The diagram shows the side view of a 14 cm by 8 cm rectangular block $PQRS$, placed on a ramp, $VS$, tilted at an acute angle of $\theta^\circ$.
The ramp is placed on a horizontal surface $ST$ and $d$ is the perpendicular distance from $Q$ to $ST$. $\angle VTS = 90^\circ$.

1. Construct horizontal and vertical lines to form right-angled triangles

Let $R'$ be the projection of $R$ onto the horizontal line $ST$. Let $Q'$ be the projection of $Q$ onto the horizontal line passing through $R$. Then $d = RR' + QQ'$.

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

In $\triangle SRR'$, $\angle RSR' = \theta$. Since $SR = 14$ cm, $RR' = 14\sin\theta$.
In $\triangle RQQ'$, $\angle QRQ' = \theta$ (since $RQ$ is perpendicular to $RS$ and $RQ'$ is parallel to $ST$). Since $QR = 8$ cm, $QQ' = 8\cos\theta$.
Therefore, $d = RR' + QQ' = 14\sin\theta + 8\cos\theta = 8\cos\theta + 14\sin\theta$.

1. Expand the target form $R\cos(\theta - \alpha)$

$R\cos(\theta - \alpha) = R(\cos\theta\cos\alpha + \sin\theta\sin\alpha) = (R\cos\alpha)\cos\theta + (R\sin\alpha)\sin\theta$

2. Equate coefficients with the expression for $d$

From part (a), $d = 8\cos\theta + 14\sin\theta$.
Comparing coefficients:
$R\cos\alpha = 8$
$R\sin\alpha = 14$

3. Calculate the value of $R$

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

4. Calculate the value of $\alpha$

$\tan\alpha = \frac{14}{8} = \frac{7}{4}$
$\alpha = \tan^{-1}\left(\frac{7}{4}\right) \approx 60.255^\circ \approx 60.3^\circ$ (to 1 d.p.)
So, $d = 2\sqrt{65}\cos(\theta - 60.3^\circ)$

1. Recall the result from part (b) and set up the equation

From part (b), $d = \sqrt{260}\cos(\theta - 60.255^\circ)$.
Equating this to $10\sqrt{2}$:
$\sqrt{260}\cos(\theta - 60.255^\circ) = 10\sqrt{2}$

2. Isolate the cosine term

$\cos(\theta - 60.255^\circ) = \frac{10\sqrt{2}}{\sqrt{260}}$
$\cos(\theta - 60.255^\circ) = \sqrt{\frac{200}{260}} = \sqrt{\frac{10}{13}}$

3. Find the possible values for the angle

$\theta - 60.255^\circ = \pm \cos^{-1}\left(\sqrt{\frac{10}{13}}\right)$
$\theta - 60.255^\circ = \pm 27.696^\circ$

4. Solve for the smallest positive value of theta

For the smallest $\theta$, we take the negative value:
$\theta = 60.255^\circ - 27.696^\circ$
$\theta = 32.559^\circ \approx 32.6^\circ \text{ (to 1 d.p.)}$