竞技跳台跳水比赛包括一名或两名运动员在不同高度处向水池中进行旋转、转体和翻腾动作。 图中显示了跳台设计的一个横截面，它由一个梯形塔 $ABCD$ 组成。$AB$ 是运动员通往跳台的楼梯。 已知 $\angle BAD = \theta$（其中 $\theta$ 是可变的），且 $\angle ADC = 90^\circ$，$AB$ 和 $BC$ 的长度分别为 $12\text{ m}$ 和 $4\text{ m}$。

1. Find the lengths of CD and AD in terms of theta

Draw a horizontal line from $B$ to meet $CD$ at $E$. Then $AB$ is the hypotenuse of right-angled triangle $ABE$ (or draw a vertical line from $B$ to $AD$ at $F$).
Let $F$ be the foot of the perpendicular from $B$ to $AD$.
In $\triangle ABF$, $\angle AFB = 90^\circ$ and $\angle BAF = \theta$.
$BF = 12\sin\theta$ and $AF = 12\cos\theta$.
Since $BCDF$ is a rectangle, $CD = BF = 12\sin\theta$ and $FD = BC = 4$.
Thus, $AD = AF + FD = 12\cos\theta + 4$.

2. Express the perimeter L in terms of theta

$L = AB + BC + CD + AD$
$L = 12 + 4 + 12\sin\theta + (12\cos\theta + 4)$
$L = 20 + 12\sin\theta + 12\cos\theta$

1. Determine the expression for $L$ from the context of part (a)

From the diagram, drawing a perpendicular from $B$ to $AD$ gives a right-angled triangle with hypotenuse $AB = 12$.
The vertical height is $12\sin\theta$, so $CD = 12\sin\theta$.
The horizontal distance is $12\cos\theta$, so $AD = 12\cos\theta + 4$.
The perimeter $L = AB + BC + CD + AD = 12 + 4 + 12\sin\theta + (12\cos\theta + 4) = 20 + 12\sin\theta + 12\cos\theta$.

2. Apply the R-formula to $12\cos\theta + 12\sin\theta$

Let $12\cos\theta + 12\sin\theta = R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Comparing coefficients:
$R\cos\alpha = 12$
$R\sin\alpha = 12$
$R = \sqrt{12^2 + 12^2} = \sqrt{288} = 12\sqrt{2}$
$\tan\alpha = \frac{12}{12} = 1 \implies \alpha = 45^\circ$

3. State the final expression for $L$

$L = 20 + 12\sqrt{2}\cos(\theta - 45^\circ)$

1. State the expression for $L$ from part (b)

From parts (a) and (b), $L = 20 + 12\sin\theta + 12\cos\theta = 20 + 12\sqrt{2}\cos(\theta - 45^\circ)$.

2. Find the maximum value of $L$

The maximum value of $\cos(\theta - 45^\circ)$ is $1$.
Maximum $L = 20 + 12\sqrt{2}(1) = 20 + 12\sqrt{2} \approx 37.0\text{ m}$

3. Find the corresponding value of $\theta$

$\cos(\theta - 45^\circ) = 1$
$\theta - 45^\circ = 0^\circ$
$\theta = 45^\circ$