WAEC 2008 · Paper 2 · Q8✱✱

The magnitude of a force xi+15jx\mathbf i + 15\mathbf j is 17 N17\text{ N}.

  1. (a)

    Find the possible values of xx.

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

  2. (b)

    Find the directions of the forces, correct to the nearest degree.

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

Worked solution (try it first)

(a)

  1. The magnitude is x2+152\sqrt{x^2 + 15^2}, so x2+225=172=289x^2 + 225 = 17^2 = 289.
  2. Subtract 225: x2=64x^2 = 64.
  3. Take both square roots: x=8x = 8 or x=−8x = -8.

(b)

  1. For 8i+15j8\mathbf i + 15\mathbf j: the force points up (north) and to the right (east).
  2. Its angle from north is tan⁡−1815\tan^{-1}\frac{8}{15}.
  3. tan⁡−1815≈28.07∘\tan^{-1}\frac{8}{15} \approx 28.07^\circ, so the direction is N28∘E\text{N}28^\circ\text{E}, a bearing of 028∘028^\circ.
  4. For −8i+15j-8\mathbf i + 15\mathbf j: the force points north and to the left (west), at the same 28∘28^\circ from north.
  5. Its direction is N28∘W\text{N}28^\circ\text{W}, a bearing of 360∘−28∘=332∘360^\circ - 28^\circ = 332^\circ.

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