The internal bisectors of the $\angle B$ and $\angle C$ of the $\triangle A B C$, intersect at $O$. If $\angle A=100^{\circ}$, then the measure of $\angle \mathrm{BOC}$ is :

$\angle \mathrm{OBC}=\frac{1}{2} \angle \mathrm{ABC} ;$

$\angle \mathrm{OCB}=\frac{1}{2} \angle \mathrm{ACB}$

From $\Delta \mathrm{OBC}, \angle \mathrm{OBC}+\angle \mathrm{OCB}+\angle \mathrm{BOC}=180^{\circ}$

$\frac{1}{2}(\angle \mathrm{ABC}+\angle \mathrm{ACB})+\angle \mathrm{BOC}=180^{\circ}$

$\Rightarrow \frac{1}{2}\left(180^{\circ}-\angle \mathrm{BAC}\right)+\angle \mathrm{BOC}=180^{\circ}$

$\Rightarrow \frac{1}{2}\left(180^{\circ}-100\right)+\angle \mathrm{BOC}=180^{\circ}$

$\Rightarrow \angle \mathrm{BOC}=180^{\circ}-40^{\circ}=140^{\circ}$