Question

The external bisectors of$\angle B$ and $\angle C$ 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

Answer: (C)

${65}^{\circ }$

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

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

Therefore, $\angle A+\angle B+\angle C={360}^o$[ext angle sum prop.]

$\Rightarrow {130}^0+\angle B+\angle C={360}^o\Rightarrow \angle B+\angle C={360}^0-{130}^0\Rightarrow \angle B+\angle C={230}^o$

In $△BOC$

$\frac{\angle B+\angle C}{2}+\angle BOC={180}^o\Rightarrow \frac{230}{2}+\angle BOC={180}^o\angle BOC={180}^o-{115}^o\Rightarrow \angle BOC={65}^o$