The diagram shows an area that is enclosed by glass panels at $AB$, $BC$ and $AD$. $AB = 10\text{ m}$, $AD = 30\text{ m}$, angle $DAB = \text{angle } BCD = 90^\circ$. The glass panel $AD$ makes an acute angle $\theta$ with $CD$.

1. Express the length of the glass panels in terms of the given sides.

The total length of the glass panels is $L = AB + BC + AD$.
Given $AB = 10$ and $AD = 30$,
$L = 10 + BC + 30 = 40 + BC$.

2. Find the length of BC by projecting AD and AB onto the vertical axis.

Draw a horizontal line through $A$ and a vertical line through $D$ to form a right-angled triangle with $AD$. The vertical height of $A$ above $CD$ is $AD \sin\theta = 30\sin\theta$.
Since $\angle DAB = 90^\circ$, the angle $AB$ makes with the vertical is $\theta$.
The vertical drop from $A$ to $B$ is $AB \cos\theta = 10\cos\theta$.
Therefore, the vertical height of $B$ above $CD$ (which is the length of $BC$) is:
$BC = 30\sin\theta - 10\cos\theta$.

3. Substitute BC into the expression for L to complete the proof.

$L = 40 + (30\sin\theta - 10\cos\theta)$
$L = 40 + 30\sin\theta - 10\cos\theta$ (Shown)

1. Set up the R-formula equation for the trigonometric part

Let $30\sin\theta - 10\cos\theta = R\sin(\theta - \alpha)$.
Expanding the right side gives:
$30\sin\theta - 10\cos\theta = R(\sin\theta\cos\alpha - \cos\theta\sin\alpha)$
Comparing coefficients:
$R\cos\alpha = 30$
$R\sin\alpha = 10$

2. Calculate the value of R

$R = \sqrt{30^2 + 10^2}$
$R = \sqrt{900 + 100} = \sqrt{1000} = 10\sqrt{10} \approx 31.623$

3. Calculate the value of alpha

$\tan\alpha = \frac{10}{30} = \frac{1}{3}$
Since $0^\circ < \alpha < 90^\circ$,
$\alpha = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.435^\circ$

4. State the final expression for L

$L = 40 + 31.6\sin(\theta - 18.4^\circ)$ (to 3 s.f. and 1 d.p. respectively)

1. Find the values of $R$ and $\alpha$ from part (ii)

From part (ii), $30\sin\theta - 10\cos\theta = R\sin(\theta - \alpha)$.
$R = \sqrt{30^2 + 10^2} = \sqrt{1000} = 10\sqrt{10}$
$\tan\alpha = \frac{10}{30} = \frac{1}{3} \implies \alpha \approx 18.435^\circ$
So, $L = 40 + \sqrt{1000}\sin(\theta - 18.435^\circ)$

2. Set up the equation for $L = 65$ and isolate the sine term

$40 + \sqrt{1000}\sin(\theta - 18.435^\circ) = 65$
$\sqrt{1000}\sin(\theta - 18.435^\circ) = 25$
$\sin(\theta - 18.435^\circ) = \frac{25}{\sqrt{1000}}$

3. Solve for $\theta$

$\theta - 18.435^\circ = \sin^{-1}\left(\frac{25}{\sqrt{1000}}\right)$
$\theta - 18.435^\circ \approx 52.239^\circ$
$\theta \approx 70.674^\circ \approx 70.7^\circ$ (to 1 d.p.)
(The other solution $\theta - 18.435^\circ = 180^\circ - 52.239^\circ = 127.761^\circ \implies \theta = 146.2^\circ$ is rejected as $\theta$ is acute)

1. Determine the maximum value of the expression for L

From (ii), $L = 40 + \sqrt{1000}\sin(\theta - 18.43^\circ) = 40 + 10\sqrt{10}\sin(\theta - 18.43^\circ)$.
The maximum value of $\sin(\theta - 18.43^\circ)$ is 1.
Maximum $L = 40 + 10\sqrt{10} \approx 71.62\text{ m}$.

2. Compare the maximum length with 90 m and conclude

Since $71.62\text{ m} < 90\text{ m}$, it is not possible for the length of the glass panels to be 90 m.