Four. (Full marks 20 points) As shown in Figure 8, point $O$ represents the sun, $\triangle A B C$ represents a triangular sunshade, and points $A$ and $B$ are two fixed points on the ground in the north-south direction. The sunlight $O C D$ from the due west direction casts the shadow of the sunshade onto the ground, forming the shadow $\triangle A B D$. It is known that the sunlight $O C D$ makes an acute angle $\theta$ with the ground.
(1) At what angle should the sunshade make with the ground to maximize the area of the shadow $\triangle A B D$?
(2) When $A C=3, B C=4, A B=5, \theta=30^{\circ}$, find the maximum area of the shadow $\triangle A B D$.
(1) As shown in Figure 17, draw the perpendicular line $O H$ from $O$ to the ground, connect $H D$ to intersect $A B$ at $E$, and connect $C E$. Then $H D$ is the projection of the oblique line $O D$ on the ground, and $\angle C D E = \theta$. Given that $A B$ is in the north-south direction and $C D$ is in the east-west direction, we have
$A B \perp C D$. (Implicit condition)

According to the inverse of the Three Perpendiculars Theorem, $D E \perp A B$. By the property theorem of line perpendicular to a plane, $A B \perp O H$. This implies that $A B$ is perpendicular to the intersecting lines $O H$ and $O D$, so $A B$ is perpendicular to the plane $C D E$, and thus,
$A B \perp C E$.
Hence, $C E$ is the altitude of $\triangle A B C$ on side $A B$, and $D E$ is the altitude of $\triangle A B D$ on side $A B$. The dihedral angle between the sunshade and the ground is $\angle C E D$.

In $\triangle A B D$, to maximize the area, with $A B$ and $\theta$ fixed, $D E$ should be maximized.
In $\triangle C E D$, we have
$$
\begin{aligned}
D E & =\frac{C E}{\sin \theta} \cdot \sin \angle C E D=\frac{2 S_{\triangle A B C}}{A B \sin \theta} \cdot \sin \angle C E D \\
& \leqslant \frac{2 S_{\triangle A B C}}{A B \sin \theta} \text { (a constant). }
\end{aligned}
$$

Therefore, when $\angle D C E = 90^{\circ}$ (the sunshade is perpendicular to the sunlight), $D E$ is maximized. Thus, when $\angle C E D = 90^{\circ} - \theta$, the area of the shadow $\triangle A B D$ is maximized.
(2) When $A C = 3$, $B C = 4$, $A B = 5$, and $\theta = 30^{\circ}$, $\triangle A B C$ is a right triangle, and $S_{\text {SMIX: }} = \frac{1}{2} A C \cdot B C = 6$.
Substituting (1) into the area formula, we get the maximum area as
$$
S_{\triangle A W B} = \frac{1}{2} A B \cdot D E = \frac{S_{\triangle A B C}}{\sin 30^{\circ}} = 12.
$$