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2008 AMC 12B Problems/Problem 4

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Problem

On circle $O$ , points $C$ and $D$ are on the same side of diameter $\overline{AB}$ , $\angle AOC = 30^\circ$ , and $\angle DOB = 45^\circ$ . What is the ratio of the area of the smaller sector $COD$ to the area of the circle?

$\textbf{(A)}\ \frac {2}{9} \qquad \textbf{(B)}\ \frac {1}{4} \qquad \textbf{(C)}\ \frac {5}{18} \qquad \textbf{(D)}\ \frac {7...$

Solution

$\angle COD = \angle AOB - \angle AOC - \angle BOD = 180^\circ - 30^\circ - 45^\circ = 105^\circ$ .

Since a circle has $360^\circ$ , the desired ratio is $\frac{105^\circ}{360^\circ}=\frac{7}{24} \Rightarrow D$ .

See Also

2008 AMC 12B (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

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