Example 5 (1998 National High School League Question) 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. $\frac{\pi}{2}-\arctan \sqrt{2}$
D. $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$
Solution: Choose D. Reason: As shown in Figure 7-5, construct $E E_{1} \perp$ plane $B C D$, then $E E_{1} \perp B G$. Since $A B C D$ is a regular tetrahedron, then $A C \perp B D$. Also, $E F / / A C, F G / / B D$, so $E F \perp F G$, hence $E_{1} F \perp F G$, thus $\angle E F E_{1}$ is the plane angle of the dihedral angle $E-F G-B$, denoted as $\alpha$.

Assume the edge length of the regular tetrahedron $A B C D$ is 1, then its height $A H=$ $\sqrt{A B^{2}-B H^{2}}=\sqrt{1^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\frac{\sqrt{6}}{3}, E E_{1}=\frac{1}{2} A H=$ $\frac{\sqrt{6}}{6}$.
Also, $E F=\frac{1}{2} A C=\frac{1}{2}$, then $\alpha=\arcsin \left(\frac{\sqrt{6}}{6} / \frac{1}{2}\right)=\arcsin \frac{\sqrt{6}}{3}$, thus the size of the dihedral angle $C-F G-E$ is $\pi-\arcsin \frac{\sqrt{6}}{3}=\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$.