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

To find the size of ∠S, we will use the sine rule:

$\frac{S}{\sin(S)} = \frac{QR}{\sin(Q)}$

From the triangle, we know:

• QR = 750 m
• ∠Q = 60°
• PR is approximately 1010.27 m

Using the sine rule:

$\frac{S}{\sin(120°)} = \frac{750}{\sin(60°)}$

Solving this gives:

$S = \frac{750 \times \sin(120°)}{\sin(60°)} \\ = 750 \times \frac{\sqrt{3}/2}{\sqrt{3}/2} \\ = 750 \\ \therefore S = 120°$

We will again use the sine rule:

$\frac{PS}{\sin(R_1)} = \frac{PR}{\sin(S)}$

Where:

• PS = unknown
• R1 = 40.5°
• PR is approximately 1010.27 m
• ∠S is 120°

Applying the sine rule:

$PS = \frac{PR \times \sin(R_1)}{\sin(S)}$

• Substituting the known values:

$PS = \frac{1010.27 imes \sin(40.5°)}{\sin(120°)}$

Calculating:

$PS = \frac{1010.27 imes 0.6533}{0.8660} \\ \approx 787.41 m$

To calculate the area of triangle QPR, we can use the formula:

$\text{Area} = \frac{1}{2} \times QR \times PR \times \sin(Q)$

• Where:
  • QR = 750 m
  • PR = 1010.27 m
  • ∠Q = 60°

Substituting the values:

$\text{Area} = \frac{1}{2} \times 750 \times 1010.27 \times \sin(60°)$

Solving this gives:

$\text{Area} \approx \frac{1}{2} \times 750 \times 1010.27 \times 0.8660 \\ \approx 389711.43 m^2$