Question

The external bisectors of $\angle B$ and $\angle C$ of $△ABC$, meet in O. If $\angle A$ is equal to ${50}^{\circ }$, then the magnitude of $\angle BOC$ is

• A. ${140}^{\circ }$
• B. ${105}^{\circ }$
• C. ${65}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${65}^{\circ }$

Let external angles of a $△ABC$ be $\angle A$, $\angle B$, & $\angle C$,

Exterior $\angle A$=$180-50={130}^o$

As we know that, sum of exterior angles of any closed figure is 360.

Therefore,

$\angle A$+$\angle B$+$\angle C$=360

=> ${130}^o$+$\angle B$+$\angle C$=${360}^o$

=> $\angle B$+$\angle C$=${230}^o$

In $△BOC$,

$\frac{\angle B+\angle C}{2}+\angle BOC={360}^o$

=> $\frac{230}{2}+\angle BOC={360}^o$

Therefore,

$\angle BOC={65}^o$