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0580/42 May/Jun 2024, Question 2: Alternate Segment Theorem

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 9 marks

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The question, in short

Use parallel-line angle facts, a regular 12-gon, a circle with tangent AMB, and a cyclic quadrilateral to work out the unknown angles a, b, f, g and k.

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

Part (a) · 2 marks

Using the two parallel lines cut by the straight line, work out the sizes of angles a and b.

  1. 1a = 180 − 38 = 142 (angles on a straight line)
  2. 2b = a (alternate angles between parallel lines)
  3. 3So a = 142, b = 142

Answera = 142, b = 142

Part (b) · 2 marks

Work out the size of each interior angle in a regular polygon with 12 sides.

  1. 1Exterior angle = 360 ÷ 12 = 30°
  2. 2Interior angle = 180 − 30 = 150°

Answer150

Part (c) · 3 marks

With AMB tangent at M, work out the sizes of angles f and g.

  1. 1f = 56 (alternate segment theorem)
  2. 2Angle MOP = 2 × 56 = 112 (angle at centre is twice angle at circumference)
  3. 3Triangle OMP is isosceles (OM = OP)
  4. 4g = (180 − 112) ÷ 2 = 34

Answerf = 56, g = 34

Part (d) · 2 marks

Using the cyclic quadrilateral and the given 24° and 27° angles, work out the size of angle k.

  1. 1Angle opposite k = 180 − 24 − 27 = 129 (angles in a triangle)
  2. 2k = 180 − 129 = 51 (opposite angles of a cyclic quadrilateral)

Answer51

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Also searched as: 0580/42/M/J/24 · 0580_s24_qp_42 · 0580 s24 42 · question 2.
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