$23 \cdot 33$ As shown in the figure, in the rectangular prism $\angle D H G=45^{\circ}, \angle F H B=60^{\circ}$, then the cosine value of $\angle B H D$ is
(A) $\frac{\sqrt{3}}{6}$.
(B) $\frac{\sqrt{2}}{6}$.
(C) $\frac{\sqrt{6}}{3}$.
(D) $\frac{\sqrt{6}}{4}$.
(E) $\frac{\sqrt{6}-\sqrt{2}}{4}$.
(33rd American High School Mathematics Examination, 1982)
[Solution] Let $H F=1$, then $B H=\frac{H F}{\cos 60^{\circ}}=2$.
Therefore, $B F^{2}=B H^{2}-H F^{2}=3$, so $B F=\sqrt{3}$.
Thus, $B F=D G=G H=\sqrt{3}, D H=\sqrt{6}$.
Also, since $D C=H C=\sqrt{3}$, we have $\triangle D C B \cong \triangle H C B$, so $D B=H B=$
2.

Therefore, $\triangle D B H$ is an isosceles triangle, and $\cos \theta=\frac{\sqrt{6}}{4}$. Hence, the answer is (D).