In triangle PQR, PQ = 41.6 m, PR = 70 m, and angle PQR = 105°. Calculate angle QPR and the area of triangle PQR.

Step 1: Calculate angle $\angle PRQ$.
We use the Sine Rule:

$$\frac{PQ}{\sin(\angle PRQ)} = \frac{PR}{\sin(\angle PQR)}$$

Substitute the given values: $PQ = 41.6 \text{ m}$, $PR = 70 \text{ m}$, $\angle PQR = 105^\circ$.

$$\frac{41.6}{\sin(\angle PRQ)} = \frac{70}{\sin 105^\circ}$$

Rearrange to solve for $\sin(\angle PRQ)$:

$$\sin(\angle PRQ) = \frac{41.6 \times \sin 105^\circ}{70}$$

Calculate $\sin 105^\circ \approx 0.9659258$:

$$\sin(\angle PRQ) = \frac{41.6 \times 0.9659258}{70}$$ $$\sin(\angle PRQ) = \frac{40.20161328}{70}$$ $$\sin(\angle PRQ) \approx 0.57430876$$

Now, find $\angle PRQ$:

$$\angle PRQ = \arcsin(0.57430876)$$ $$\angle PRQ \approx 35.05^\circ$$

Part (i): Calculate angle QPR.
The sum of angles in a triangle is $180^\circ$.

$$\angle QPR + \angle PQR + \angle PRQ = 180^\circ$$ $$\angle QPR = 180^\circ - \angle PQR - \angle PRQ$$

Substitute the values:

$$\angle QPR = 180^\circ - 105^\circ - 35.05^\circ$$ $$\angle QPR = 39.95^\circ$$

Rounding to one decimal place:

$$\angle QPR \approx \boxed{40.0^\circ}$$

Part (ii): Calculate the area of triangle PQR.
The area of a triangle can be calculated using the formula:

$$\text{Area} = \frac{1}{2}ab\sin C$$

Using sides $PQ$, $PR$ and the included angle $\angle QPR$:

$$\text{Area} = \frac{1}{2} \times PQ \times PR \times \sin(\angle QPR)$$

Substitute the values: $PQ = 41.6 \text{ m}$, $PR = 70 \text{ m}$, $\angle QPR = 39.95^\circ$.

$$\text{Area} = \frac{1}{2} \times 41.6 \text{ m} \times 70 \text{ m} \times \sin(39.95^\circ)$$

Calculate $\sin(39.95^\circ) \approx 0.64207$:

$$\text{Area} = \frac{1}{2} \times 41.6 \times 70 \times 0.64207$$ $$\text{Area} = 20.8 \times 70 \times 0.64207$$ $$\text{Area} = 1456 \times 0.64207$$ $$\text{Area} \approx 934.78 \text{ m}^2$$

Rounding to one decimal place:

$$\text{Area} \approx \boxed{934.8 \text{ m}^2}$$

Part (iii): Calculate the shortest distance from Q to PR.
The shortest distance from a vertex to the opposite side is the altitude (height).
Let $h$ be the shortest distance from Q to PR. The base is $PR$.
The area of a triangle can also be calculated as:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Area} = \frac{1}{2} \times PR \times h$$

We know the Area from Part (ii) and $PR = 70 \text{ m}$.

$$934.78 \text{ m}^2 = \frac{1}{2} \times 70 \text{ m} \times h$$ $$934.78 = 35 \times h$$

Rearrange to solve for $h$:

$$h = \frac{934.78}{35}$$ $$h \approx 26.708 \text{ m}$$

Rounding to one decimal place:

$$h \approx \boxed{26.7 \text{ m}}$$