In the given figure, BO and CO are the bisectors of ∠B and ∠C respectively.

(c) 115°

In $\Delta A B C$, we have:

$\angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle]

$\Rightarrow 50^{\circ}+\angle B+\angle C=180^{\circ}$

$\Rightarrow \angle B+\angle C=130^{\circ}$

$\Rightarrow \frac{1}{2} \angle B+\frac{1}{2} \angle C=65^{\circ} \quad \ldots(i)$

$\ln \Delta O B C$, we have:

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

$\Rightarrow \frac{1}{2} \angle B+\frac{1}{2} \angle C+\angle B O C=180^{\circ} \quad$ [Using $\left.(i)\right]$

$\Rightarrow 65^{\circ}+\angle B O C=180^{\circ}$

$\Rightarrow \angle B O C=115^{\circ}$