图中显示了一个尺寸为 $14\text{ cm} \times 8\text{ cm}$ 的矩形块 $PQRS$ 放置在倾斜角为锐角 $\theta^\circ$ 的斜面 $VS$ 上的侧视图。
斜面放置在水平面 $ST$ 上，且 $d$ 是从 $Q$ 到 $ST$ 的垂直距离。$\angle VTS = 90^\circ$。

1. Identify the relevant right-angled triangles to express the vertical distance $d$.

Let the vertical line from $Q$ meet the horizontal line through $R$ at $A$, and let the vertical line from $R$ meet $ST$ at $B$. Then $d = QA + RB$.

2. Find the lengths of the segments using trigonometry.

In $\triangle RSB$, $\angle RSB = \theta$. Since $RS = PQ = 14$, $RB = 14\sin\theta$.
In $\triangle QRA$, $\angle QRA = 90^\circ - \angle ARS = 90^\circ - (90^\circ - \theta) = \theta$.
Since $QR = 8$, $QA = 8\cos\theta$.
Therefore, $d = QA + RB = 8\cos\theta + 14\sin\theta$.

1. Expand the target form and compare coefficients with the given expression

$R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$
Comparing with $8\cos\theta + 14\sin\theta$:
$R\cos\alpha = 8$
$R\sin\alpha = 14$

2. Calculate the value of R

$R = \sqrt{8^2 + 14^2}$
$R = \sqrt{64 + 196} = \sqrt{260} = 2\sqrt{65} \approx 16.1245$

3. Calculate the value of alpha

$\tan\alpha = \frac{14}{8} = 1.75$
$\alpha = \tan^{-1}(1.75) \approx 60.255^\circ$

4. State the final expression

$d = \sqrt{260}\cos(\theta - 60.3^\circ)$

1. Recall or calculate the values of $R$ and $\alpha$ from part (b)

From part (b), $d = R\cos(\theta - \alpha)$.
$R = \sqrt{8^2 + 14^2} = \sqrt{260}$
$\tan\alpha = \frac{14}{8} \implies \alpha = \tan^{-1}(1.75) \approx 60.255^\circ$
So, $d = \sqrt{260}\cos(\theta - 60.255^\circ)$

2. Set up the equation with the given value of $d$

$\sqrt{260}\cos(\theta - 60.255^\circ) = 10\sqrt{2}$

3. Isolate the cosine term and find the basic angle

$\cos(\theta - 60.255^\circ) = \frac{10\sqrt{2}}{\sqrt{260}} = \sqrt{\frac{200}{260}} = \sqrt{\frac{10}{13}}$
Basic angle $= \cos^{-1}\left(\sqrt{\frac{10}{13}}\right) \approx 28.716^\circ$

4. Solve for the smallest positive value of $\theta$

$\theta - 60.255^\circ = -28.716^\circ \text{ or } 28.716^\circ$
$\theta = 60.255^\circ - 28.716^\circ = 31.539^\circ \text{ or } \theta = 60.255^\circ + 28.716^\circ = 88.971^\circ$
The smallest value of $\theta$ is $31.5^\circ$ (to 1 d.p.).