imc

Because all the central angles of a circle add up to $360^{\circ },$ $$\begin{array}{rl}\angle BOC+\angle AOB+\angle AOC& =360\\ 120+140+\angle AOC& =360\\ \angle AOC& =100.\end{array}$$ 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, $\angle ABC=\boxed{50}$