Question

In the given figure, $BO$ and $CO$ are the bisectors of $\angle B$ and $\angle C$ respectively. If $\angle A={50}^{\circ }$ then $\angle BOC=?$

$\left(a\right){130}^{\circ }\left(b\right){100}^{\circ }\left(c\right){115}^{\circ }\left(d\right){120}^{\circ }$

Answer 1

The correct option is (c): ${115}^{\circ }$

In $△ABC,$ we have:

$\angle A+\angle B+\angle C={180}^{\circ }$ [Angle sum property]

$\Rightarrow {50}^{\circ }+\angle B+\angle C={180}^{\circ }$

$\Rightarrow \angle B+\angle C={130}^{\circ }$

$\Rightarrow \frac{1}{2}\angle B+\frac{1}{2}\angle C={65}^{\circ }...\left(i\right)$

In $△OBC,$ we have:

$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$

$\Rightarrow \frac{1}{2}\angle B+\frac{1}{2}\angle C+\angle BOC={180}^{\circ }$

Using eq.(i), we have

$\Rightarrow {65}^{\circ }+\angle BOC={180}^{\circ }$

$\Rightarrow \angle BOC={115}^{\circ }$