The diagram shows a pair of similar triangles $YUX$ and $VUW$. Point $Y$ lies on the straight line $VU$ such that $VY = 12\text{ cm}$ and $YU = 20\text{ cm}$. Angle $UVW = \theta$ where $0^\circ \le \theta \le 90^\circ$ and $UW$ is perpendicular to $VW$.

1. Identify the lengths of the sides of the large triangle VUW

In $\triangle VUW$, $\angle UVW = \theta$ and $\angle UWV = 90^\circ$.
The hypotenuse $VU = VY + YU = 12 + 20 = 32\text{ cm}$.
Using trigonometric ratios:
$\sin\theta = \frac{UW}{VU} \implies UW = 32\sin\theta$
$\cos\theta = \frac{VW}{VU} \implies VW = 32\cos\theta$

2. Identify the lengths of the sides of the small triangle YUX

Since $\triangle YUX$ is similar to $\triangle VUW$, $\angle UYX = \angle UVW = \theta$ and $\angle UXY = 90^\circ$.
The hypotenuse $YU = 20\text{ cm}$.
Using trigonometric ratios:
$\sin\theta = \frac{UX}{YU} \implies UX = 20\sin\theta$
$\cos\theta = \frac{YX}{YU} \implies YX = 20\cos\theta$

3. Calculate the length of XW and the perimeter of the trapezium VYXW

The length $XW = UW - UX = 32\sin\theta - 20\sin\theta = 12\sin\theta$.
The perimeter of trapezium $VYXW$ is $L = VY + YX + XW + VW$.
$L = 12 + 20\cos\theta + 12\sin\theta + 32\cos\theta$
$L = 12\sin\theta + 52\cos\theta + 12$ (Shown)

1. Set up the R-formula equation for the trigonometric part of $L$.

Let $12\sin\theta + 52\cos\theta = R\sin(\theta + \alpha)$.
$R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$
Comparing coefficients:
$R\cos\alpha = 12$
$R\sin\alpha = 52$

2. Calculate the value of $R$.

$R = \sqrt{12^2 + 52^2}$
$R = \sqrt{144 + 2704} = \sqrt{2848} \approx 53.366$

3. Calculate the value of $\alpha$ and write the final expression.

$\tan\alpha = \frac{52}{12}$
$\alpha = \tan^{-1}\left(\frac{13}{3}\right) \approx 77.005^\circ \approx 77.0^\circ$ (to 1 d.p.)
$L = 12 + 53.4\sin(\theta + 77.0^\circ)$

1. Find the maximum value of $L$ using the $R$ value from part (b)

From part (b), $L = 12 + \sqrt{2848}\sin(\theta + 77.005^\circ)$.
The maximum value of $\sin(\theta + 77.005^\circ)$ is $1$.
Maximum $L = 12 + \sqrt{2848} = 12 + 4\sqrt{178} \approx 65.4$

2. Find the corresponding value of $\theta$

For maximum $L$, $\sin(\theta + 77.005^\circ) = 1$
$\theta + 77.005^\circ = 90^\circ$
$\theta = 90^\circ - 77.005^\circ = 12.995^\circ \approx 13.0^\circ$

1. Find the values of $R$ and $\alpha$ from part (b) to use in the equation

$12\sin\theta + 52\cos\theta = R\sin(\theta + \alpha)$
$R = \sqrt{12^2 + 52^2} = \sqrt{2848}$
$\tan\alpha = \frac{52}{12} \implies \alpha = 77.005^\circ$
$L = 12 + \sqrt{2848}\sin(\theta + 77.005^\circ)$

2. Set up the equation for $L = 30$ and isolate the sine term

$12 + \sqrt{2848}\sin(\theta + 77.005^\circ) = 30$
$\sqrt{2848}\sin(\theta + 77.005^\circ) = 18$
$\sin(\theta + 77.005^\circ) = \frac{18}{\sqrt{2848}} \approx 0.33729$

3. Solve for $\theta$ in the given domain $0^\circ \le \theta \le 90^\circ$

Basic angle $= \sin^{-1}(0.33729) \approx 19.712^\circ$
Since $0^\circ \le \theta \le 90^\circ$, the angle $\theta + 77.005^\circ$ must be in the second quadrant.
$\theta + 77.005^\circ = 180^\circ - 19.712^\circ = 160.288^\circ$
$\theta = 160.288^\circ - 77.005^\circ = 83.283^\circ \approx 83.3^\circ$