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JAMB 1983 · UME · Q20

PQRSPQRS is a desk of dimensions 2 m×0.8 m2\text{ m} \times 0.8\text{ m} which is inclined at 30∘30^\circ to the horizontal (PQ=2 mPQ = 2\text{ m} is horizontal and the 0.8 m0.8\text{ m} edges slope up at 30∘30^\circ). Find the inclination of the diagonal PRPR to the horizontal.

Worked solution (try it first)
  1. The sloping edge QR=0.8QR = 0.8 m rises at 30∘30^\circ, so RR is 0.8sin⁡30∘=0.40.8\sin30^\circ = 0.4 m above the horizontal edge PQPQ.
  2. The diagonal of the desk top, by Pythagoras: PR=22+0.82PR = \sqrt{2^2 + 0.8^2}
    =4.64= \sqrt{4.64}
    ≈2.154\approx 2.154 m.
  3. The angle θ\theta of PRPR to the horizontal has sin⁡θ=heightPR\sin\theta = \frac{\text{height}}{PR}
    =0.42.154= \frac{0.4}{2.154}
    ≈0.1857\approx 0.1857.
  4. So θ≈10.70∘=10∘42′\theta \approx 10.70^\circ = 10^\circ42', option E.

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