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2008 AMC 12B Problem 4

Problem 4 of 25EasierGeometry

On circle O,O, points CC and DD are on the same side of diameter AB‾,\overline{AB}, ∠AOC=30∘,\angle AOC = 30^\circ, and ∠DOB=45∘.\angle DOB = 45^\circ. What is the ratio of the area of the smaller sector CODCOD to the area of the circle?

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Solution

Since ∠AOC,\angle AOC, ∠COD,\angle COD, and ∠DOB\angle DOB fill the straight angle over diameter AB‾,\overline{AB}, ∠COD=180∘−30∘−45∘=105∘. \begin{aligned} \angle COD &= 180^\circ - 30^\circ - 45^\circ \\ &= 105^\circ. \end{aligned} The sector’s share of the circle is 105∘360∘=724.\dfrac{105^\circ}{360^\circ} = \dfrac{7}{24}. Thus, the correct answer is D.
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Tagged: sector · angle sum

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