WAEC 2009 · Paper 2 · Q5

  1. (a)

    In the diagram, the two circles intersect at XX and YY. The centre OO of the smaller circle is on the circumference of the bigger circle. AA and BB are any two points on the major arcs, one on each circle. Find an equation connecting aa and bb.

    a°b°ABXYO
    Model answer

    Join OXOX and OYOY. In the small circle, ∠XOY=2b∘\angle XOY = 2b^\circ (angle at the centre). AXOYAXOY is a cyclic quadrilateral of the big circle, so a+2b=180a + 2b = 180.

  2. (b)

    In the diagram, ∠QPR=∠PTR=90∘\angle QPR = \angle PTR = 90^\circ, ∣PR∣=8|PR| = 8 cm and ∣QP∣=6|QP| = 6 cm. Find ∣TR∣|TR|.

    6 cm8 cmPQRT
Worked solution (try it first)

(a)

  1. Join OXOX and OYOY.
  2. In the smaller circle (centre OO), the angle at the centre is twice the angle at the circumference: ∠XOY=2b∘\angle XOY = 2b^\circ.
  3. AA, XX, OO and YY all lie on the bigger circle, so AXOYAXOY is a cyclic quadrilateral and ∠XAY+∠XOY=180∘\angle XAY + \angle XOY = 180^\circ.
  4. So a+2b=180a + 2b = 180.

(b)

  1. In triangle PQRPQR, tan⁡R=68=0.75\tan R = \frac{6}{8} = 0.75, so R=36.87∘R = 36.87^\circ.
  2. In the right-angled triangle PTRPTR, ∣TR∣=8cos⁡R=8×0.8=6.4|TR| = 8\cos R = 8 \times 0.8 = 6.4 cm.

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