jmc

We can see that $\angle AOC=360^{\circ }-(110^{\circ }+100^{\circ })=150^{\circ }.$ Now, $△AOC$ and $△AOB$ are both isosceles triangles. That means that $\angle OAC=\frac{1}{2}\cdot (180^{\circ }-150^{\circ })=15^{\circ }$ and $\angle OAB=\frac{1}{2}\cdot (180^{\circ }-110^{\circ })=35^{\circ }.$ Therefore, our answer is $\angle BAC=\angle OAB+\angle OAC=15^{\circ }+35^{\circ }=\boxed{50^{\circ }}.$