2007 AMC 10B Problem 4

Problem 4 of 25EasierGeometry

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?

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Solution

Since

OA=OB=OC,

triangles

AOB, BOC,

and

COA

are isosceles. The base angles give

\angle ABO=\dfrac{180^\circ-140^\circ}{2}=20^\circ

and

\angle OBC=\dfrac{180^\circ-120^\circ}{2}=30^\circ.

Therefore

\angle ABC=20^\circ+30^\circ=50^\circ.

Thus, the correct answer is D.