The diagram below shows farmland in the form of a cyclic quadrilateral PQRS - NSC Technical Mathematics - Question 6 - 2021 - Paper 2

Using the sine rule to find the size of ∠S: $\frac{S}{\text{sin}(S)} = \frac{QR}{\text{sin}(Q)}$ This gives us: $S = \frac{QR \cdot \text{sin}(S)}{\text{sin}(Q)}$

Substituting the values:

• $QR = 750 m$
• $\text{sin}(Q) = \text{sin}(60°) = \frac{\sqrt{3}}{2}$

Equating this gives, $S = \frac{750 \cdot \text{sin}(120°)}{\text{sin}(60°)}$ Calculating this:

• $\text{sin}(120°) = \frac{\sqrt{3}}{2}$
  So, $S = \frac{750 \cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} \approx 120°$.

To find PS, we again use the sine rule: $\frac{PS}{\text{sin}(R_1)} = \frac{PR}{\text{sin}(S)}$ Rearranging gives: $PS = \frac{PR \cdot \text{sin}(R_1)}{\text{sin}(S)}$

Substituting the known values:

• $PR \approx 1050 m$
• $\text{sin}(R_1) = \text{sin}(40.5°)$
• $\text{sin}(S) = \text{sin}(120°)$

Performing the calculations yields: $PS \approx \frac{1050 \cdot \text{sin}(40.5°)}{\text{sin}(120°)}$ This leads to: $PS \approx 787.41 m$.

We can use the area formula for a triangle: $\text{Area} = \frac{1}{2} \cdot QR \cdot PR \cdot \text{sin}(Q)$ Substituting the values:

• $QR = 750 m$
• $PR \approx 1050 m$
• $\text{sin}(Q) = \text{sin}(60°)$

Calculating the area gives: $\text{Area} = \frac{1}{2} \cdot (750) \cdot (1200) \cdot \text{sin}(60°)$ $\approx 389711.43 m^2.$