Example 2 As shown in Figure 3, let $E$, $F$, and $G$ be the midpoints of the edges $AB$, $BC$, and $CD$ of a regular tetrahedron $ABCD$, respectively. Then the size of the dihedral angle $C-FG-E$ is ( ).
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(C) $\frac{\pi}{2}-\operatorname{arctg} \sqrt{2}$
(D) $\pi-\operatorname{arcctg} \frac{\sqrt{2}}{2}$
Solution: As shown in Figure 4, let the edge length of the regular tetrahedron be $2a$, $CE$ is the height of the equilateral $\triangle ABC$ with side length $2a$, i.e., $CE=\sqrt{3}a$, $EG$ is the height of the isosceles $\triangle ECD$ with legs of length $\sqrt{3}a$ and base of length $2a$, i.e., $EG=\sqrt{2}a$.
From the equilateral $\triangle GCF$, we know $\alpha=60^{\circ}$.
By the converse of the Pythagorean theorem, we have
$$
\beta=45^{\circ}, \gamma=90^{\circ} \text {. }
$$

By the trihedral angle formula, we get
$$
\begin{array}{l}
\cos \theta=\frac{\cos 90^{\circ}-\cos 60^{\circ} \cdot \cos 45^{\circ}}{\sin 60^{\circ} \cdot \sin 45^{\circ}} \\
=\frac{0-\frac{1}{2} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}}=-\frac{1}{\sqrt{3}} .
\end{array}
$$

And $\sin \theta=\sqrt{1-\left(-\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\sqrt{2}}{\sqrt{3}}$, so, $\operatorname{ctg} \theta=\frac{\cos \theta}{\sin \theta}=-\frac{\sqrt{2}}{2}$.
Therefore, $\theta=\pi-\operatorname{arcctg} \frac{\sqrt{2}}{2}$.