Question

In the given figure, BO and CO are the bisectors of the exterior angles meeting each other at O. If $\angle A={70}^{\circ }$ find $\angle BOC$ in degrees.

Answer 1

$\because sumofanglesonastraightline={180}^{\circ }$

$=>\angle ABC={180}^{\circ }-2\theta$

$and\angle ACB={180}^{\circ }-2\alpha$

$In△ABC$

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

$=>{70}^{\circ }+180-2\theta +180-2\alpha =180$

$=>{250}^{\circ }-(2\theta +2\alpha )=0$

$=>2(\theta +\alpha )={250}^{\circ }$

$=>\theta +\alpha ={125}^{\circ }-\left(1\right)$

$In△BOC$

$=>\angle CBO+\angle BCO+\angle BOC={180}^{\circ }$

$=>\theta +\alpha +\angle BOC={180}^{\circ }$

$=>{125}^{\circ }+\angle BOC={180}^{\circ }-from\left(1\right)$

$\therefore \angle BOC={55}^{\circ }$