2007 AMC 10B Problem 4

Question

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?

35

40

45

50

60

Each radius is equal, so each of the triangles

AOB, BOC, COA

is isosceles

\angle ABC=\angle ABO+\angle OBC,

where each part is a base angle of an isosceles triangle

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.

Problem 4 in other years