$7 \cdot 64$ intercepts a sector with a central angle of acute angle $\theta$ from a circle with a radius of 6. The radius of the circumcircle of this sector is
(A) $3 \cos \theta$.
(B) $3 \sec \theta$.
(C) $3 \cos \frac{1}{2} \theta$.
(D) $3 \sec \frac{1}{2} \theta$.
(E) 3 .
(24th American High School Mathematics Examination, 1973)
[Solution] $C$ is the center of the circumcircle of sector $P O Q$, i.e., the intersection of the perpendicular bisectors $S C$ and $R C$ of $O P$ and $O Q$.
In $\triangle O R C$, we have
$$
\sec \frac{\theta}{2}=\frac{O C}{3} \text { or } O C=3 \sec \frac{\theta}{2} \text {. }
$$

Therefore, the answer is $(D)$.