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JAMB 1989 · UME · Q41

The pilot of an aeroplane flying 10 km above the ground towards a landmark sees the landmark at angles of depression of 35∘35^\circ and then 55∘55^\circ. Find the distance between the two points of observation.

Worked solution (try it first)
  1. The angle of depression from the plane equals the angle of elevation from the landmark (alternate angles).
  2. So each horizontal distance dd has tan⁡θ=10d\tan\theta = \frac{10}{d}.
  3. So d=10tan⁡θ=10cot⁡θd = \frac{10}{\tan\theta} = 10\cot\theta: first 10cot⁡35∘10\cot35^\circ, then 10cot⁡55∘10\cot55^\circ.
  4. The plane flew the difference: 10(cot⁡35∘−cot⁡55∘)10(\cot35^\circ - \cot55^\circ), option D.

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