The diagram shows a figure $ABCD$, in which $BC=13\ \text{m}$, $CD=6\ \text{m}$, angle $BAD = 90^\circ$ and angle $ADC =$ angle $ABC = \theta$.

1. Express the lengths of $AB$ and $AD$ in terms of $\theta$ by drawing perpendiculars from $C$.

Let $E$ be the foot of the perpendicular from $C$ to $AB$, and $F$ be the foot of the perpendicular from $C$ to $AD$.
In $\triangle BCE$, $\angle BEC = 90^\circ$ and $\angle EBC = \theta$:
$BE = 13\cos\theta$
$CE = 13\sin\theta$

In $\triangle CDF$, $\angle CFD = 90^\circ$ and $\angle FDC = \theta$:
$CF = 6\sin\theta$
$DF = 6\cos\theta$

2. Find the lengths of $AB$ and $AD$.

Since $AECF$ forms a rectangle, $AE = CF$ and $AF = CE$.
$AB = AE + BE = CF + BE = 6\sin\theta + 13\cos\theta$
$AD = AF + DF = CE + DF = 13\sin\theta + 6\cos\theta$

3. Calculate the total perimeter $L$.

$L = AB + BC + CD + AD$
$L = (6\sin\theta + 13\cos\theta) + 13 + 6 + (13\sin\theta + 6\cos\theta)$
$L = 19 + 19\cos\theta + 19\sin\theta$ (shown)

1. Find the value of $R$ for the expression $19\cos\theta + 19\sin\theta$

$R = \sqrt{19^2 + 19^2} = \sqrt{722} = 19\sqrt{2}$

2. Find the value of $\alpha$

$\tan\alpha = \frac{19}{19} = 1$
$\alpha = \tan^{-1}(1) = 45^\circ$

3. Express $L$ in the required form

$L = 19 + 19\sqrt{2}\cos(\theta - 45^\circ)$

4. Find the maximum possible perimeter

The maximum value of $\cos(\theta - 45^\circ)$ is $1$.
Maximum perimeter $= 19 + 19\sqrt{2}(1) = 19 + 19\sqrt{2} \approx 45.9\text{ m}$

1. Express $L$ using the $R$-formula result from part (b)

From part (b), $19\cos\theta + 19\sin\theta = R\cos(\theta - \alpha)$.
$R = \sqrt{19^2 + 19^2} = 19\sqrt{2}$
$\tan\alpha = \frac{19}{19} = 1 \implies \alpha = 45^\circ$
So, $L = 19 + 19\sqrt{2}\cos(\theta - 45^\circ)$

2. Set up the equation for $L = 45$ and isolate the cosine term

$19 + 19\sqrt{2}\cos(\theta - 45^\circ) = 45$
$19\sqrt{2}\cos(\theta - 45^\circ) = 26$
$\cos(\theta - 45^\circ) = \frac{26}{19\sqrt{2}} \approx 0.96761$

3. Solve for $\theta$

Basic angle $= \cos^{-1}\left(\frac{26}{19\sqrt{2}}\right) \approx 14.62^\circ$
$\theta - 45^\circ = -14.62^\circ \text{ or } 14.62^\circ$
$\theta = 45^\circ - 14.62^\circ \text{ or } 45^\circ + 14.62^\circ$
$\theta = 30.38^\circ \text{ or } 59.62^\circ$
$\theta \approx 30.4^\circ \text{ or } 59.6^\circ$ (to 1 d.p.)