The diagram below represents two observers at P and Q who are equidistant from point R - NSC Technical Mathematics - Question 6 - 2022 - Paper 2

We can find RQ using the sine rule. In triangle PQR:

$\frac{RQ}{\sin(33.9°)} = \frac{481.1}{\sin(112.2°)}$

So, $RQ = \frac{481.1 \cdot \sin(33.9°)}{\sin(112.2°)}$

Calculating this gives: $RQ ≈ 289.81 m$

To find h, we can use the tangent function. From triangle QRS:

$\tan(23.5°) = \frac{h}{RQ}$

Substituting the value of RQ we found earlier: $h = RQ \cdot \tan(23.5°$

Thus, $h ≈ 289.81 \cdot \tan(23.5° ≈ 126 m$

The area of triangle ΔPQR can be calculated using the formula:

$\text{Area} = \frac{1}{2} \cdot base \cdot height$

Where the base is 481.1 m and the height can be found using the sine:

$\text{Area} = \frac{1}{2} \cdot 481.1 \cdot 289.81 \cdot \sin(33.9°)$

Calculating this results in: $\text{Area} ≈ 38,882.53 m^2$