A L-shaped structure, OPQ can be rotated about O. OP and PQ measures 5 m and 1 m respectively. OP makes an acute angle, $\theta$, with the ground. The shortest distance from Q to the wall is $L$ m.

1. Find the horizontal distance from O to P.

Let the horizontal distance from O to P be $x_1$.
In the right-angled triangle formed by OP and the ground,
$\cos\theta = \frac{x_1}{5} \implies x_1 = 5\cos\theta$

2. Find the horizontal distance from Q to P and subtract it to find L.

Let the horizontal distance from Q to P be $x_2$.
Since PQ is perpendicular to OP, the angle PQ makes with the vertical is $\theta$.
In the right-angled triangle formed by PQ and its horizontal/vertical components,
$\sin\theta = \frac{x_2}{1} \implies x_2 = \sin\theta$
Since Q is to the left of P, the horizontal distance from O to Q (which is the distance $L$ to the wall at O) is:
$L = x_1 - x_2 = 5\cos\theta - \sin\theta$ (Shown)

1. Expand the target form and compare coefficients with the given expression

$R\cos(\theta+\alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$
Comparing with $L = 5\cos\theta - \sin\theta$:
$R\cos\alpha = 5$
$R\sin\alpha = 1$

2. Calculate the value of R

$R = \sqrt{5^2 + 1^2} = \sqrt{26} \approx 5.0990$

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

$\tan\alpha = \frac{1}{5}$
$\alpha = \tan^{-1}(0.2) \approx 11.3099^\circ \approx 11.3^\circ$ (to 1 d.p.)
$L = \sqrt{26}\cos(\theta + 11.3^\circ)$

1. Identify the minimum value of the trigonometric expression

From part (b), $L = \sqrt{26}\cos(\theta + 11.31^\circ)$.
The minimum value of a cosine function is $-1$.
Minimum value of $L = -\sqrt{26}$

2. Find the corresponding value of theta

For the minimum value, $\cos(\theta + 11.31^\circ) = -1$
$\theta + 11.31^\circ = 180^\circ$
$\theta = 180^\circ - 11.31^\circ = 168.69^\circ \approx 168.7^\circ$

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

From part (b), $L = \sqrt{26}\cos(\theta + 11.31^\circ)$.
Given $L = 3$, we have:
$\sqrt{26}\cos(\theta + 11.3099^\circ) = 3$
$\cos(\theta + 11.3099^\circ) = \frac{3}{\sqrt{26}}$

2. Solve for the acute angle $\theta$

$\theta + 11.3099^\circ = \cos^{-1}\left(\frac{3}{\sqrt{26}}\right)$
$\theta + 11.3099^\circ \approx 53.960^\circ$
$\theta = 53.960^\circ - 11.3099^\circ = 42.650^\circ \approx 42.7^\circ$

1. Identify the condition for $L$ to reach its theoretical maximum of $R$

For $L = R\cos(\theta + \alpha)$ to reach its maximum value of $R$, we must have $\cos(\theta + \alpha) = 1$.
This occurs when $\theta + \alpha = 0^\circ$, which means $\theta = -\alpha$.

2. Explain why this condition is impossible given the context

The question states that $\theta$ is an acute angle, so $0^\circ < \theta < 90^\circ$.
Since $\alpha > 0^\circ$, $\theta$ would have to be negative to satisfy $\theta = -\alpha$.
Therefore, $\theta$ cannot take the value required to make $L = R$.