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0580/43 May/Jun 2024, Question 11: Arc Length

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 9 marks

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The question, in short

For a circle with centre O, use the relationship between the minor and major arcs PQ to prove x = 108, then work with sector OAB where triangle OAB has half the sector's area to derive and solve an equation in y.

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

Part (a) · 3 marks

The minor arc PQ measures 3/7 of the major arc PQ. Show that the angle x at the centre equals 108.

  1. 1Minor arc : major arc = 3 : 7
  2. 2Minor arc is 3/10 of the circumference
  3. 3x = 3/10 × 360
  4. 4x = 108

Answer3/10 × 360 = 108

Part (bi) · 2 marks

Triangle OAB's area is half the sector's area, with obtuse angle y. Show that 360 sin y = πy.

  1. 1Area of triangle = 1/2 r² sin y
  2. 2Area of sector = y/360 × πr²
  3. 31/2 r² sin y = 1/2 × y/360 × πr²
  4. 4Multiply by 720/r²: 360 sin y = πy

Answer1/2 r² sin y = 1/2 × y/360 × πr² leading to 360 sin y = πy

Part (bii) · 3 marks

Fill in the missing table values for 360 sin y and πy at y = 108.6 and 108.7, to two decimal places.

  1. 1π × 108.6 = 341.18
  2. 2360 × sin 108.7° = 341.00
  3. 3π × 108.7 = 341.49

Answer108.6: πy = 341.18 or 341.22; 108.7: 360 sin y = 341.00, πy = 341.49 or 341.54

Part (biii) · 1 mark

State the value of y, to one decimal place, that solves 360 sin y = πy.

  1. 1At 108.6, 360 sin y − πy = +0.02; at 108.7 it is −0.49
  2. 2The sign changes and 108.6 is much closer to the root
  3. 3y = 108.6

Answer108.6

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Also searched as: 0580/43/M/J/24 · 0580_s24_qp_43 · 0580 s24 43 · question 11.
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