2008 AMC 12B Problems/Problem 4

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?

[asy] unitsize(6mm); defaultpen(linewidth(0.7)+fontsize(8pt));  pair C = 3*dir (30); pair D = 3*dir (135); pair A = 3*dir (0); pair B = 3*dir(180); pair O = (0,0); draw (Circle ((0, 0), 3)); label ("\(C\)", C, NE); label ("\(D\)", D, NW); label ("\(B\)", B, W); label ("\(A\)", A, E); label ("\(O\)", O, S); label ("\(45^\circ\)", (-0.3,0.1), WNW); label ("\(30^\circ\)", (0.5,0.1), ENE); draw (A--B); draw (O--D); draw (O--C); [/asy]

$\textbf{(A)}\ \frac {2}{9} \qquad \textbf{(B)}\ \frac {1}{4} \qquad \textbf{(C)}\ \frac {5}{18} \qquad \textbf{(D)}\ \frac {7}{24} \qquad \textbf{(E)}\ \frac {3}{10}$

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 (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions

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