A competitive platform diving competition involves one or two divers performing spins, twists and somersaults into a pool from a series of heights. The diagram shows the cross-section of a design of the diving platform, which consists of a trapezium-shaped tower $ABCD$. $AB$ is the stairway which the divers use to reach the platform. Given that $\angle BAD = \theta$, where $\theta$ can vary and $\angle ADC = 90^\circ$, the length of $AB$ and $BC$ are $12\text{ m}$ and $4\text{ m}$ respectively.

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 $\theta$

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