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Difference between revisions of "2007 AMC 10B Problems/Problem 4"

(Created page with '==Problem 4== The point <math>O</math> is the center of the circle circumscribed about <math>\triangle ABC,</math> with <math>\angle BOC=120^\circ</math> and <math>\angle AOB=14…')
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Revision as of 16:25, 27 May 2011

Problem 4

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 &= 120. \end{align*}

Therefore, the measure of $\text{arc}AC$ is also $120^\circ.$ Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts, $\angle ABC = \boxed{\textbf{(E)} 60}$

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