2007 AMC 12B Problem 3

The angles around $O$ sum to $360^\circ,$ so $$\begin{aligned} \angle AOC&=360^\circ-140^\circ-120^\circ \\ &=100^\circ. \end{aligned}$$ By the inscribed angle theorem, $\angle ABC$ subtends the same arc $AC$ as the central angle $\angle AOC,$ so $$\angle ABC=\tfrac12\angle AOC=50^\circ.$$ Thus, the correct answer is D.