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WAEC 2013 · Paper 2 · Q7

In the diagram, PQSPQS is a right-angled triangle, ∠PSQ=90∘\angle PSQ = 90^\circ, ∣PS∣=3 cm|PS| = 3\text{ cm}, ∣PQ∣=5 cm|PQ| = 5\text{ cm}, ∠PRS=61∘\angle PRS = 61^\circ and ∠QPR=x∘\angle QPR = x^\circ. Calculate, correct to one decimal place, the:

5 cm3 cm61°x°PSQR
  1. (a)

    value of xx;

  2. (b)

    area of △PQR\triangle PQR.

Worked solution (try it first)

(a)

  1. In triangle PQSPQS, PSPS is next to ∠QPS\angle QPS and PQPQ is the hypotenuse, so cos⁡∠QPS=35\cos\angle QPS = \frac35.
  2. So ∠QPS=cos⁡−1(0.6)=53.13∘\angle QPS = \cos^{-1}(0.6) = 53.13^\circ.
  3. In triangle PRSPRS, the angles add up to 180∘180^\circ: ∠RPS=180∘−90∘−61∘\angle RPS = 180^\circ - 90^\circ - 61^\circ
    =29∘= 29^\circ.
  4. x∘=∠QPS−∠RPSx^\circ = \angle QPS - \angle RPS
    =53.13∘−29∘= 53.13^\circ - 29^\circ
    =24.13∘= 24.13^\circ.
  5. So x=24.1x = 24.1 to one decimal place.

(b)

  1. Pythagoras in triangle PQSPQS: ∣QS∣=52−32=4 cm|QS| = \sqrt{5^2 - 3^2} = 4\text{ cm}.
  2. Area of △PQS=12×3×4\triangle PQS = \frac12 \times 3 \times 4
    =6 cm2= 6\text{ cm}^2.
  3. In triangle PRSPRS, ∣RS∣=3tan⁡29∘=1.6629 cm|RS| = 3\tan29^\circ = 1.6629\text{ cm}.
  4. Area of △PRS=12×3×1.6629\triangle PRS = \frac12 \times 3 \times 1.6629
    =2.4944 cm2= 2.4944\text{ cm}^2.
  5. Area of △PQR\triangle PQR is the difference: 6−2.4944=3.50566 - 2.4944 = 3.5056.
  6. The area is 3.5 cm23.5\text{ cm}^2 to one decimal place.

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