$22 \cdot 9$ dihedral angle $\alpha-A B-\beta$ is $60^{\circ}, P \in \alpha, Q \in \beta$, and $P, Q$ are not on $A B$, the angle between $P Q$ and $A B$ is $45^{\circ}$, and $P Q=7 \sqrt{2},$ the distance from $Q$ to $A B$ is 3, then the distance from $P$ to $A B$ is 
(A) 5.
(B) $\sqrt{37}$.
(C) 8.
(D) $\sqrt{79}$.
(China Sichuan Province High School Mathematics Competition, 1993)
[Solution] As shown in the figure, draw $Q T \perp A B$ at $T$, draw $Q R / /$ $A B$, then draw $P R \perp Q R$ at $R$, and $R S \perp A B$ at $S$, connect $P S$, then $P S \perp A S$.
$$
\begin{array}{lc}
\therefore & \angle P S R=60^{\circ} . \\
\text { Also } & \angle P Q R=45^{\circ}, P Q=7 \sqrt{2}, \\
\therefore & P R=\frac{\sqrt{2}}{2} P Q=7 .
\end{array}
$$

By the cosine rule,
$$
P R^{2}=P S^{2}+S R^{2}-2 P S \cdot S R \cos \angle P S R \text {. }
$$

Also, $S R=Q T=3, P R=7, \angle P S R=60^{\circ}$, substituting these into the above equation and simplifying, we get
$$
P S^{2}-3 P S-40=0,
$$

Solving this, we get $P S=8$ (discarding the negative value).
Therefore, the answer is $(C)$.