搬运工需要将一台冰箱送给客户。该冰箱可以建模为尺寸为 80 cm × 80 cm × 185 cm 的长方体。

1. Identify the vertical components of the tilted cuboid's edges

Let the vertical projection of the edge $AD$ (length 185 cm) be $h_1$ and the vertical projection of the edge $DC$ (length 80 cm) be $h_2$.
The total height $h$ of point $C$ above the ground is $h = h_1 + h_2$.

2. Express each vertical component in terms of $\theta$

From the right-angled triangle formed by $AD$ and the horizontal, the vertical height is $h_1 = 185\sin\theta$.
The angle the edge $DC$ makes with the vertical is $\theta$ (since $AD \perp DC$).
From the right-angled triangle formed by $DC$, the vertical height is $h_2 = 80\cos\theta$.
Therefore, $h = 185\sin\theta + 80\cos\theta$.

1. Set up the R-formula equation

Let $185\sin\theta + 80\cos\theta = R\sin(\theta + \alpha) = R(\sin\theta\cos\alpha + \cos\theta\sin\alpha)$.
Comparing coefficients:
$R\cos\alpha = 185$
$R\sin\alpha = 80$

2. Calculate R

$R = \sqrt{185^2 + 80^2} = \sqrt{34225 + 6400} = \sqrt{40625} \approx 201.556$

3. Calculate alpha and write the final expression

$\tan\alpha = \frac{80}{185}$
$\alpha = \tan^{-1}\left(\frac{80}{185}\right) \approx 23.385^\circ$
$h = 201.56\sin(\theta + 23.38^\circ)$ (to 2 d.p.)

1. Substitute $h = 180$ into the R-formula expression from part (b)

From part (b), $R = \sqrt{185^2 + 80^2} = \sqrt{40625}$ and $\tan\alpha = \frac{80}{185} \implies \alpha \approx 23.385^\circ$.

$\sqrt{40625}\sin(\theta + 23.385^\circ) = 180$

2. Solve for the angle $(\theta + \alpha)$

$\sin(\theta + 23.385^\circ) = \frac{180}{\sqrt{40625}}$
$\theta + 23.385^\circ = \sin^{-1}\left(\frac{180}{\sqrt{40625}}\right)$
$\theta + 23.385^\circ \approx 63.260^\circ$

3. Find the value of $\theta$

$\theta = 63.260^\circ - 23.385^\circ = 39.875^\circ \approx 39.9^\circ$ (to 1 d.p.)