Question

In the following figure, OB and OC are the angle bisectors. Find $\angle BOC$

• A. ${100}^{\circ }$
• B. ${90}^{\circ }$
• C. ${80}^{\circ }$
• D. ${110}^{\circ }$

Answer 1

The correct option is D

${110}^{\circ }$

In $\Delta ABC$, $\angle ABC=\angle ACB$

So, $\frac{\left(\angle ABC\right)}{2}=\frac{\left(\angle ABC\right)}{2}$

$\angle BOC=\angle BCO$

Therefore, $\Delta BOC$ is also isosceles.

Also, $\angle A+\angle ABC+\angle ACB={180}^{\circ }$

Let $\angle ABC=\angle ACB=x$

So, $\angle A+x+x={180}^{\circ }$

$2x={180}^{\circ }-\angle A={180}^{\circ }-{40}^{\circ }={140}^{\circ }$

$x={70}^{\circ }$

$\angle OBC=\angle OCB=\frac{{70}^{\circ }}{2}={35}^{\circ }$

$\angle BOC={180}^{\circ }-(\angle OBC+\angle OCB)={180}^{\circ }-{70}^{\circ }={110}^{\circ }$