imc

Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since $O$ is the center, $OC$ and $OA$ are radii and they are congruent. Thus, $△COA$ is an isosceles triangle. Also, note that $\angle COB$ and $\angle COA$ are supplementary, then $\angle COA=180-50=130^{\circ }$. Since $△COA$ is isosceles, then $\angle OCA\cong \angle OAC$. They also sum to $50^{\circ }$, so each angle is $\boxed{25}$.