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

To find the distance RQ, we can use the sine rule in triangle PQR:

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

Rearranging gives:

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

Calculating this yields:

$RQ \approx 289.81 \, m$

Using the tangent function in triangle QRS, we have:

$\tan(23.5°) = \frac{h}{RQ}$

This leads to:

$h = RQ \cdot \tan(23.5°$

Substituting the value of RQ found earlier:

$h = 289.81 \cdot \tan(23.5° \approx 126 \, m$

The area of triangle QPR can be determined using the formula:

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

Here, the base is RQ and the height is h. Thus:

$Area = \frac{1}{2} \cdot 289.81 \cdot 126$

Calculating gives:

$Area \approx 18,882.53 \, m^2$