2007 AMC 10B Problem 4

The angle $\angle ABC$ has its vertex on the circle and opens onto arc $AC$ (the arc not containing $B$). Its measure is half the central angle $\angle AOC$ that subtends the same arc.

The three central angles at $O$ fill the full turn:
$$\angle AOC = 360^\circ - 120^\circ - 140^\circ = 100^\circ.$$
Hence
$$\angle ABC = \tfrac{1}{2}\cdot 100^\circ = 50^\circ.$$

The answer is $\boxed{\textbf{(D)}\ 50}$.

The inscribed angle theorem links every angle of a triangle to the central angle over the opposite arc. Given two central angles, the third comes for free from $360^\circ$, and then each triangle angle is half of the central angle opposite it: $\angle A = 60^\circ$, $\angle B = 50^\circ$, $\angle C = 70^\circ$, which do sum to $180^\circ$.

Triangles $OAB$, $OBC$, $OCA$ are isosceles (all legs are radii). Their base angles are $20^\circ$, $30^\circ$, $40^\circ$ respectively, and $\angle ABC$ is made of the base angles from triangles $OAB$ and $OBC$: $20^\circ + 30^\circ = 50^\circ$.

• Halving one of the given central angles (getting 60 or 70) instead of first finding the central angle that subtends arc AC.
• Forgetting the factor of one half and answering $100$, which is not offered, then guessing.