Question

In a triangle ABC, the angle bisectors of angle ABC and ACB meet at point O. If $\angle$BAC $={32}^o$, find $\angle$ BOC.

Answer 1

Let the $\angle OBC=x$ and $\angle OCB=y$

$\angle BOC=180-(x+y)$

Exterior $\angle BOC=360-(180-(x+y\left)\right)=180+(x+y)$

For the quadilateral $ABOC$, we have

$32+180+(x+y)+x+y=360$

$2(x+y)=148$

$(x+y)=74$

$\angle BOC=180-74={106}^o$