Example 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: Let the body be a unit cube. Establish a spatial rectangular coordinate system as shown in Figure 9, then
$$
\begin{array}{l}
D(0,0,0), \\
A(1,0,0), \\
D_{1}(0,0,1), \\
B(1,1,0), \\
A_{1}(1,0,1).
\end{array}
$$

Thus, $A D_{1}=(-1,0,1), A B=(0,1,0)$,
$$
A_{1} B=(0,1,-1), A_{1} D_{1}=(-1,0,0).
$$

Let the normal vectors of the planes $A B D_{1}$ and $A_{1} B D_{1}$ be
$$
\boldsymbol{n}_{1}=\left(x_{1}, y_{1}, z_{1}\right), \boldsymbol{n}_{2}=\left(x_{2}, y_{2}, z_{2}\right).
$$

Since $\boldsymbol{n}_{1} \perp A B, \boldsymbol{n}_{1} \perp A D_{1}, \boldsymbol{n}_{2} \perp A_{1} B, \boldsymbol{n}_{2} \perp A_{1} D_{1}$, we have
$$
\begin{array}{l}
y_{1}=0, -x_{1}+z_{1}=0, \\
y_{2}-z_{2}=0, -x_{2}=0.
\end{array}
$$

Let $x_{1}=a$, then $z_{1}=a$;
Let $y_{2}=b$, then $z_{2}=b$.
Thus, $n_{1}=(a, 0, a), n_{2}=(0, b, b)$, and let the angle between $\boldsymbol{n}_{1}$ and $\boldsymbol{n}_{2}$ be $\alpha$.
Therefore, $\cos \alpha=\frac{\boldsymbol{n}_{1} \cdot \boldsymbol{n}_{2}}{\left|\boldsymbol{n}_{1}\right| \cdot\left|\boldsymbol{n}_{2}\right|}$
$$
=\frac{(a, 0, a) \cdot(0, b, b)}{\sqrt{2} a \cdot \sqrt{2} b}=\frac{1}{2}.
$$

Thus, $\alpha=60^{\circ}$.
It is also easy to see that the dihedral angle $A-B D_{1}-A_{1}$ is $\theta=\alpha$.

Therefore, the dihedral angle $A-D_{1} B-A_{1}$ is $60^{\circ}$.