WAEC 2010 · Paper 2 · Q3✱

  1. (a)

    In a right-angled triangle, sin⁡x=35\sin x = \frac35. Evaluate 5cos⁡2x−35\cos^2 x - 3.

  2. (b)

    The angle of elevation of the top of a vertical pole from a point 63 m63\text{ m} east of the base of the pole is 30∘30^\circ. From another point due west of the pole, the angle of elevation of the top is 60∘60^\circ. (i) Draw a sketch diagram to illustrate the information. (ii) Calculate, correct to three significant figures, the distance of the second point from the base of the pole.

    Model answer
    63 mx30°60°TBAC

    The pole TBTB stands between the two points: AA is 63 m63\text{ m} east of BB, CC is x mx\text{ m} west of BB.

Worked solution (try it first)

(a)

  1. Draw a right-angled triangle with opposite side 3 and hypotenuse 5.
  2. By Pythagoras the adjacent side is 25−9=4\sqrt{25 - 9} = 4.
  3. So cos⁡x=45\cos x = \frac45 and cos⁡2x=1625\cos^2 x = \frac{16}{25}.
  4. Substitute: 5×1625−3=165−35 \times \frac{16}{25} - 3 = \frac{16}{5} - 3
    =15= \frac15.

(b)(i)

  1. Sketch the pole TBTB with AA 63 m63\text{ m} east of the foot BB (elevation 30∘30^\circ) and CC west of BB (elevation 60∘60^\circ), both on level ground.

(ii)

  1. From AA: the height is ∣TB∣=63tan⁡30∘=36.37 m|TB| = 63 \tan 30^\circ = 36.37\text{ m}.
  2. From CC: tan⁡60∘=∣TB∣x\tan 60^\circ = \dfrac{|TB|}{x}, so x=63tan⁡30∘tan⁡60∘x = \dfrac{63 \tan 30^\circ}{\tan 60^\circ}.
  3. tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt3} and tan⁡60∘=3\tan 60^\circ = \sqrt3, so x=633=21.0x = \frac{63}{3} = 21.0.
  4. The second point is 21.0 m21.0\text{ m} from the base of the pole.

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