Example 6 As shown in Figure 16-9, in the tetrahedron $S-ABC$, $\triangle ABC$ is an equilateral triangle with a side length of 4, the plane $SAC \perp$ plane $ABC$, $SA=SC=2\sqrt{3}$, and $M, N$ are the midpoints of $AB, SB$ respectively.
(1) Find the size of the dihedral angle $N-CM-B$;
(2) Find the distance from point $B$ to the plane $CMN$.
Solution 1 (1) As shown in Figure 16-10, take the midpoint $D$ of $AC$, and connect $SD$, $DB$.
Since $SA = SC$, $AB = BC$. Therefore, $AC \perp SD$ and $AC \perp BD$, so $AC \perp$ plane $SDB$. Since $AC \subset$ plane $ABC$, plane $SDB \perp$ plane $ABC$.
Draw $NE \perp BD$ at $E$, so $NE \perp$ plane $ABC$.
Draw $EF \perp CM$ at $F$, connect $NF$. By the three perpendiculars theorem, $NF \perp CM$.
Therefore, $\angle NFE$ is the plane angle of the dihedral angle $N-CM-B$.
Since plane $SAC \perp$ plane $ABC$, $SD \perp AC$, so $SD \perp$ plane $ABC$.
Since $NE \perp$ plane $ABC$, $NE \parallel SD$.
Since $SN = NB$, $NE$ is the midline of $\triangle SBD$ and $ED = EB$.
So $NE = \frac{1}{2} SD = \frac{1}{2} \sqrt{SA^2 - AD^2} = \frac{1}{2} \sqrt{12 - 4} = \sqrt{2}$.
In the equilateral $\triangle ABC$, $EF = \frac{1}{4} MB = \frac{1}{2}$.
In the right $\triangle NEF$, $\tan \angle NFE = \frac{EN}{EF} = 2\sqrt{2}$.
Therefore, the size of the dihedral angle $N-CM-B$ is $\arctan 2\sqrt{2}$.
(2) In the right $\triangle NEF$, $NF = \sqrt{EF^2 + EN^2} = \frac{3}{2}$.

So $S_{\triangle M} = \frac{1}{2} CM \cdot NF = \frac{3}{2} \sqrt{3}$, $S_{\triangle CMB} = \frac{1}{2} BM \cdot CM = 2\sqrt{3}$.
Let the distance from point $B$ to plane $CMN$ be $h$.
Since $V_{B-CMN} = V_{V-CMB}$, and $NE \perp$ plane $CMB$,
So $\frac{1}{3} S_{\angle M M M} \cdot h = \frac{1}{3} S_{\angle CMB} \cdot NE$, so $h = \frac{S_{\triangle CMB} \cdot NE}{S_{\triangle CMN}} = \frac{4\sqrt{2}}{3}$.
Therefore, the distance from point $B$ to plane $CMN$ is $\frac{4\sqrt{2}}{3}$.
Solution 2 (1) Take the midpoint $O$ of $AC$, and connect $OS$, $OB$.
Since $SA = SC$, $AB = BC$, so $AC \perp SO$ and $AC \perp BO$.
So $AC \perp$ plane $SOB$.
Since plane $SAC \perp$ plane $ABC$, plane $SAC \cap$ plane $ABC = AC$,
So $SO \perp$ plane $ABC$. So $SO \perp BO$.
As shown in Figure 16-11, establish a spatial rectangular coordinate system $O-xyz$.
Then $A(2,0,0)$, $B(0,2\sqrt{3},0)$, $C(-2,0,0)$,
$$
S(0,0,2\sqrt{2}), M(1,\sqrt{3},0), N(0,\sqrt{3},\sqrt{2})
$$

So $\overrightarrow{CM} = (3,\sqrt{3},0)$, $\overrightarrow{MN} = (-1,0,\sqrt{2})$.
Let $\boldsymbol{n} = (x,y,z)$ be a normal vector of plane $CMN$,
Then $\left\{\begin{array}{l}\overrightarrow{CM} \cdot \boldsymbol{n} = 3x + \sqrt{3}y = 0 \\ \overrightarrow{MN} \cdot \boldsymbol{n} = -x + \sqrt{2}z = 0\end{array}\right.$
Take $z = 1$. Then $x = \sqrt{2}$, $y = -\sqrt{6}$.
Then $\boldsymbol{n} = (\sqrt{2},-\sqrt{6},1)$.
Also, $\overrightarrow{OS} = (0,0,2\sqrt{2})$ is a normal vector of plane $ABC$,
So $\cos \langle \boldsymbol{n}, \overrightarrow{OS} \rangle = \frac{\boldsymbol{n} \cdot \overrightarrow{OS}}{|\boldsymbol{n}| \cdot |\overrightarrow{OS}|} = \frac{1}{3}$.
So the size of the dihedral angle $V-CM-B$ is $\arccos \frac{1}{3}$.
(2) Since $\overrightarrow{MB} = (-1,\sqrt{3},0)$
And $\boldsymbol{n} = (\sqrt{2},-\sqrt{6},1)$ is a normal vector of plane $CMN$,
So the distance from point $B$ to plane $CMN$ is $d = |\overrightarrow{MB}| \cos \langle \boldsymbol{n}, \overrightarrow{MB} \rangle = \frac{|\boldsymbol{n} \cdot \overrightarrow{MB}|}{|\boldsymbol{n}|} = \frac{4\sqrt{2}}{3}$.