Example 19 (1988 Beijing Competition Question) In the cube $A C_{1}$, find the size of the dihedral angle $C_{1}-D_{1} B-C$.
Solve: As shown in Figure 7-17, connect $C_{1} 、 D$ intersecting $C D_{1}$ at $O$. It is easy to know that $O C_{1} \perp$ plane $B C D_{1}$, connect $B O$.

In the tetrahedron $C_{1}-B O D_{1}$, $O C_{1} \perp$ plane $B O D_{1} ; B C_{1} \perp$ $C_{1} D_{1}$. Let the dihedral angle $C_{1}-B D_{1}-O$ be $G$, then by the conclusion 5(2) of this chapter, we have
$$
\sin ^{2} \varphi=\sin ^{2} \angle C_{1} B O+\sin ^{2} \angle C_{1} D_{1} O .
$$

And $\sin \angle C_{1} B O=\frac{O C_{1}}{B C_{1}}=\frac{1}{2}$,
$$
\sin \angle C_{1} D_{1} O=\sin 45^{\circ}=\frac{\sqrt{2}}{2} \text {. }
$$

Therefore, we find $\sin \varphi=\frac{\sqrt{3}}{2}$, which means the dihedral angle $C_{1}-D_{1} B-C$ is $60^{\circ}$.