如图所示为一块梯形纸板，其中 $\angle BAD$ 和 $\angle ABC$ 均为直角。$CD = 20\text{ cm}$，$BC = 8\text{ cm}$ 且 $\angle ADC = \theta$ 弧度。

1. Construct a right-angled triangle to find the unknown lengths.

Draw a perpendicular line from $C$ to $AD$, meeting $AD$ at $E$.
Since $\angle BAD = 90^\circ$ and $\angle ABC = 90^\circ$, $ABCE$ is a rectangle.
Therefore, $AE = BC = 8$ and $AB = CE$.

2. Use trigonometry to find the lengths of $CE$ and $DE$.

In the right-angled $\triangle CDE$:
$\sin\theta = \frac{CE}{CD} \implies CE = 20\sin\theta$
$\cos\theta = \frac{DE}{CD} \implies DE = 20\cos\theta$

3. Express the lengths of $AB$ and $AD$ in terms of $\theta$.

$AB = CE = 20\sin\theta$
$AD = AE + DE = 8 + 20\cos\theta$

4. Calculate the perimeter of the trapezium.

$P = AB + BC + CD + AD$
$P = 20\sin\theta + 8 + 20 + (8 + 20\cos\theta)$
$P = 20\cos\theta + 20\sin\theta + 36$ (Shown)

1. Find the value of $R$

Let $20\cos\theta + 20\sin\theta = R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Comparing coefficients:
$R\cos\alpha = 20$
$R\sin\alpha = 20$
$R = \sqrt{20^2 + 20^2} = \sqrt{800} = 20\sqrt{2}$

2. Find the value of $\alpha$ in radians

$\tan\alpha = \frac{20}{20} = 1$
Since $\theta$ is in radians, $\alpha = \tan^{-1}(1) = \frac{\pi}{4}$

3. Write the final expression for $P$

$P = 20\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) + 36$

1. Express $P$ using the result from part (ii) to find its maximum value

From part (ii), $20\cos\theta + 20\sin\theta = R\cos(\theta - \alpha)$.
$R = \sqrt{20^2 + 20^2} = \sqrt{800} = 20\sqrt{2}$
$\tan\alpha = \frac{20}{20} = 1 \implies \alpha = \frac{\pi}{4}$ rad.
So, $P = 20\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) + 36$.
The maximum value of $P$ occurs when $\cos\left(\theta - \frac{\pi}{4}\right) = 1$.
Maximum $P = 20\sqrt{2}(1) + 36 = 20\sqrt{2} + 36 \approx 64.3$ cm.

2. Find the corresponding value of $\theta$

For maximum $P$, $\cos\left(\theta - \frac{\pi}{4}\right) = 1$.
$\theta - \frac{\pi}{4} = 0$
$\theta = \frac{\pi}{4} \approx 0.785$ rad.

1. Equate the R-formula expression for perimeter to 60

From part (ii), $P = 20\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) + 36$.
$20\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) + 36 = 60$

2. Isolate the cosine term

$20\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) = 24$
$\cos\left(\theta - \frac{\pi}{4}\right) = \frac{24}{20\sqrt{2}} = \frac{6}{5\sqrt{2}} \approx 0.8485$

3. Solve for $\theta$

Basic angle $\alpha = \cos^{-1}\left(\frac{6}{5\sqrt{2}}\right) \approx 0.55759$
Since $\theta$ is an acute angle in the trapezium, we consider the principal and fourth quadrant solutions for the argument:
$\theta - \frac{\pi}{4} = \pm 0.55759$
$\theta = \frac{\pi}{4} \pm 0.55759$
$\theta \approx 0.78540 \pm 0.55759$
$\theta \approx 1.34299 \text{ or } 0.22781$