The diagram below shows two horizontal lines $AB$ and $CD$, where $L$ is the vertical distance between the lines $AB$ and $CD$.
A rectangle $PQRS$ touches lines $AB$ and $CD$ at $P$ and $R$ respectively.
It is given that $PQ = 6\text{ cm}$, $SP = 8\text{ cm}$ and $\angle QRD = \theta^\circ$.

1. Find the vertical distance from $R$ to $Q$.

Let $U$ be the foot of the perpendicular from $Q$ to the line $CD$.
In right-angled $\triangle RQU$, $QR = SP = 8\text{ cm}$.
The vertical distance $QU = QR \sin\theta = 8\sin\theta$.

2. Find the vertical distance from $Q$ to $P$.

Let $V$ be the foot of the perpendicular from $P$ to the horizontal line passing through $Q$.
Since $PQ \perp QR$ and $QR$ makes an angle of $\theta$ with the horizontal, $PQ$ makes an angle of $90^\circ - \theta$ with the horizontal (or $\theta$ with the vertical).
In right-angled $\triangle PVQ$, $PQ = 6\text{ cm}$.
The vertical distance $PV = PQ \sin(90^\circ - \theta) = 6\cos\theta$.

3. Sum the vertical components to find $L$.

The total vertical distance $L$ between the parallel lines $AB$ and $CD$ is the sum of the vertical distances $QU$ and $PV$.
$L = QU + PV$
$L = 8\sin\theta + 6\cos\theta$ (Shown)

1. Expand $R\sin(\theta + \alpha)$ and equate coefficients

$8\sin\theta + 6\cos\theta = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$
Comparing coefficients:
$R\cos\alpha = 8$
$R\sin\alpha = 6$

2. Find the value of $R$

$R = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

3. Find the value of $\alpha$

$\tan\alpha = \frac{6}{8} = 0.75$
$\alpha = \tan^{-1}(0.75) \approx 36.8698^\circ \approx 36.87^\circ$ (to 2 d.p.)
So, $8\sin\theta + 6\cos\theta = 10\sin(\theta + 36.87^\circ)$

1. State the maximum value of L using the R-formula expression

From part (b), $L = 10\sin(\theta + 36.87^\circ)$.
The maximum value of $\sin(\theta + 36.87^\circ)$ is $1$.
Therefore, the maximum value of $L$ is $10(1) = 10$.

2. Find the corresponding value of theta

For $L$ to be maximum, $\sin(\theta + 36.87^\circ) = 1$.
$\theta + 36.87^\circ = 90^\circ$
$\theta = 90^\circ - 36.87^\circ = 53.13^\circ \approx 53.1^\circ$ (to 1 d.p.)

1. Determine the range of the angle in the sine function

From part (b), $L = 10\sin(\theta + 36.87^\circ)$.
Given $0^\circ \le \theta < 90^\circ$, the angle $(\theta + 36.87^\circ)$ ranges from $36.87^\circ$ to $< 126.87^\circ$.

2. Find the minimum value of L and the corresponding theta

In the interval $[36.87^\circ, 126.87^\circ)$, the sine function reaches its minimum at the lowest boundary, which is $36.87^\circ$.
Thus, the minimum value of $L$ occurs when $\theta = 0^\circ$.
Minimum $L = 10\sin(0^\circ + 36.87^\circ) = 10(0.6) = 6$.