Example 1. As shown in Figure 6, given the dihedral angle $P-A C-Q$
$$
\begin{array}{l}
=120^{\circ}, \angle B A C=60^{\circ}, \angle D C A=45^{\circ} . \\
B \in P, D \in Q .
\end{array}
$$

Find the angle formed by the skew lines $A B$ and $C D$.
Solve: Translate line $AB$ within plane $P$ so that points $A$ and $C$ coincide. This results in three rays originating from point $C$. Let the angle between rays $AB$ and $CD$ be $\theta$, the angle between $AB$ and $CA$ be $\theta_{1}=120^{\circ}$, and the angle between $CD$ and $CA$ be $\theta_{2}=45^{\circ}$. By the theorem of three rays,
$$
\begin{array}{c}
\cos \theta=\cos \theta_{1} \cos \theta_{2}+\sin \theta_{1} \sin \theta_{2} \cos 120^{\circ} \\
=\cos 120^{\circ} \cos 45^{\circ}+\sin 120^{\circ} \sin 45^{\circ} \cos 120^{\circ} \\
=-\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{8}\right) .
\end{array}
$$

Thus, the angle between rays $AB$ and $CD$ is
$$
\arccos \left(-\frac{2 \sqrt{2}+\sqrt{6}}{8}\right) .
$$
Since the angle between skew lines is acute,
the angle between lines $AB$ and $CD$ is
$$
\arccos \left(\frac{2 \sqrt{2}+\sqrt{6}}{8}\right) .
$$