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4024/21 May/Jun 2023, Question 3: Combining Circle Theorems

Cambridge O Level Mathematics 4024 · Paper 21 (Calculator) · Calculator · 7 marks

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The question, in short

Parallel lines PQ and RS cut by straight line ABCD, with BE = CE and angle ABE = 110°; find angle ECQ with reasons. Then a circle with diameter UY, tangent ZX, angle VUX = 35°, angle XZY = a°; express b in terms of a.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 3 marks

Using PQ parallel to RS, BE = CE and angle ABE = 110°, work out angle ECQ, justifying each step.

  1. 1Angle CBE = 180 − 110 = 70° (angles on a straight line add to 180°)
  2. 2Angle BCE = 70° (base angles of isosceles triangle BCE, since BE = CE)
  3. 3Angle BCQ = 110° (corresponding angles are equal, PQ parallel to RS)
  4. 4Angle ECQ = 110 − 70 = 40°

AnswerAngle ECQ = 40 and correct reasons for each step, e.g. angle CBE = 70 (sum of angles on a straight line equal 180), angle BCE = 70 (base angles in an isosceles triangle are equal), angle BCQ = 110 (corresponding angles are equal)

Part (b) · 4 marks

With UY a diameter, ZX tangent at X, angle VUX = 35° and angle XZY = a°, find b in simplest form.

  1. 1Angle OXZ = 90° (tangent is perpendicular to the radius)
  2. 2Angle XOZ = 180 − 90 − a = 90 − a
  3. 3Angle XUO = (90 − a)/2 = 45 − a/2 (angle at centre is twice angle at circumference)
  4. 4Angle VUY + angle VWY = 180° (opposite angles of cyclic quadrilateral UVWY)
  5. 5b + 35 + 45 − a/2 = 180
  6. 6b = 100 + a/2

Answer100 + a/2 or (200 + a)/2

Stuck on a step? Read the free lesson: Combining Circle Theorems

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Also searched as: 4024/21/M/J/23 · 4024_s23_qp_21 · 4024 s23 21 · question 3.
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