$7 \cdot 66$ As shown in the figure, points $A, B, C, D$ lie on a circle with a diameter of 1, and point $X$ is on the diameter $A D$. If $B X=C X, 3 \angle B A C=\angle B X C=36^{\circ}$, then $A X$ equals
(A) $\cos 6^{\circ} \cos 12^{\circ} \sec 18^{\circ}$.
(B) $\cos 6^{\circ} \sin 12^{\circ} \csc 18^{\circ}$.
(C) $\cos 6^{\circ} \sin 12^{\circ} \sec 18^{\circ}$.
(D) $\sin 6^{\circ} \sin 12^{\circ} \csc 18^{\circ}$.
(E) $\sin 6^{\circ} \sin 12^{\circ} \sec 18^{\circ}$.
(44th American High School Mathematics Examination, 1993)
[Solution] Since $2 \angle B A C \neq \angle B X C$, $X$ is not the center. Let the center be $O$, and connect $O B, O C$, then
$\triangle O X B \cong \triangle O X C$. Therefore, we can get
$$
\begin{array}{l}
\angle B X D=\angle C X D, \\
\angle B A X=\angle C A X .
\end{array}
$$

Since $\angle A B D=90^{\circ}, A D=1$, then $A B={ }^{\circ} A D \cos \angle B A D=\cos \frac{1}{2} \angle B A C=\cos 6^{\circ}$.
Noting that $\angle A X B=162^{\circ}, \angle A B X=12^{\circ}$, by the Law of Sines, we have
$$
\frac{A B}{\sin \angle A X B}=\frac{A X}{\sin \angle A B X} .
$$

Thus, $\frac{\cos 6^{\circ}}{\sin 162^{\circ}}=\frac{A X}{\sin 12^{\circ}}$.
Since $\sin 162^{\circ}=\sin 18^{\circ}$,
we have
$$
A X=\frac{\cos 6^{\circ} \sin 12^{\circ}}{\sin 18^{\circ}}=\cos 6^{\circ} \sin 12^{\circ} \csc 18^{\circ} \text {. }
$$

Therefore, the answer is $(B)$.