2012-II-41
答案:C
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連結 $OB$。
$\begin{array}{rcl}
\angle BOQ & =
連結 $OB$。
$\begin{array}{rcl}
\angle BOQ & =
$\begin{array}{cl}
= & C^{20}_5 – C
$\begin{array}{cl}
= & \dfrac{3}{6}\tim
$\begin{array}{rcl}
m_1 & = & \df
$\begin{array}{cl}
= & 360\times
$\left\{ \
四分
$\begin{arr