Question

The bisects of exterior angles at B and C of $\Delta ABC$, meet at O. If $\angle A=x^{\circ }$, then $\angle BOC$=

Answer 1

In $\Delta ABC,\angle A=x^{\circ }$

and bisectors of $\angle B$ and $\angle C$ meet at O.

$\therefore \angle BOC={90}^{\circ }+\frac{1}{2}\angle A$

=${90}^{\circ }+\frac{1}{2}x={90}^{\circ }+\frac{x}{2}$