Question

The external bisectors of $\angle$ B and $\angle$ C meet in O. If $\angle$ A is equal to ${50}^0$, then the magnitude of $\angle$ BOC is:

• A. ${140}^0$
• B. ${105}^0$
• C. ${65}^0$
• D. ${60}^0$

Answer 1

The correct option is C ${65}^0$

Given $ABC$ is a triangle and external bisector of $\angle B$ and $\angle C$ meet at $O$,

And $\angle A={50}^o$

Then Ex $\angle PBC=\angle A+\angle C$...........................(1)

Then Ex $\angle QCB=\angle B+\angle A$...........................(2)

$\angle PBC=2\angle OBC$ ( BO is bisector of angle B)

$\angle QCB=2\angle BCO$ ( CO is bisector of angle C)

Add (1) anbd (2) we get

$\angle PBC+\angle QCB=\angle A+\angle C+\angle B+\angle A$

$\Rightarrow 2\angle OBC+2\angle BCO=\angle A+{180}^0$

In triangle ABC

$\angle A+\angle B+\angle C={180}^0$ and $\angle A={50}^0$ (given)

Then $\angle 2\angle OBC+2\angle BCO=\angle A+{180}^0$

$\Rightarrow \angle OBC+\angle BCO=\frac{1}{2}\angle +{90}^0=\frac{50}{2}+90={115}^0$

$\Delta BOC$

$\angle BOC+\angle OBC+\angle BCO={180}^0$

$\Rightarrow \angle BOC=180-115={65}^0$