The diagram shows the front view of a rectangular block $ABCE$, with dimensions 14 cm by 8 cm, placed on a ramp, $VE$, tilted at an acute angle of $\theta^\circ$ and $\angle VTE = 90^\circ$. The ramp is placed on a horizontal surface $ET$ and $d$ is the perpendicular distance from $B$ to $ET$.

1. Identify the horizontal and vertical components of the rectangle's sides by drawing a horizontal line through C and a vertical line from B.

Let $F$ be the foot of the perpendicular from $C$ to the horizontal line $ET$. Let $G$ be the foot of the perpendicular from $B$ to the horizontal line passing through $C$.
The total vertical distance $d$ is the sum of the vertical distance from $E$ to $C$ (which is $CF$) and the vertical distance from $C$ to $B$ (which is $BG$).
$d = CF + BG$

2. Express the vertical segments in terms of $\theta$ using trigonometry.

In the right-angled triangle $CFE$, the angle $\angle CEF = \theta$ (since $CE$ lies on the ramp $VE$ which is inclined at $\theta$).
$CF = CE \sin\theta = 14 \sin\theta$

In the right-angled triangle $BGC$, the angle $\angle BCG = 90^\circ - \angle FCE = 90^\circ - (90^\circ - \theta) = \theta$.
$BG = BC \cos\theta = 8 \cos\theta$

Therefore, $d = CF + BG = 14 \sin\theta + 8 \cos\theta$.

1. Identify the form and expand $R \sin(\theta + \alpha)$

$d = 14 \sin\theta + 8 \cos\theta$
$R \sin(\theta + \alpha) = R \sin\theta \cos\alpha + R \cos\theta \sin\alpha$

2. Find the value of $R$

Comparing coefficients:
$R \cos\alpha = 14$
$R \sin\alpha = 8$
$R = \sqrt{14^2 + 8^2} = \sqrt{196 + 64} = \sqrt{260} = 2\sqrt{65} \approx 16.125$

3. Find the value of $\alpha$ and write the final expression

$\tan\alpha = \frac{8}{14} = \frac{4}{7}$
$\alpha = \tan^{-1}\left(\frac{4}{7}\right) \approx 29.744^\circ \approx 29.7^\circ$ (to 1 d.p.)
$d = 2\sqrt{65} \sin(\theta + 29.7^\circ)$

1. State the maximum value of $d$ using the $R$ value from part (b)

From part (b), $d = \sqrt{260} \sin(\theta + 29.745^\circ)$.
The maximum value of $\sin(\theta + \alpha)$ is $1$.
Maximum value of $d = \sqrt{260} \approx 16.1$ (to 3 s.f.)

2. Find the corresponding value of $\theta$

For maximum $d$, $\sin(\theta + 29.745^\circ) = 1$
$\theta + 29.745^\circ = 90^\circ$
$\theta = 90^\circ - 29.745^\circ = 60.255^\circ \approx 60.3^\circ$ (to 1 d.p.)

1. Set up the equation using the $R$-formula result from part (b)

From part (b), $R = \sqrt{14^2 + 8^2} = \sqrt{260}$ and $\tan \alpha = \frac{8}{14} \implies \alpha \approx 29.745^\circ$.

Equating $d$ to 13:
$\sqrt{260} \sin(\theta + 29.745^\circ) = 13$

2. Solve for the basic angle

$\sin(\theta + 29.745^\circ) = \frac{13}{\sqrt{260}}$
$\text{Basic angle} = \sin^{-1}\left(\frac{13}{\sqrt{260}}\right) \approx 53.727^\circ$

3. Find the smallest positive value of $\theta$

For the smallest value of $\theta$, we take the first quadrant solution:
$\theta + 29.745^\circ = 53.727^\circ$
$\theta = 53.727^\circ - 29.745^\circ = 23.982^\circ \approx 24.0^\circ \text{ (to 1 d.p.)}$