Example 11 (2004 National High School Competition Question) As shown in Figure 7-10, 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 $\qquad$ .
Solution: Fill in $60^{\circ}$. Reason: As shown in Figure 7-10, connect $D_{1} C$, draw $C E \perp B D_{1}$ at $E$, extend $C E$ to intersect $A_{1} B$ at $F$, then $F E \perp B D_{1}$. Connect $A E$, by symmetry, $A E \perp B D_{1}$, thus $\angle F E A$ is the plane angle of the dihedral angle $A-B D_{1} A_{1}$. Connect $A C$, let $A B=1$, then $A C=A D_{1}=\sqrt{2}$, $B D_{1}=\sqrt{3}$.

In the right triangle $\triangle A B D_{1}$, $A E=\frac{A B \cdot A D_{1}}{B D_{1}}=\frac{\sqrt{2}}{\sqrt{3}}$. In $\triangle A E C$, $\cos \angle A E C=\frac{A E^{2}+C E^{2}-A C^{2}}{2 A E \cdot C E}=\frac{2 A E^{2}-A C^{2}}{2 A E^{2}}=$ $\frac{\frac{3}{4}-2}{\frac{4}{3}}=-\frac{1}{2}$, so $\angle A E C=120^{\circ}$. Since $\angle F E A$ is the supplementary angle of $\angle A E C$, hence $\angle F E A=60^{\circ}$.