The point $O$ is the center of the circle circumscribed about $\triangle ABC,$ with $\angle BOC=120^\circ$ and $\angle AOB=140^\circ,$ as shown. What is the degree measure of $\angle ABC?$
Because all the central angles of a circle add up to $360^\circ,$ \begin{align} \angle BOC + \angle AOB + \angle AOC &= 360\\ 120 + 140 + \angle AOC &= 360\\ \angle AOC &= 100. \end{align} Therefore, the measure of $\text{arc}AC$ is also $100^\circ.$ Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts, $\angle ABC = \fbox{50}$