac{\pi}{2}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.