The diagram shows two triangles DAB and DBC. It is given that BD = 5 m and BC = 2 m. Angle DAB = angle DBC = angle DEC = $90^{\circ}$. Angle ADB = $\theta$ and $\theta$ can vary.

1. Find the length of $AD$ in terms of $\theta$.

In the right-angled triangle $ABD$,
$\cos\theta = \frac{AD}{BD}$
$AD = 5\cos\theta$

2. Find the horizontal distance from $A$ to $E$ by analysing the angles around $B$.

Draw a horizontal line through $B$ to the right, meeting $CE$ at $F$.
Since $AB$ is vertical and $AD$ is horizontal, $\angle ABD = 90^{\circ} - \theta$.
The angle between $BD$ and the horizontal through $B$ is $\theta$.
Since $\angle DBC = 90^{\circ}$, the angle between $BC$ and the horizontal is $90^{\circ} - \theta$.
In the right-angled triangle $BFC$, the horizontal distance $BF$ (which equals $AE$) is:
$BF = BC \cos(90^{\circ} - \theta) = 2\sin\theta$
So, $AE = 2\sin\theta$.

3. Express $DE$ as the difference between $AD$ and $AE$.

From the diagram, $E$ lies on $AD$, so:
$DE = AD - AE$
$DE = 5\cos\theta - 2\sin\theta$
This is in the form $a\cos\theta - b\sin\theta$ with $a = 5$ and $b = 2$.

1. Equate the expression from part (i) to the R-formula form and expand

From (i), $DE = 5\cos\theta - 2\sin\theta$.
Let $5\cos\theta - 2\sin\theta = R\cos(\theta + \alpha)$.
$5\cos\theta - 2\sin\theta = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$

2. Compare coefficients to find $R$ and $\alpha$

Comparing coefficients:
$R\cos\alpha = 5$
$R\sin\alpha = 2$

$R = \sqrt{5^2 + 2^2} = \sqrt{29} \approx 5.385$
$\tan\alpha = \frac{2}{5}$
$\alpha = \tan^{-1}(0.4) \approx 21.801^{\circ}$

3. State the final expression

$DE = \sqrt{29}\cos(\theta + 21.8^{\circ})$

1. Express CE in terms of trigonometric ratios of theta

Let $F$ be the foot of the perpendicular from $B$ to $CE$. Then $CE = CF + FE$. Since $ABED$ is a rectangle, $FE = AB = 5\sin\theta$. In $\triangle BCF$, $\angle CBF = \theta$ (since $\angle ABD = 90^\circ - \theta$ and $\angle DBC = 90^\circ$). Thus $CF = 2\cos\theta$. Therefore, $CE = 5\sin\theta + 2\cos\theta$.

2. Convert the expression for CE into the R-formula form

We want to express $5\sin\theta + 2\cos\theta$ in the form $R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$.
Comparing coefficients:
$R\cos\alpha = 5$
$R\sin\alpha = 2$
$R = \sqrt{5^2 + 2^2} = \sqrt{29}$
$\tan\alpha = \frac{2}{5} \implies \alpha = \tan^{-1}(0.4) \approx 21.801^{\circ}$
Thus, $CE = \sqrt{29}\sin(\theta + 21.8^{\circ})$.

1. Write down the formula for the area of triangle CDE using the expressions for DE and CE.

Area of $\triangle CDE = \frac{1}{2} \times DE \times CE$
From (ii) and (iii), $DE = R\cos(\theta + \alpha)$ and $CE = R\sin(\theta + \alpha)$.
Area $= \frac{1}{2} [R\cos(\theta + \alpha)][R\sin(\theta + \alpha)]$
Area $= \frac{1}{2} R^2 \sin(\theta + \alpha) \cos(\theta + \alpha)$

2. Apply the double angle identity for sine.

Using $\sin(2A) = 2\sin A \cos A$, we have $\sin(\theta + \alpha) \cos(\theta + \alpha) = \frac{1}{2}\sin(2\theta + 2\alpha)$.
Area $= \frac{1}{2} R^2 \left[ \frac{1}{2}\sin(2\theta + 2\alpha) \right] = \frac{1}{4} R^2 \sin(2\theta + 2\alpha)$

3. Determine the value of $R^2$ and hence $p$.

From (i) and (ii), $DE = 5\cos\theta - 2\sin\theta$, so $R = \sqrt{5^2 + 2^2} = \sqrt{29}$.
$R^2 = 29$
Area $= \frac{1}{4} (29) \sin(2\theta + 2\alpha) = \frac{29}{4} \sin(2\theta + 2\alpha)$
Comparing with $p\sin(2\theta + 2\alpha)$, we get $p = \frac{29}{4}$.