Example 3 As shown in Figure 5, in a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $E$ and $F$ are the midpoints of edges $A B$ and $B C$ respectively. Find the size of the dihedral angle $B-F B_{1}-E$.
Solution: As shown in Figure 6, by the Pythagorean theorem, we have
$$
\begin{array}{l}
B_{1} E=B_{1} F=\frac{\sqrt{5}}{2} a, \\
E F=\frac{\sqrt{2}}{2} a .
\end{array}
$$

In the right triangle $\triangle E B B_{1}$, we have
$$
\cos \gamma=\frac{2}{\sqrt{5}}.
$$

Since $\triangle E B_{1} B \cong \triangle F B_{1} B$, we know $\beta=\gamma$.
Therefore, $\cos \beta=\frac{2}{\sqrt{5}}, \sin \beta=\frac{1}{\sqrt{5}}$.
By the cosine rule, we get
$$
\cos \alpha=\frac{\left(\frac{\sqrt{5}}{2} a\right)^{2}+\left(\frac{\sqrt{5}}{2} a\right)^{2}-\left(\frac{\sqrt{2}}{2} a\right)^{2}}{2 \times \frac{\sqrt{5}}{2} a \times \frac{\sqrt{5}}{2} a}=\frac{4}{5}.
$$

Thus, $\sin \alpha=\frac{3}{5}$.
By the trihedral angle formula, we get
$$
\cos \theta=\frac{\frac{2}{\sqrt{5}}-\frac{4}{5} \times \frac{2}{\sqrt{5}}}{\frac{3}{5} \times \frac{1}{\sqrt{5}}}=\frac{2}{3}.
$$

Therefore, $\theta=\arccos \frac{2}{3}$.