Question

In $△ABC$, bisector of $\angle A$ and $\angle B$ intersect at point $O$. If $\angle C={70}^o$ what is the value of $\angle AOB$?

Answer 1

In a triangle, the sum of interior angle $={180}^{\circ }$

Thus, in $△ABC$ ,

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

$\angle C={70}^{\circ }$,

Hence,

$\angle A+\angle B={130}^{\circ }$

Bisector of $\angle Aand\angle B$ intersect at point O then

$\frac{\angle A}{2}+\frac{\angle B}{2}={65}^{\circ }$

$\angle OAB+\angle OBA={65}^{\circ }$

Now, in $△OAB$,

$\angle OAB+\angle OBA+\angle AOB={180}^{\circ }$

$\Rightarrow \angle AOB={115}^{\circ }$