图中显示了一个靠在垂直墙上的长方形 $ABCD$。 $AB$ 为 10 米，$AD$ 为 3 米，且边 $BC$ 与地面成 $\theta^\circ$ 的角。
$EC$ 为 $C$ 到墙的水平距离。 只要 $EC$ 介于 7.5 米和 9.5 米之间，长方形 $ABCD$ 就会保持靠在墙上。

1. Find the horizontal distance from the wall to point B.

Let the corner where the wall meets the floor be $O$. In the right-angled triangle $AOB$, the angle $OBA$ and the angle $\theta$ that $BC$ makes with the floor are complementary because $\angle ABC = 90^\circ$ and the floor is a straight line.
Therefore, $\angle OBA = 180^\circ - 90^\circ - \theta = 90^\circ - \theta$.
The horizontal distance $OB$ is given by:
$OB = AB \cos(90^\circ - \theta) = 10 \sin \theta$

2. Find the horizontal distance from B to C and sum them to find EC.

The horizontal distance from $B$ to $C$ is the projection of $BC$ onto the horizontal floor.
Since $ABCD$ is a rectangle, $BC = AD = 3$ m.
Horizontal projection of $BC = BC \cos \theta = 3 \cos \theta$.
The total horizontal distance $EC$ from the wall to $C$ is the sum of the horizontal distance to $B$ and the horizontal projection of $BC$:
$EC = OB + 3 \cos \theta = 10 \sin \theta + 3 \cos \theta$
Comparing with $a \sin \theta + b \cos \theta$, we have $a = 10$ and $b = 3$.

1. Identify the expression for $EC$ from part (i) and set up the R-formula equation

From part (i), $EC = 10 \sin \theta + 3 \cos \theta$.
Let $10 \sin \theta + 3 \cos \theta = R \sin (\theta + \alpha) = R \sin \theta \cos \alpha + R \cos \theta \sin \alpha$.

2. Find the value of $R$

Comparing coefficients:
$R \cos \alpha = 10$
$R \sin \alpha = 3$
$R = \sqrt{10^2 + 3^2} = \sqrt{109}$

3. Find the value of $\alpha$

$\tan \alpha = \frac{3}{10} = 0.3$
$\alpha = \tan^{-1}(0.3) \approx 16.699^\circ \approx 16.7^\circ$ (to 1 d.p.)

4. State the final expression

$EC = \sqrt{109} \sin (\theta + 16.7^\circ)$

1. Set up the inequality using the condition given and the result from part (ii).

From part (ii), $EC = \sqrt{109} \sin (\theta + 16.7^\circ)$.
Given $7.5 \leqslant EC \leqslant 9.5$,
$7.5 \leqslant \sqrt{109} \sin (\theta + 16.699^\circ) \leqslant 9.5$

2. Solve for the lower bound of $\theta$.

$\sqrt{109} \sin (\theta + 16.699^\circ) \geqslant 7.5$
$\sin (\theta + 16.699^\circ) \geqslant \frac{7.5}{\sqrt{109}}$
$\theta + 16.699^\circ \geqslant \sin^{-1}\left(\frac{7.5}{\sqrt{109}}\right)$
$\theta + 16.699^\circ \geqslant 45.919^\circ$
$\theta \geqslant 29.220^\circ$

3. Solve for the upper bound of $\theta$.

$\sqrt{109} \sin (\theta + 16.699^\circ) \leqslant 9.5$
$\sin (\theta + 16.699^\circ) \leqslant \frac{9.5}{\sqrt{109}}$
$\theta + 16.699^\circ \leqslant \sin^{-1}\left(\frac{9.5}{\sqrt{109}}\right)$
$\theta + 16.699^\circ \leqslant 65.496^\circ$
$\theta \leqslant 48.797^\circ$

4. State the final range of values for $\theta$.

$29.2^\circ \leqslant \theta \leqslant 48.8^\circ \text{ (to 1 d.p.)}$