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WAEC 2022 · Paper 1 · Q20

Given that P=(−4,−5)P = (-4, -5) and Q=(2,3)Q = (2, 3), express PQ→\overrightarrow{PQ} in the form (k,θ)(k, \theta), where kk is the magnitude and θ\theta the bearing.

Worked solution (try it first)
  1. PQ→=Q−P\overrightarrow{PQ} = Q - P
    =(2−(−4), 3−(−5))= (2 - (-4),\ 3 - (-5))
    =(6,8)= (6, 8): 6 east and 8 north.
  2. The magnitude is 62+82=100=10\sqrt{6^2 + 8^2} = \sqrt{100} = 10 units.
  3. A bearing is measured clockwise from north, so tan⁡θ=eastnorth\tan\theta = \dfrac{\text{east}}{\text{north}}
    =68= \dfrac{6}{8} and θ=36.9∘≈037∘\theta = 36.9^\circ \approx 037^\circ.
  4. So PQ→=(10 units,037∘)\overrightarrow{PQ} = (10 \text{ units}, 037^\circ), option C.

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