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Maths · Further Trigonometry

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The Cosine Rule

Using the cosine rule to find a side from two sides and the included angle, or an angle from three sides.

  • Higher
  • 9 key terms
  • All boards

Learning Objectives

  1. 1Use the cosine rule to find a missing side.
  2. 2Use the cosine rule to find a missing angle.
  3. 3Know when to use the cosine rule rather than the sine rule.
  4. 4Use exact values of \(\cos 60^\circ\) and \(\cos 120^\circ\) without a calculator.

When the sine rule does not work

The sine rule needs a matching pair, a side and the angle opposite to it. If you know two sides and the angle between them, or all three sides, there is no matching pair, and you use the cosine rule instead. The cosine rule is like Pythagoras' theorem with an extra term that corrects for the angle not being a right angle. If the angle is \(90^\circ\), then \(\cos 90^\circ = 0\), and the rule becomes Pythagoras' theorem.

Finding a side

Use the cosine rule when you know two sides and the angle between them.

  • The rule

    \(a^2 = b^2 + c^2 - 2bc\cos A\).

  • Substitute

    Write the numbers in before you simplify.

  • Take the square root

    At the end, if the question asks for \(a\).

  • Check

    The side opposite the largest angle is the longest.

Finding a side

In triangle \(ABC\), \(b = 5\) cm, \(c = 8\) cm and angle \(A = 60^\circ\). Work out the length of \(a\).

Show the solutionHide the solution
  1. 1 Choose the rule Two sides and the angle between them are known, so use the cosine rule.
  2. 2 Substitute \(a^2 = 5^2 + 8^2 - 2 \times 5 \times 8 \times \cos 60^\circ\).
  3. 3 Exact value \(\cos 60^\circ = \dfrac{1}{2}\), so \(a^2 = 25 + 64 - 40 = 49\).
  4. 4 Square root \(a = 7\) cm.

Answer\(a = 7\) cm

Finding an angle

Rearrange the rule to find an angle when you know all three sides.

  • The rule

    \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\).

  • The side \(a\)

    The angle \(A\) is opposite the side \(a\), which comes last in the numerator with a minus sign.

  • Negative cosine

    If \(\cos A\) is negative the angle is obtuse.

  • Check

    The angles should add up to \(180^\circ\).

Finding an angle

A triangle has sides of length 3 cm, 5 cm and 7 cm. Work out the largest angle.

Show the solutionHide the solution
  1. 1 The largest angle It is opposite the longest side, 7 cm. Call it \(A\), with \(a = 7\).
  2. 2 Substitute \(\cos A = \dfrac{3^2 + 5^2 - 7^2}{2 \times 3 \times 5} = \dfrac{9 + 25 - 49}{30}\).
  3. 3 Simplify \(\cos A = \dfrac{-15}{30} = -\dfrac{1}{2}\).
  4. 4 Exact value \(\cos 120^\circ = -\dfrac{1}{2}\), so \(A = 120^\circ\).

Answer\(120^\circ\)

Test yourself

  1. 1

    What is the cosine rule for the side \(a\)?

    Show answerHide answer

    \(a^2 = b^2 + c^2 - 2bc\cos A\).

  2. 2

    When do you use the cosine rule?

    Show answerHide answer

    With two sides and the included angle, or with three sides.

  3. 3

    What is \(\cos 60^\circ\)?

    Show answerHide answer

    \(\dfrac{1}{2}\).

  4. 4

    What is \(\cos 90^\circ\)?

    Show answerHide answer

    0.

  5. 5

    What does the rule become if \(A = 90^\circ\)?

    Show answerHide answer

    Pythagoras' theorem.

Exam technique: choosing a rule

Pick the rule from the information you have.

  • Matching pair

    If you know a side and its opposite angle, use the sine rule.

  • Two sides and the angle between

    Use the cosine rule for the third side.

  • Three sides

    Use the cosine rule for an angle.

  • Order of working

    Square, multiply, subtract, in that order, and write each stage.

Summary and exam focus

  • \(a^2 = b^2 + c^2 - 2bc\cos A\) finds a side.
  • \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\) finds an angle.
  • Use the exact values \(\cos 60^\circ = \dfrac{1}{2}\) and \(\cos 120^\circ = -\dfrac{1}{2}\) in non-calculator questions.
  • The sine rule needs a matching pair, and the cosine rule does not.

Exam focus

In triangle \(ABC\), \(b = 6\) cm, \(c = 10\) cm and angle \(A = 120^\circ\). Work out the length of \(a\). (3 marks) (3 marks)

\(a^2 = 6^2 + 10^2 - 2 \times 6 \times 10 \times \cos 120^\circ = 36 + 100 + 60 = 196\), because \(\cos 120^\circ = -\dfrac{1}{2}\). So \(a = 14\) cm.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Cosine rule
A rule linking three sides and one angle in any triangle.
Included angle
The angle between two given sides.
Hypotenuse
The longest side of a right-angled triangle.
Obtuse angle
An angle between \(90^\circ\) and \(180^\circ\).
Acute angle
An angle less than \(90^\circ\).
Exact value
A value written with fractions and surds.
Subject
The letter on its own on one side of a formula.
Square root
The number that multiplies by itself to give another.
Substitute
Replace letters with numbers.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Work out 3 marks Core

Diagram NOT accurately drawn. Work out the length of \(BC\). (3 marks)

A triangle diagram showing a triangle with two sides and the angle between them.

Mark scheme — 3 marks available

  • \(x^2 = 8^2 + 15^2 - 2 \times 8 \times 15 \times \cos 60^\circ\) — M1
  • \(64 + 225 - 120 = 169\) — M1
  • \(13\) — A1

Model answer

\(x^2 = 8^2 + 15^2 - 2 \times 8 \times 15 \times \cos 60^\circ = 64 + 225 - 120 = 169\), so \(x = 13\) cm.

2. Exam question Work out 3 marks Core

A triangle has sides of length 5 cm, 7 cm and 8 cm. Work out the size of the angle opposite the side of length 7 cm. (3 marks)

Mark scheme — 3 marks available

  • \(\cos A = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8}\) — M1
  • \(\dfrac{1}{2}\) — A1
  • \(60\) — A1

Model answer

\(\cos A = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8} = \dfrac{40}{80} = \dfrac{1}{2}\), so \(A = 60^\circ\).

3. Exam question Work out 3 marks Core

Diagram NOT accurately drawn. Two ships leave a port \(P\) at the same time. Ship \(Q\) sails 50 km and ship \(R\) sails 80 km. The angle between their paths is \(60^\circ\). Work out the distance \(QR\) between the ships. (3 marks)

A triangle diagram showing two ships leaving a port P.

Mark scheme — 3 marks available

  • \(x^2 = 50^2 + 80^2 - 2 \times 50 \times 80 \times \cos 60^\circ\) — M1
  • \(2500 + 6400 - 4000 = 4900\) — M1
  • \(70\) — A1

Model answer

\(x^2 = 50^2 + 80^2 - 2 \times 50 \times 80 \times \cos 60^\circ = 2500 + 6400 - 4000 = 4900\), so \(x = 70\) km.

4. Exam question Work out 3 marks Core

In triangle \(ABC\), \(AB = AC = 6\) cm and angle \(A = 120^\circ\). Work out the length of \(BC\). Give your answer as a surd in its simplest form. (3 marks)

Mark scheme — 3 marks available

  • \(BC^2 = 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos 120^\circ\) — M1
  • \(108\) — M1
  • \(6\sqrt{3}\) — A1

Model answer

\(BC^2 = 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos 120^\circ = 36 + 36 + 36 = 108\), so \(BC = \sqrt{108} = 6\sqrt{3}\) cm.

5. Exam question Work out 5 marks Stretch

In triangle \(ABC\), \(AB = 5\) cm, \(BC = 8\) cm and \(AC = 7\) cm. (a) Show that angle \(ABC = 60^\circ\). (3 marks) (b) Work out the exact area of triangle \(ABC\). (2 marks)

Mark scheme — 5 marks available

  • (a) \(\cos B = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8}\) — M1
  • (a) \(\dfrac{40}{80} = \dfrac{1}{2}\) — M1
  • (a) \(B = 60^\circ\), with the conclusion stated — C1
  • (b) \(\dfrac{1}{2} \times 5 \times 8 \times \sin 60^\circ\) — M1
  • (b) \(10\sqrt{3}\) — A1

Model answer

(a) \(\cos B = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8} = \dfrac{40}{80} = \dfrac{1}{2}\), so \(B = 60^\circ\). (b) Area \(= \dfrac{1}{2} \times 5 \times 8 \times \sin 60^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\) cm\(^2\).

6. Exam question Work out 3 marks Stretch

A triangle has sides of length 6 cm, 6 cm and \(6\sqrt{3}\) cm. Work out the size of the largest angle. (3 marks)

Mark scheme — 3 marks available

  • \(\cos A = \dfrac{6^2 + 6^2 - (6\sqrt{3})^2}{2 \times 6 \times 6}\) — M1
  • \(-\dfrac{1}{2}\) — A1
  • \(120\) — A1

Model answer

The largest angle is opposite the longest side. \(\cos A = \dfrac{6^2 + 6^2 - (6\sqrt{3})^2}{2 \times 6 \times 6} = \dfrac{36 + 36 - 108}{72} = -\dfrac{1}{2}\), so \(A = 120^\circ\).

7. Multiple choice 1 mark Easier

Which is the cosine rule for the side \(a\)?

  1. A \(a^2 = b^2 + c^2 + 2bc\cos A\)
  2. B \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)
  3. C \(a = b + c - 2bc\cos A\)
  4. D \(a^2 = b^2 + c^2 - 2bc\cos A\) Correct

Why: The cosine rule takes \(2bc\cos A\) away from the sum of the squares.

8. Multiple choice 1 mark Easier

When do you use the cosine rule to find a side?

  1. A When you know two angles and a side
  2. B When you know a side and its opposite angle
  3. C When you know two sides and the angle between them Correct
  4. D When the triangle has a right angle

Why: Two sides and the included angle is the cosine rule situation.

9. Multiple choice 1 mark Easier

What is \(\cos 60^\circ\)?

  1. A \(\dfrac{\sqrt{3}}{2}\)
  2. B \(\dfrac{1}{2}\) Correct
  3. C \(\dfrac{\sqrt{2}}{2}\)
  4. D 1

Why: This is an exact value.

10. Multiple choice 1 mark Core

What does the cosine rule become when \(A = 90^\circ\)?

  1. A Pythagoras' theorem Correct
  2. B The sine rule
  3. C The area formula
  4. D \(a = b + c\)

Why: \(\cos 90^\circ = 0\), so \(a^2 = b^2 + c^2\).

11. Multiple choice 1 mark Core

In triangle \(ABC\), \(b = 3\), \(c = 8\) and \(A = 60^\circ\). What is \(a^2\)?

  1. A 73
  2. B 97
  3. C 25
  4. D 49 Correct

Why: \(a^2 = 9 + 64 - 2 \times 3 \times 8 \times \frac{1}{2} = 73 - 24 = 49\).

12. Multiple choice 1 mark Core

In triangle \(ABC\), \(b = 6\), \(c = 10\) and \(A = 120^\circ\). What is \(a\)?

  1. A 8
  2. B \(\sqrt{76}\)
  3. C 14 Correct
  4. D 16

Why: \(a^2 = 36 + 100 - 120 \times (-\frac{1}{2}) = 136 + 60 = 196\), so \(a = 14\).

13. Multiple choice 1 mark Core

A triangle has sides 5, 7 and 8. What is \(\cos\) of the angle opposite the side of length 7?

  1. A \(-\dfrac{1}{2}\)
  2. B \(\dfrac{1}{2}\) Correct
  3. C \(\dfrac{1}{7}\)
  4. D \(\dfrac{3}{5}\)

Why: \(\cos A = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8} = \dfrac{40}{80} = \dfrac{1}{2}\).

14. Multiple choice 1 mark Stretch

A triangle has sides 3, 5 and 7. What is the largest angle?

  1. A \(120^\circ\) Correct
  2. B \(60^\circ\)
  3. C \(150^\circ\)
  4. D \(90^\circ\)

Why: \(\cos A = \dfrac{9 + 25 - 49}{30} = -\dfrac{1}{2}\), so \(A = 120^\circ\).

15. Multiple choice 1 mark Stretch

A triangle has sides 7, 8 and 13. What is the angle opposite the longest side?

  1. A \(60^\circ\)
  2. B \(90^\circ\)
  3. C \(150^\circ\)
  4. D \(120^\circ\) Correct

Why: \(\cos A = \dfrac{49 + 64 - 169}{112} = \dfrac{-56}{112} = -\dfrac{1}{2}\), so \(A = 120^\circ\).