Question

The bisectors of $\angle B$ and $\angle C$ of a $△ABC$ meet at O. So that $\angle BOC={90}^o+\frac{\angle A}{2}$

Answer 1

In $△ABC$

$\angle A+\angle B+\angle C={180}^0$

The line OB and OC are the bisectors of angle B and C

$\Rightarrow \angle A+2\angle OBC+2\angle OCB={180}^0\Rightarrow \angle OBC+\angle OCB=90-\frac{\angle A}{2}$

In $△BOC$

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

$\Rightarrow BOC=180-(\angle OBC+OCB)$

$\therefore \angle BOC=180-(90-\frac{\angle A}{2})={90}^o+\frac{\angle A}{2}$