The diagram above shows the side view of a bus stop shelter $BCD$ such that $BC = 4\text{ m}$, $CD = 1\text{ m}$, angle $BCD = 90^\circ$ and angle $CBA = \theta$. $AB$ is a concrete pavement under the shelter such that $DA$ is perpendicular to $AB$.

1. Construct perpendiculars to form right-angled triangles

Draw a line from $C$ perpendicular to $AB$, meeting $AB$ at $E$.
Draw a line from $D$ perpendicular to $CE$, meeting $CE$ at $F$.

2. Find the length of $BE$ using triangle $BCE$

In right-angled $\triangle BCE$, $\angle CBE = \theta$.
$\cos\theta = \frac{BE}{BC} = \frac{BE}{4}$
$BE = 4\cos\theta$

3. Determine the angle $FCD$ and find the length of $FD$

$\angle BCE = 90^\circ - \theta$.
Since $\angle BCD = 90^\circ$, $\angle FCD = 90^\circ - \angle BCE = 90^\circ - (90^\circ - \theta) = \theta$.
In right-angled $\triangle CDF$, $\sin\theta = \frac{FD}{CD} = \frac{FD}{1}$.
$FD = \sin\theta$

4. Express $AB$ as the sum of segments

Since $DA \perp AB$, $CE \perp AB$, and $DF \perp CE$, the quadrilateral $FDAE$ is a rectangle.
Therefore, $EA = FD = \sin\theta$.
$AB = BE + EA = 4\cos\theta + \sin\theta$ (shown).

1. Expand the R-formula and compare coefficients with the given expression

Let $4\cos\theta + \sin\theta = R\cos(\theta - \alpha)$.
Using the addition formula, $R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.
Comparing coefficients:
$R\cos\alpha = 4$
$R\sin\alpha = 1$

2. Calculate the value of R

$R = \sqrt{4^2 + 1^2}$
$R = \sqrt{17} \approx 4.123$

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

$\tan\alpha = \frac{1}{4}$
$\alpha = \tan^{-1}(0.25) \approx 14.036^\circ$
$\alpha \approx 14.0^\circ$ (to 1 d.p.)

$AB = \sqrt{17}\cos(\theta - 14.0^\circ)$

1. Determine the R-formula expression from part (ii)

From (i), $AB = 4\cos\theta + \sin\theta$.
Using $R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$, we have:
$R\cos\alpha = 4$
$R\sin\alpha = 1$
$R = \sqrt{4^2 + 1^2} = \sqrt{17}$
$\tan\alpha = \frac{1}{4} \implies \alpha \approx 14.036^\circ$
So $AB = \sqrt{17}\cos(\theta - 14.036^\circ)$.

2. State the maximum value of AB

The maximum value of $\cos(\theta - 14.036^\circ)$ is $1$.
Maximum value of $AB = \sqrt{17}$

3. Find the corresponding value of theta

For maximum $AB$, $\cos(\theta - 14.036^\circ) = 1$
$\theta - 14.036^\circ = 0^\circ$
$\theta = 14.036^\circ \approx 14.0^\circ$ (to 1 d.p.)