$20 \cdot 67$ As shown, $A$ is a point outside the circle $\odot O$ with radius 1, $O A=2, A B$ is a tangent to $\odot O$, $B$ is the point of tangency, chord $B C / / O A$, and $A C$ is connected. Then the area of the shaded part is equal to
(A) $\frac{2 \pi}{9}$.
(B) $\frac{\pi}{6}$.
(C) $\frac{\pi}{6}+\frac{\sqrt{3}}{8}$.
(D) $\frac{\pi}{4}-\frac{\sqrt{3}}{8}$.
(China Junior High School Mathematics League, 1996)
[Solution] Connect $O B, O C$.
$$
\because B C / / O A, \quad \therefore \quad S_{\triangle O B C}=S_{\triangle A B C} \text {, }
$$

Thus $S_{\text {shaded }}=S_{\text {sector } O B C}$.
$\because A B$ is the tangent of the circle, $\therefore O B \perp A B$.
In the right triangle $\triangle A O B$, $A O=2, O B=1$,
$\therefore \angle A O B=60^{\circ}$, hence $\angle B O C=60^{\circ}$.
$\therefore \quad S_{\text {shaded }}=S_{\text {sector } O B C}=\pi \cdot \frac{1}{6}=\frac{\pi}{6}$.
Therefore, the answer is (B).