2007 AMC 10B Problem 4

We want the angle at $B$. An angle whose corner is on the circle is half of the center angle that opens onto the same arc. Angle $ABC$ opens onto arc $AC$, so we need the center angle $AOC$.

1. The three angles around the center $O$ fill a full turn of $360^\circ$:
$$\angle AOB + \angle BOC + \angle AOC = 360^\circ.$$
2. Two of them are given, so
$$\angle AOC = 360^\circ - 140^\circ - 120^\circ = 100^\circ.$$
3. Angle $ABC$ has its corner $B$ on the circle and looks at the same arc $AC$, so it is half as big:
$$\angle ABC = \tfrac{1}{2} \times 100^\circ = 50^\circ.$$

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

Each angle of a triangle drawn in a circle is half of the center angle across from it. Here that gives $\angle A = 60^\circ$, $\angle B = 50^\circ$, $\angle C = 70^\circ$, and these add to $180^\circ$ as they should. When two center angles are given, the third always comes from $360^\circ$.

Use only isosceles triangles. The segments $OA$, $OB$, $OC$ are all radii, so they are equal.

1. Triangle $OAB$ has two equal sides and top angle $140^\circ$. Its two bottom angles share $180^\circ - 140^\circ = 40^\circ$, so each is $20^\circ$.
2. Triangle $OBC$ has top angle $120^\circ$, so each of its bottom angles is $(180^\circ - 120^\circ) \div 2 = 30^\circ$.
3. The angle at $B$ in triangle $ABC$ is made of one bottom angle from each: $20^\circ + 30^\circ = 50^\circ$.

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 $\boxed{\textbf{(D)}\ 50}$.