Example 9. Given the dihedral angle $C-A B=D=120^{\circ}$, $\angle C B A=60^{\circ}, \angle D A B=30^{\circ}, A B=\sqrt{37} \mathrm{~cm}$ (Figure 17). Find the distance $d$ between the skew lines $B C, A D$.
Let the angle between $BC$ and $AD$ be $\theta$. By the triple product formula,
$$
\begin{aligned}
\cos \theta= & \cos 60^{\circ} \cos 150^{\circ} \\
& +\sin 60^{\circ} \sin 150^{\circ} \cos 120^{\circ} \\
= & -\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{8}=-\frac{3 \sqrt{3}}{8} .
\end{aligned}
$$

Thus, $\sin \theta=\frac{\sqrt{37}}{8}$.
Therefore, $d=\frac{\sqrt{37} \sin 60^{\circ} \sin 150^{\circ} \sin 120^{\circ}}{\sin \theta}$
$=3(\text{cm})$