上图显示了两根长度分别为 4 m 和 9 m 的连杆 $AB$ 和 $BC$，且 $\angle ABC = 90^\circ$。在点 $A$ 和 $B$ 处固定有滚轮，使得 $A$ 能够沿墙壁移动，而 $B$ 能够沿天花板移动。$C$ 距离垂直墙壁的水平距离为 $d$ m。

1. Define horizontal distances from the wall using right-angled triangles

Let the horizontal distance from the wall to $B$ be $x_1$ and the horizontal distance from $B$ to $C$ be $x_2$. Then $d = x_1 + x_2$.

2. Express $x_1$ in terms of $\theta$

In the right-angled triangle formed by $A$, $B$, and the wall, the angle at $A$ is $\theta$. The side opposite to $\theta$ is $x_1$.
$\sin\theta = \frac{x_1}{4} \implies x_1 = 4\sin\theta$

3. Determine the relevant angle for the second triangle

Draw a vertical line down from $B$. The angle between $AB$ and the vertical is $\theta$ (alternate interior angles). Since $\angle ABC = 90^\circ$, the angle between $BC$ and the horizontal is $\theta$.

4. Express $x_2$ in terms of $\theta$ and find $d$

In the right-angled triangle formed by $B$, $C$, and the horizontal/vertical lines, the horizontal side $x_2$ is adjacent to the angle $\theta$ (if we use the angle with the horizontal) or opposite to $90^\circ - \theta$.
$\cos\theta = \frac{x_2}{9} \implies x_2 = 9\cos\theta$
Therefore, $d = x_1 + x_2 = 4\sin\theta + 9\cos\theta$ (Shown).

1. Expand $R\cos(\theta-\alpha)$ and equate coefficients with the expression for $d$

From part (a), $d = 9\cos\theta + 4\sin\theta$.
$R\cos(\theta-\alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$
Comparing coefficients:
$R\cos\alpha = 9$
$R\sin\alpha = 4$

2. Calculate the value of $R$

$R = \sqrt{9^2 + 4^2}$
$R = \sqrt{81 + 16} = \sqrt{97} \approx 9.85$

3. Calculate the value of $\alpha$

$\tan\alpha = \frac{4}{9}$
$\alpha = \tan^{-1}\left(\frac{4}{9}\right) \approx 23.962^\circ \approx 24.0^\circ$ (to 1 d.p.)

4. State the final expression

$d = \sqrt{97}\cos(\theta - 24.0^\circ)$

1. Express $d$ using the $R$-formula from part (b) and set up the equation

From part (b), $d = 4\sin\theta + 9\cos\theta = R\cos(\theta - \alpha)$.
$R = \sqrt{4^2 + 9^2} = \sqrt{97}$
$\tan\alpha = \frac{4}{9} \implies \alpha \approx 23.962^\circ$
So, $\sqrt{97}\cos(\theta - 23.962^\circ) = 6$

2. Solve the trigonometric equation for $\theta$

$\cos(\theta - 23.962^\circ) = \frac{6}{\sqrt{97}}$
$\theta - 23.962^\circ = \cos^{-1}\left(\frac{6}{\sqrt{97}}\right)$
$\theta - 23.962^\circ \approx 52.468^\circ$
$\theta = 52.468^\circ + 23.962^\circ = 76.430^\circ \approx 76.4^\circ$ (to 1 d.p.)

1. State the maximum value of $d$ using the $R$ value from part (b)

From part (b), $d = \sqrt{97}\cos(\theta - 23.962^\circ)$.
The maximum value of $\cos(\theta - 23.962^\circ)$ is $1$.
Maximum $d = \sqrt{97} \approx 9.85$ m

2. Find the corresponding value of $\theta$

For maximum $d$, $\cos(\theta - 23.962^\circ) = 1$
$\theta - 23.962^\circ = 0^\circ$
$\theta = 23.962^\circ \approx 24.0^\circ$