图示为两个直角三角形 $ABC$ 和 $DEC$。$\angle BAC = \angle CDE = \theta$，$AB = 3.5\text{ cm}$ 且 $DE = 6\text{ cm}$。

1. Express the length of $BC$ in terms of $\theta$ using triangle $ABC$.

In $\triangle ABC$, $\angle BCA = 90^\circ$.
$\sin \theta = \frac{BC}{AB}$
$BC = 3.5 \sin \theta$

2. Express the length of $CD$ in terms of $\theta$ using triangle $DEC$.

In $\triangle DEC$, $\angle ECD = 90^\circ$.
$\cos \theta = \frac{CD}{DE}$
$CD = 6 \cos \theta$

3. Combine $BC$ and $CD$ to find the total length of $BD$.

$BD = BC + CD$
$BD = 3.5 \sin \theta + 6 \cos \theta$
$BD = 6 \cos \theta + 3.5 \sin \theta \text{ (Shown)}$

1. Find the value of $R$

Let $6 \cos \theta + 3.5 \sin \theta = R \cos(\theta - \alpha) = R \cos \theta \cos \alpha + R \sin \theta \sin \alpha$.
Comparing coefficients:
$R \cos \alpha = 6$
$R \sin \alpha = 3.5$
$R = \sqrt{6^2 + 3.5^2}$
$R = \sqrt{36 + 12.25} = \sqrt{48.25} \approx 6.9462$

2. Find the value of $\alpha$

$\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{3.5}{6}$
$\alpha = \tan^{-1}\left(\frac{3.5}{6}\right) \approx 30.256^\circ$

3. Write the final expression

$BD = 6.95 \cos(\theta - 30.3^\circ)$ (to 3 s.f. and 1 d.p. respectively)

1. Identify the maximum value of BD from the R-formula form

From part (b), $BD = R \cos(\theta - \alpha)$.
$R = \sqrt{6^2 + 3.5^2} = \sqrt{36 + 12.25} = \sqrt{48.25} \approx 6.9462$
$\tan \alpha = \frac{3.5}{6} \implies \alpha \approx 30.256^\circ$
So $BD = \sqrt{48.25} \cos(\theta - 30.256^\circ)$.
The maximum value of $\cos(\theta - \alpha)$ is $1$.
Maximum length of $BD = \sqrt{48.25} \approx 6.95\text{ cm}$ (to 3 s.f.).

2. Find the corresponding value of theta

For maximum $BD$, $\cos(\theta - 30.256^\circ) = 1$.
$\theta - 30.256^\circ = 0^\circ$
$\theta = 30.256^\circ \approx 30.3^\circ$ (to 1 d.p.).

1. Recall the $R$-formula expression for $BD$ from part (b)

From part (b), $BD = R \cos(\theta - \alpha)$ where $R = \sqrt{6^2 + 3.5^2} = \sqrt{48.25}$ and $\alpha = \tan^{-1}\left(\frac{3.5}{6}\right) \approx 30.256^\circ$.
So, $BD = \sqrt{48.25} \cos(\theta - 30.256^\circ)$.

2. Set up the equation for $BD = 6$ and solve for $\theta$

$\sqrt{48.25} \cos(\theta - 30.256^\circ) = 6$
$\cos(\theta - 30.256^\circ) = \frac{6}{\sqrt{48.25}}$
$\theta - 30.256^\circ = \cos^{-1}\left(\frac{6}{\sqrt{48.25}}\right)$
$\theta - 30.256^\circ = 30.256^\circ$
$\theta = 60.512^\circ \approx 60.5^\circ$ (to 1 d.p.)