2007 AMC 12B Problems/Problem 3

Problem

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 $\angle ABC$ ?

$\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60$

Solution

Since triangles $ABO$ and $BOC$ are isosceles, $\angle ABO=20^o$ and $\angle OBC=30^o$ . Therefore, $\angle ABC=50^o$ , or $\mathrm{(D)}$ .

Alternative Solution

$\angle AOC = 100^{\circ} \implies \angle ABC =\frac{\angle AOC}{2} =50^{ \circ},$ or $\mathrm{(D)}.$

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 12 Problems and Solutions

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2007 AMC 12B Problems/Problem 3

Contents

Problem

Solution

Alternative Solution

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