图中显示了一个长为 5 m、宽为 2 m 的矩形木板 $ABCD$ 靠在垂直墙壁上。$AD$ 与水平地面成锐角 $\theta$。 $E$ 是垂直墙壁上的一点，使得 $EB$ 平行于地面；$F$ 是地面上的一点，使得 $AF$ 垂直于 $FD$。点 $B$ 在地面上方垂直高度为 $h$ m 处。

1. Identify the vertical components of $h$

Let the vertical line from $B$ meet the horizontal line through $A$ at $G$, and let the horizontal line through $A$ meet the vertical line from $D$ at $H$. Then $h$ is the sum of the vertical distance from $B$ to $A$ and the vertical distance from $A$ to the ground. Alternatively, draw a horizontal line through $A$ meeting the vertical from $B$ at $P$, and note the vertical distance from $A$ to the ground is the same as the vertical distance from $D$ to the horizontal line through $A$. Let's use the standard projection method: $h$ is the sum of the vertical projection of $BA$ and the vertical projection of $AD$.

2. Express the vertical projections in terms of $\theta$

The angle $AD$ makes with the horizontal is $\theta$. The vertical projection of $AD$ is $AD \sin\theta = 5\sin\theta$.
The angle $BA$ makes with the vertical is $\theta$ (since $AB \perp AD$). Thus, the vertical projection of $BA$ is $AB \cos\theta = 2\cos\theta$.
Therefore, the total height $h = 5\sin\theta + 2\cos\theta$.

1. Expand $R\sin(\theta + \alpha)$ and equate coefficients to the expression for $h$.

From part (a), $h = 5\sin\theta + 2\cos\theta$.
Let $5\sin\theta + 2\cos\theta = R\sin(\theta + \alpha) = R(\sin\theta\cos\alpha + \cos\theta\sin\alpha)$.
Comparing coefficients:
$R\cos\alpha = 5$
$R\sin\alpha = 2$

2. Calculate the value of $R$.

$R = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}$

3. Calculate the value of $\alpha$.

$\tan\alpha = \frac{R\sin\alpha}{R\cos\alpha} = \frac{2}{5}$
$\alpha = \tan^{-1}\left(\frac{2}{5}\right) \approx 21.801^\circ \approx 21.8^\circ$ (to 1 d.p.)
So, $h = \sqrt{29}\sin(\theta + 21.8^\circ)$.

1. Find the greatest possible value of $h$ using the $R$-formula result from part (b)

From part (b), $h = \sqrt{29}\sin(\theta + 21.801^\circ)$.
The maximum value of the sine function is $1$.
Therefore, the greatest possible value of $h$ is $\sqrt{29}$ (or $5.39$ to 3 s.f.).

2. Find the value of $\theta$ at which the maximum occurs

The maximum occurs when $\sin(\theta + 21.801^\circ) = 1$.
$\theta + 21.801^\circ = 90^\circ$
$\theta = 90^\circ - 21.801^\circ = 68.199^\circ \approx 68.2^\circ$ (to 1 d.p.)

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

From part (b), $h = \sqrt{29}\sin(\theta + 21.801^\circ)$.
Equating this to $\sqrt{15}$:
$\sqrt{29}\sin(\theta + 21.801^\circ) = \sqrt{15}$

2. Solve for $\theta$

$\sin(\theta + 21.801^\circ) = \sqrt{\frac{15}{29}}$
$\theta + 21.801^\circ = \sin^{-1}\left(\sqrt{\frac{15}{29}}\right)$
$\theta + 21.801^\circ \approx 45.989^\circ$
$\theta \approx 45.989^\circ - 21.801^\circ = 24.188^\circ$
Since $\theta$ is acute, $\theta = 24.2^\circ$ (to 1 d.p.).