Die diagram hieronder toon plaasgrond in die formaat van 'n koordevierehoek PQRS - NSC Technical Mathematics - Question 6 - 2021 - Paper 1

To find the size of angle ∠S, we utilize the sine rule:

$\frac{PQ}{\sin S} = \frac{QR}{\sin Q}$ Given that:

• PQ = 1200 m
• QR = 750 m
• ∠Q = 60°

Rearranging for ∠S:

$\sin S = \frac{PQ \times \sin Q}{QR} = \frac{1200 \times \sin(60°)}{750}$

Calculating this gives:

$S \approx 120°$

Again applying the sine rule:

$\frac{PS}{\sin R1} = \frac{PR}{\sin S}$ We have:

• PR = 1 050 m
• ∠S = 120°
• ∠R1 = 40.5°

Rearranging:

$PS = \frac{PR \times \sin R1}{\sin S}$ Substituting the values:

$PS = \frac{1050 \times \sin(40.5°)}{\sin(120°)}$

This results in:

$PS \approx 787.41 \, ext{m}$

To compute the area of triangle ΔQPR:

$\, ext{Area} = \frac{1}{2} \times QR \times PR \times \sin Q$ Substituting the known values:

$= \frac{1}{2} \times (750) \times (1200) \times \sin(60°)$

Calculating the area gives approximately:

$\text{Area} \approx 389711.43 \, ext{m}^2$