Stuck on this problem: In the figure, $O$ is the center of the circle, $\angle BAO = 30^\circ$ and $\angle BCO = 40^\circ$. Find the shaded angle $ \angle AOC $.
Connect BO and note that BO = AO = CO since they are all radii
Then in triangle ABO, AO = BO so angle BAO = angle OBA = 30
Similarly, in triangle CBO , CO = BO so angle BCO = angle OBC = 40
And angle ABC =angle OBA + angle OBC = 30 + 40 = 70
But angle ABC is an inscribed angle interceptin the same arc as central angle AOC
So AOC = 2 * ABC = 2 * 70 = 140°