2007 AMC 10B Problems/Problem 4

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 $\angle ABC?$

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

Solution

Because all the central angles of a circle add up to $360^\circ,$

$\begin{align*} \angle BOC + \angle AOB + \angle AOC &= 360\\ 120 + 140 + \angle AOC &= 360\\ \angle AOC &= 100. \end{align*}$

Therefore, the measure of $\text{arc}AC$ is also $100^\circ.$ Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts, $\angle ABC = \boxed{\textbf{(D)} 50}$

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2007 AMC 10B Problems/Problem 4

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