WAEC 2019 · Paper 2 · Q8

  1. (a)

    Solve the inequality 13x−14(x+2)≥3x−113\frac13x - \frac14(x + 2) \ge 3x - 1\frac13.

    Show the answer

    x≤27x \le \frac27

  2. (b)

    In the diagram, ABCABC is a right-angled triangle on a horizontal ground and ∣AD∣|AD| is a vertical tower. ∠BAC=90∘\angle BAC = 90^\circ, ∠ACB=35∘\angle ACB = 35^\circ, ∠ABD=52∘\angle ABD = 52^\circ and ∣BC∣=66 m|BC| = 66\text{ m}. Find, correct to two decimal places, the: (i) height of the tower; (ii) angle of elevation of the top of the tower from CC.

    66 m52°35°ABCD
    A 3-D sketch: triangle ABC lies on the ground and AD is vertical.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. Multiply every term by 12: 4x−3(x+2)≥36x−164x - 3(x + 2) \ge 36x - 16.
  2. So x−6≥36x−16x - 6 \ge 36x - 16, and −35x≥−10-35x \ge -10.
  3. Divide by −35-35, which reverses the sign: x≤1035=27x \le \frac{10}{35} = \frac27.

(b)(i)

  1. In triangle ABCABC the right angle is at AA, so BC=66BC = 66 m is the hypotenuse.
  2. ABAB is opposite the 35∘35^\circ angle at CC: ∣AB∣=66sin⁡35∘≈37.86|AB| = 66\sin 35^\circ \approx 37.86 m.
  3. The tower ADAD stands at AA, and from BB its top is at 52∘52^\circ: ∣AD∣=∣AB∣tan⁡52∘|AD| = |AB|\tan 52^\circ
    ≈37.86×1.2799\approx 37.86 \times 1.2799
    ≈48.45\approx 48.45 m.

(ii)

  1. ∣AC∣=66cos⁡35∘≈54.06|AC| = 66\cos 35^\circ \approx 54.06 m.
  2. From CC: tan⁡θ=48.4554.06\tan\theta = \frac{48.45}{54.06}
    ≈0.8962\approx 0.8962, so θ≈41.87∘\theta \approx 41.87^\circ.

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