Example 4 As shown in Figure 7, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the degree of the dihedral angle $A-B D_{1}-A_{1}$ is
Solution: As shown in Figure 8, let the edge length of the cube be 1, then $B A_{1}=\sqrt{2}, B D_{1}=\sqrt{3}$.
By the theorem of three perpendiculars,
$B A_{1} \perp A_{1} D_{1}, \cos \alpha=\frac{\sqrt{2}}{\sqrt{3}}$.
Also, $B A \perp A D_{1}, \cos \beta=\frac{1}{\sqrt{3}}$, so,
$\sin \alpha=\frac{1}{\sqrt{3}}, \sin \beta=\frac{\sqrt{2}}{\sqrt{3}}$.
By the isosceles right $\triangle A_{1} A B$, we know $\cos \gamma=\frac{\sqrt{2}}{2}$.
By the trihedral angle formula, we get
$$
\cos \theta=\frac{\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{\sqrt{3}} \times \frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}} \times \frac{\sqrt{2}}{\sqrt{3}}}=\frac{1}{2} .
$$

Thus, $\theta=60^{\circ}$.