18.8 In quadrilateral $ABCD$, $\angle A=120^{\circ}$, and $\angle B$, $\angle D$ are right angles, while $AB=13$, $AD=46$. Then $AC$ equals
(A) 60.
(B) 62.
(C) 64.
(D) 65.
(E) 72.
(49th American High School Mathematics Examination, 1998)
[Solution] From the given, we know that points $A, B, C, D$ are concyclic, and $AC$ is the diameter of the circle.

Connecting $BD$, let $BD=a$. In $\triangle ABD$, by the Law of Sines, we have:
$$
\begin{aligned}
2 R & =\frac{a}{\sin A} \\
& =\frac{\sqrt{AB^{2}+AD^{2}-2 AB \cdot AD \cdot \cos \angle BAD}}{\sin 120^{\circ}} \\
& =\frac{\sqrt{2883}}{\sqrt{3 / 2}}=62 .
\end{aligned}
$$

Therefore, the answer is $(B)$.