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0580/32 May/Jun 2021, Question 8: Angles in a Triangle

Cambridge IGCSE Mathematics 0580 · Paper 32 (Core) · Calculator · 15 marks

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The question, in short

Use a triangle and parallel line diagram with AB = BC, angle ACB = 36° and angle ABC = x°, plus a quadrilateral PQRS with parallel sides and given lengths, to find angles, names, perimeter, polygon sides, height and area.

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

Part (ai) · 1 mark

State the mathematical name for an angle of 36°.

  1. 136° is less than 90°, so it is an acute angle

AnswerAcute

Part (aii) · 1 mark

State the mathematical name for triangle ABC, given AB = BC.

  1. 1AB = BC, so triangle ABC is isosceles

AnswerIsosceles

Part (aiii) · 2 marks

Calculate the value of x, the angle at B in triangle ABC.

  1. 1Angle BAC = angle ACB = 36° (base angles of isosceles triangle)
  2. 2x = 180 − 2 × 36 = 108

Answer108

Part (aiv) · 2 marks

Find angle CBD and give a geometrical reason for it.

  1. 1Angle CBD = angle ACB = 36°
  2. 2because they are alternate angles (AC is parallel to BD)

Answer36, alternate angles

Part (bi) · 1 mark

State the mathematical name of quadrilateral PQRS with both pairs of opposite sides parallel.

  1. 1Both pairs of opposite sides are parallel, so PQRS is a parallelogram

AnswerParallelogram

Part (bii) · 1 mark

Calculate the perimeter of quadrilateral PQRS using SR = 8 cm and PS = 6.5 cm.

  1. 1Perimeter = 6.5 + 8 + 6.5 + 8 = 29 cm

Answer29

Part (biii) · 2 marks

Find angle PSR and give a reason, using angle PST = 120° on straight line TSR.

  1. 1Angle PSR = 180 − 120 = 60°
  2. 2because angles on a straight line add up to 180°

Answer60, angles on a straight line add up to 180

Part (biv) · 1 mark

Given PS and ST are sides of a regular polygon, work out how many sides it has.

  1. 1Interior angle = 120°, so exterior angle = 180 − 120 = 60°
  2. 2Number of sides = 360 ÷ 60 = 6

Answer6

Part (bv) · 2 marks

Show that the perpendicular height h of PQRS is 5.63 cm to 2 decimal places.

  1. 1sin 60° = h/6.5
  2. 2h = 6.5 × sin 60° = 5.629…
  3. 3h = 5.63 cm (2 d.p.)

Answersin 60 = h/6.5 so h = 5.629…

Part (bvi) · 2 marks

Calculate the area of quadrilateral PQRS using base SR = 8 cm and height h.

  1. 1Height h = 6.5 × sin 60° = 5.629… cm
  2. 2PQRS is a parallelogram, so area = base × height.
  3. 3Area = 8 × 5.629… = 45.03… ≈ 45.0 cm²

Answer45.0 or 45.03 to 45.04

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Also searched as: 0580/32/M/J/21 · 0580_s21_qp_32 · 0580 s21 32 · question 8.
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