Question

In $△ABC$, if $2R+r=r_2$, then $\angle B=$

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

Answer 1

The correct option is D $\frac{\pi }{2}$

We know $R=\frac{abc}{4\Delta },r=\frac{\Delta }{s}$ and $r_2=\frac{\Delta }{s-b}$

$\Rightarrow \frac{abc}{2\Delta }+\frac{\Delta }{s}=\frac{\Delta }{s-b}$

$\Rightarrow \frac{abc}{2\Delta }=\frac{\Delta }{s-b}-\frac{\Delta }{s}=\frac{\Delta b}{s(s-b)}$

$\Rightarrow \frac{ac}{2\Delta }=\frac{\Delta }{s(s-b)}\dots \dots \left(1\right)$

Now we know $\tan \frac{B}{2}=\frac{\Delta }{s(s-b)}$ and $\frac{ac}{2\Delta }=\frac{1}{\sin B}$

Therefore we have, from equation $\left(1\right)$

$\Rightarrow \frac{1}{\sin B}=\tan \frac{B}{2}$

$\Rightarrow \tan \frac{B}{2}\sin B=1$

$\Rightarrow \tan \frac{B}{2}\times \frac{2\tan \frac{B}{2}}{1+{\tan }^2\frac{B}{2}}=1$

$2{\tan }^2\frac{B}{2}=1+{\tan }^2\frac{B}{2}$

${\tan }^2\frac{B}{2}=1⟹\angle B=\frac{\pi }{2}$