NECO 2023 · Paper 1 · Q44

An aeroplane flies from a town PP on a bearing of 045∘045^\circ to a town QQ, a distance of 200 km200\text{ km}. It then changes course and flies to another town RR on a bearing of 120∘120^\circ. If RR is directly east of PP, calculate ∣PR∣|PR|, correct to the nearest km.

Worked solution (try it first)
  1. On 045∘045^\circ, QQ is 200sin⁡45∘≈141.4200\sin45^\circ \approx 141.4 km east and 200cos⁡45∘≈141.4200\cos45^\circ \approx 141.4 km north of PP.
  2. RR is due east of PP, so the second leg must come back 141.4 km south.
  3. The bearing 120∘120^\circ is 60∘60^\circ from due south, so QRcos⁡60∘=141.4QR\cos60^\circ = 141.4, giving QR≈282.8QR \approx 282.8 km.
  4. That leg goes 282.8sin⁡60∘≈244.9282.8\sin60^\circ \approx 244.9 km east.
  5. So ∣PR∣≈141.4+244.9≈386|PR| \approx 141.4 + 244.9 \approx 386 km, option B.

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