Question

In $\Delta ABC$, if $2R+r=r_2$, then $\angle B$ is equal to

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

We have,
$2R+r=r_2$
$\Rightarrow 2R=r_2-r$
$\Rightarrow 2R=4R\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2}-4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$\Rightarrow 2R=4R\sin \frac{B}{2}\cos \left(\frac{A+C}{2}\right)$
$\Rightarrow 1=2{\sin }^2\frac{B}{2}$
$\Rightarrow {\sin }^2\frac{B}{2}=\frac{1}{2}$
$\Rightarrow \sin \frac{B}{2}=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}$
$\Rightarrow \frac{B}{2}=\frac{\pi }{4}$
$\Rightarrow B=\frac{\pi }{2}$