Lemmara logoLemmaraWorked solution

4PM1/01 Jun 2020, Question 9: The Cosine Rule

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 12 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

Triangle ABC has AB = 12 cm, BC = 6 cm, AC = 2x cm, D midpoint of AC with BD = 6 cm, angles θ and φ at B. Show several results about cosines, AC, and the angles.

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

Part (a) · 2 marks

Using triangle ABD with sides 12, 6 and x, show that cos ADB equals (x² − 108)/(12x).

  1. 1In triangle ABD: AD = x, BD = 6, AB = 12
  2. 2cos ADB = (6² + x² − 12²)/(2 × 6 × x)
  3. 3= (x² − 108)/(12x)

Answercos ADB = (x² − 108)/(12x) shown

Part (b) · 4 marks

Deduce that AC = 6√6 cm, using the cosine result or another method.

  1. 1In triangle BDC: cos BDC = (6² + x² − 6²)/(2 × 6x) = x²/(12x)
  2. 2∠ADB = 180° − ∠BDC ⇒ cos ADB = −cos BDC
  3. 3(x² − 108)/(12x) = −x²/(12x)
  4. 42x² = 108 ⇒ x = 3√6
  5. 5AC = 2x = 6√6 cm

AnswerAC = 6√6 cm shown

Part (c) · 4 marks

Show that sin(θ° + φ°) = sin φ° for the two angles at B.

  1. 1Sine rule in triangle ABC: sin(θ° + φ°)/(2x) = sin BCD/12 ⇒ sin(θ° + φ°) = x sin BCD/6
  2. 2Sine rule in triangle BDC: sin φ°/x = sin BCD/6 ⇒ sin φ° = x sin BCD/6
  3. 3Therefore sin(θ° + φ°) = sin φ°

Answersin(θ° + φ°) = sin φ° shown

Part (d) · 2 marks

Hence show that θ = 180 − 2φ.

  1. 1sin(θ + φ)° = sin φ° and θ ≠ 0, so θ + φ ≠ φ
  2. 2Hence θ + φ = 180 − φ
  3. 3θ = 180 − 2φ

Answerθ = 180 − 2φ shown

Stuck on a step? Read the free lesson: The Cosine Rule

Practise more the cosine rule questions

Real 4PM1 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on the cosine rule

Also searched as: 4PM1 1 June 2020 · 4PM1/01 · 4pm1 jun20 1 · question 9.
Lemmara is an independent study tool, not affiliated with or endorsed by Pearson Edexcel. The worked solutions are Lemmara’s own.