Example 2 As shown in Figure 19-4, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ is the midpoint of $B C$, and $F$ is on $A A_{1}$, with $A_{1} F: F A=1: 2$. Find the dihedral angle formed by plane $B_{1} E F$ and the base $A_{1} B_{1} C_{1} D_{1}$.
(1985 National High School League Question)
Let $E E_{1} \perp B_{1} C_{1}$ at $E_{1}$, set $\angle F B_{1} A_{1}=\theta_{1}, \angle E B_{1} E_{1}=$ $\theta_{2}$, then $\tan \theta_{1}=\frac{1}{3}, \tan \theta_{2}=2$. It is easy to know that $\angle A_{1} B_{1} E_{1}$ is the projection angle of $\angle F B E$ on the plane $A_{1} C_{1}$, i.e., $\angle A_{1} B_{1} E_{1}=\alpha=90^{\circ}$. By Corollary 2 of Property 9, we have $\tan ^{2} \varphi=\frac{\left(\frac{1}{3}\right)^{2}+2^{2}-2 \cdot \frac{1}{3} \cdot 2}{12}=\frac{37}{9}$, hence the dihedral angle between the plane $B_{1} E F$ and the base is $\arctan \frac{\sqrt{37}}{3}$.