Question

12(b). The bearing of Q from P is 150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P: (i) represent this information in a diagram; (ii) calculate the distance between Q and R, correct to two decimal places; (iii) find the bearing of R from Q, correct to the nearest degree.

Accepted Answer

(i) Sketch P with PQ=24km at bearing 150°, and PR=32km at bearing 195° (since bearing of P from R is 015°, bearing of R from P = 015+180=195°).
(ii) Angle QPR = 195°−150° = 45°.
By cosine rule: QR² = 24²+32²−2(24)(32)cos45° = 576+1024−1536×0.7071 ≈ 513.9
QR ≈ 22.67km
(iii) By sine rule: sin(PQR)/32 = sin45°/22.67 ⟹ sin(PQR) ≈ 0.998 ⟹ ∠PQR ≈ 86.6°
Bearing of P from Q = 150+180 = 330°. Bearing of R from Q = 330°−86.6° ≈ 243° (to the nearest degree).

WAEC Mathematics 2014 Theory — Question 21

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