AMC12 2008 B

AMC12 2008 B · Q4

AMC12 2008 B · Q4. It mainly tests Angle chasing, Circle theorems.

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?

在圆 $O$ 上，点 $C$ 和 $D$ 位于直径 $\overline{AB}$ 的同一侧，$\angle AOC = 30^\circ$，且 $\angle DOB = 45^\circ$。较小的扇形 $COD$ 的面积与圆面积之比是多少？

(A) \frac{2}{9} \frac{2}{9}

(B) \frac{1}{4} \frac{1}{4}

(D) \frac{7}{24} \frac{7}{24}

(E) \frac{3}{10} \frac{3}{10}

Answer

Correct choice: (D)

正确答案：（D）

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 $\frac{105^\circ}{360^\circ}=\frac{7}{24} \Rightarrow D$.

$\angle COD = \angle AOB - \angle AOC - \angle BOD = 180^\circ - 30^\circ - 45^\circ = 105^\circ$. 由于圆周角总计为 $360^\circ$，所求比值为 $\frac{105^\circ}{360^\circ}=\frac{7}{24} \Rightarrow D$。

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