16. (12 points) As

shown in Figure 2, given the dihedral angle
$$
\begin{array}{l}
M-C D-N \text { is } \theta \\
\left(0^{\circ}<\theta<90^{\circ}\right), A \in \\
M, B \in N, A B \perp \\
C D, A B \text { forms a } 30^{\circ} \\
\text { angle with plane } N
\end{array}
$$

If the area of $\triangle A C D$ is $S$, for what value of $\theta$ is the area of $\triangle B C D$ maximized? What is the maximum value?
16. As shown in Figure 5, draw $A E \perp$ plane $N$, with $E$ as the foot of the perpendicular. Connect $B E$ and extend it to intersect $C D$ at point $F$, then connect $A F$.

Since $A E \perp$ plane $N$, $B E$ is the projection of $A B$ on plane $N$, and $A E \perp C D$.
Also, $A B \perp C D$, so $C D \perp$ plane $A B F$.
Therefore, $C D \perp A F, C D \perp B F$.
Thus, $\angle A F B$ is the plane angle of the dihedral angle $M-C D-N$, and $\angle A B F$ is the angle between $A B$ and plane $N$, i.e.,
$$
\angle A F B=\theta, \angle A B F=30^{\circ} \text {. }
$$

Given $S_{\triangle A C D}=\frac{1}{2} C D \cdot A F=S$,
$$
S_{\triangle B C D}=\frac{1}{2} C D \cdot B F,
$$

we have $\frac{S_{\triangle B C D}}{S}=\frac{B F}{A F}$.
In $\triangle A F B$, by the Law of Sines,
$$
\begin{array}{l}
\frac{B F}{A F}=\frac{\sin \angle B A F}{\sin 30^{\circ}} \\
=2 \sin \left[180^{\circ}-\left(\theta+30^{\circ}\right)\right] \\
=2 \sin \left(\theta+30^{\circ}\right),
\end{array}
$$

i.e., $\frac{S_{\triangle B C D}}{S}=2 \sin \left(\theta+30^{\circ}\right)$.
Since $0^{\circ}<\theta<90^{\circ}$, $\sin \left(\theta+30^{\circ}\right)$ reaches its maximum value of 1 when and only when $\theta=60^{\circ}$, i.e., $\frac{S_{\triangle B C D}}{S}$ reaches its maximum value of 2.
Therefore, when $\theta=60^{\circ}$, $S_{\triangle B C D}$ reaches its maximum value of $2 S$.