Example 3 Let $A A_{1} B_{1} B$ be the axial section of a cylinder, $C$ be a point on the circumference of the base circle, $A A_{1}=1, A B=4, \angle B A C=$ $60^{\circ}$. Find the dihedral angle formed by the plane $A_{1} C B_{1}$ and the base $A B$ of the cylinder.
Solve As shown in Figure 19-5, let $\angle A_{1} C A=\theta_{1}, \angle B_{1} C B=\theta_{2}, \angle A_{1} C B_{1}=\theta$. Given $A B=4, \angle B A C=60^{\circ}$, so $A C=2, B C=2 \sqrt{3}$. Also, since $A A_{1}=1$, we have $A_{1} C= \sqrt{5}, B_{1} C=\sqrt{13}$, thus $\cos \theta_{1}=\frac{2}{5} \sqrt{5}, \cos \theta_{2}=\frac{2}{13} \sqrt{39}$.
Therefore, $\cos \theta=\frac{A_{1} C^{2}+B_{1} C^{2}-A_{1} B_{1}^{2}}{2 A_{1} C \cdot B_{1} C}=\frac{\sqrt{65}}{65}$, from which we get $\sin \theta=\frac{8 \sqrt{65}}{65}$.
Clearly, $\angle B A C=\alpha=90^{\circ}$, substituting this into the conclusion of Property 9, we get $\cos \varphi=\frac{\cos \theta_{1} \cdot \cos \theta_{2} \cdot \sin \alpha}{\sin \theta}=\frac{\sqrt{3}}{2}$, hence $\varphi=30^{\circ}$.
Therefore, the dihedral angle between plane $A_{1} C B_{1}$ and the base is $30^{\circ}$.