In the following diagram, $BO and CO$are the bisectors of $\angle ABC and \angle ACB$ respectively.

 Find $\angle BOC.$

In the triangle ABC,

$\begin{array}{rcl}\angle BAC+\angle ABC+\angle ACB& =& {180}^o\\ \Rightarrow {80}^o+{40}^o+\angle ACB& =& {180}^o\\ \Rightarrow \angle ACB& =& {180}^o-{120}^o\\ \Rightarrow \angle ACB& =& {60}^o\\ & & \end{array}$

Since BO and CO are the bisectors of $\angle ABC and \angle ACB$ respectively.

$\therefore \angle OBC = {20}^o and \angle OCB = \frac{\angle ACB}{2}=\frac{{60}^o}{2}={30}^o$

Now in triangle OBC,

$\begin{array}{rcl}\angle BOC+\angle OBC+\angle OCB& =& {180}^o\\ \Rightarrow \angle BOC+{20}^o+{30}^o& =& {180}^o\\ \Rightarrow \angle BOC& =& {130}^o\\ & & \\ & & \end{array}$