JAMB 2003 · UME · Q29

An aeroplane flies due north from airport PP to QQ and then due east to RR. If QQ is equidistant from PP and RR, find the bearing of PP from RR.

Worked solution (try it first)
  1. PQPQ is due north and QRQR is due east, and PQ=QRPQ = QR, so triangle PQRPQR is right-angled and isosceles.
  2. So RR is 45∘45^\circ east of north from PP: the bearing of RR from PP is 045∘045^\circ.
  3. The back bearing adds 180∘180^\circ: the bearing of PP from RR is 225∘225^\circ, option D.

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