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Maths · Further Trigonometry

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The Sine Rule

Using the sine rule to find sides and angles in triangles without a right angle.

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  • 9 key terms
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Learning Objectives

  1. 1Label a triangle with the side \(a\) opposite angle \(A\), and so on.
  2. 2Use the sine rule to find a missing side.
  3. 3Use the sine rule to find a missing angle.
  4. 4Decide when the sine rule is the right method.

Triangles without a right angle

Right-angled trigonometry only works when the triangle has a right angle. Many triangles do not, and there are two rules that work in any triangle: the sine rule and the cosine rule. The sine rule is used when you know a side and its opposite angle, which is called a matching pair, together with one other side or angle. On a non-calculator paper, the angles are chosen so that the sines are exact values, such as \(30^\circ\), \(45^\circ\) and \(60^\circ\).

The sine rule

In any triangle, the ratio of a side to the sine of its opposite angle is the same for all three sides.

  • Finding a side

    \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\) , with the unknown side on the top.

  • Finding an angle

    \(\dfrac{\sin A}{a} = \dfrac{\sin B}{b}\) , with the unknown angle on the top.

  • You need

    A matching pair and one more side or angle.

  • Not for

    Two sides and the angle between them, or three sides. Use the cosine rule for these.

Finding a side

In triangle \(ABC\), angle \(A = 30^\circ\), angle \(B = 45^\circ\) and \(a = 8\) cm. Work out the length of \(b\). Give your answer in the form \(k\sqrt{2}\).

Show the solutionHide the solution
  1. 1 Choose the rule The side \(a\) and angle \(A\) are a matching pair, and \(b\) is opposite \(B\), so use the sine rule.
  2. 2 Substitute \(\dfrac{b}{\sin 45^\circ} = \dfrac{8}{\sin 30^\circ}\).
  3. 3 Exact values \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\).
  4. 4 Solve \(b = \dfrac{8 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 8\sqrt{2}\) cm.

Answer\(b = 8\sqrt{2}\) cm

Finding an angle

Turn the rule upside down so that the sine of the unknown angle is on the top.

  • Write it

    \(\dfrac{\sin B}{b} = \dfrac{\sin A}{a}\).

  • Rearrange

    \(\sin B = \dfrac{b \sin A}{a}\).

  • Inverse sine

    Find the angle from the exact values table.

  • Check

    The largest angle is opposite the longest side.

Finding an angle

In triangle \(ABC\), angle \(A = 30^\circ\), \(a = 5\) cm and \(b = 10\) cm. Work out angle \(B\).

Show the solutionHide the solution
  1. 1 Set up \(\dfrac{\sin B}{10} = \dfrac{\sin 30^\circ}{5}\).
  2. 2 Substitute \(\sin B = \dfrac{10 \times \frac{1}{2}}{5} = 1\).
  3. 3 Inverse sine \(\sin B = 1\), so \(B = 90^\circ\).
  4. 4 Check \(b = 10\) is the longest side and is opposite the largest angle.

AnswerAngle \(B = 90^\circ\)

Test yourself

  1. 1

    What does a matching pair mean in the sine rule?

    Show answerHide answer

    A side and the angle opposite to it.

  2. 2

    Which side is opposite angle \(B\)?

    Show answerHide answer

    Side \(b\).

  3. 3

    What is \(\sin 45^\circ\) as an exact value?

    Show answerHide answer

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

  4. 4

    Which side is opposite the largest angle?

    Show answerHide answer

    The longest side.

  5. 5

    When can you not use the sine rule?

    Show answerHide answer

    When you have no matching pair.

Exam technique: the sine rule

Choose the method first, and then write each step.

  • Sketch

    Draw the triangle and label the sides and angles that you know.

  • Matching pair

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

  • Show the equation

    Write the sine rule with the values substituted before rearranging.

  • Exact values

    Write the exact values, then simplify, so that every mark is clear.

Summary and exam focus

  • Each side is opposite the angle with the same letter.
  • To find a side, use \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\).
  • To find an angle, use \(\dfrac{\sin A}{a} = \dfrac{\sin B}{b}\).
  • You need a matching pair.

Exam focus

In triangle \(ABC\), angle \(A = 60^\circ\), angle \(B = 45^\circ\) and \(a = 6\sqrt{3}\) cm. Work out the length of \(b\). (3 marks) (3 marks)

Use \(\dfrac{b}{\sin 45^\circ} = \dfrac{6\sqrt{3}}{\sin 60^\circ}\). Then \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), so \(b = \dfrac{6\sqrt{3} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = 6\sqrt{2}\) cm.

Key terms

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

Sine rule
A rule linking sides and the sines of the opposite angles in any triangle.
Matching pair
A side and the angle opposite to it.
Opposite
Across the triangle, not touching.
Exact value
A value written with fractions and surds.
Acute angle
An angle less than \(90^\circ\).
Obtuse angle
An angle between \(90^\circ\) and \(180^\circ\).
Surd
A root that cannot be written as a whole number or fraction.
Subject
The letter on its own on one side of a formula.
Ratio
A comparison of two quantities.

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

Not drawn accurately. Work out the length of \(AC\). [3 marks]

A triangle diagram showing a triangle with two angles and one side.

Mark scheme — 3 marks available

  • \(\dfrac{x}{\sin 30^\circ} = \dfrac{6\sqrt{2}}{\sin 45^\circ}\) — M1
  • Uses \(\sin 30^\circ = \dfrac{1}{2}\) and \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\) — M1
  • \(6\) — A1

Model answer

\(\dfrac{x}{\sin 30^\circ} = \dfrac{6\sqrt{2}}{\sin 45^\circ}\), so \(x = \dfrac{6\sqrt{2} \times \frac{1}{2}}{\frac{\sqrt{2}}{2}} = 6\) cm.

2. Exam question Work out 3 marks Core

In triangle \(ABC\), angle \(A = 60^\circ\), \(a = 6\) cm and \(b = 2\sqrt{3}\) cm. Angle \(B\) is acute. Work out the size of angle \(B\). [3 marks]

Mark scheme — 3 marks available

  • \(\dfrac{\sin B}{2\sqrt{3}} = \dfrac{\sin 60^\circ}{6}\) or equivalent — M1
  • \(\sin B = \dfrac{1}{2}\) — M1
  • \(30\) — A1

Model answer

\(\dfrac{\sin B}{2\sqrt{3}} = \dfrac{\sin 60^\circ}{6}\), so \(\sin B = \dfrac{2\sqrt{3} \times \frac{\sqrt{3}}{2}}{6} = \dfrac{3}{6} = \dfrac{1}{2}\). So \(B = 30^\circ\).

3. Exam question Work out 3 marks Core

In triangle \(ABC\), angle \(A = 30^\circ\), angle \(B = 30^\circ\) and \(a = 9\) cm. Work out the length of \(c\). Give your answer in surd form. [3 marks]

Mark scheme — 3 marks available

  • Angle \(C = 120^\circ\) — M1
  • \(\dfrac{c}{\sin 120^\circ} = \dfrac{9}{\sin 30^\circ}\) — M1
  • \(9\sqrt{3}\) — A1

Model answer

Angle \(C = 120^\circ\). \(\dfrac{c}{\sin 120^\circ} = \dfrac{9}{\sin 30^\circ}\), so \(c = \dfrac{9 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 9\sqrt{3}\) cm.

4. Exam question Work out 3 marks Core

Not drawn accurately. The diagram shows a triangular park \(PQR\). Work out the length of \(PR\). Give your answer in surd form. [3 marks]

A triangle diagram showing a triangular park PQR.

Mark scheme — 3 marks available

  • \(\dfrac{x}{\sin 60^\circ} = \dfrac{6\sqrt{2}}{\sin 45^\circ}\) — M1
  • Uses the exact values of \(\sin 60^\circ\) and \(\sin 45^\circ\) — M1
  • \(6\sqrt{3}\) — A1

Model answer

\(\dfrac{x}{\sin 60^\circ} = \dfrac{6\sqrt{2}}{\sin 45^\circ}\), so \(x = \dfrac{6\sqrt{2} \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = 6\sqrt{3}\) m.

5. Exam question Work out 3 marks Stretch

In triangle \(ABC\), angle \(A = 30^\circ\), \(a = 6\) cm and \(b = 6\sqrt{3}\) cm. There are two possible sizes of angle \(B\). Work out both of them. [3 marks]

Mark scheme — 3 marks available

  • \(\sin B = \dfrac{\sqrt{3}}{2}\) — M1
  • \(60\) — A1
  • \(120\) — A1

Model answer

\(\sin B = \dfrac{6\sqrt{3} \times \frac{1}{2}}{6} = \dfrac{\sqrt{3}}{2}\). So \(B = 60^\circ\) or \(B = 180 - 60 = 120^\circ\).

6. Exam question Show that 3 marks Stretch

In triangle \(ABC\), angle \(A = 30^\circ\), angle \(B = 45^\circ\) and \(a = 3\sqrt{2}\) cm. Show that \(b = 6\) cm. [3 marks]

Mark scheme — 3 marks available

  • \(\dfrac{b}{\sin 45^\circ} = \dfrac{3\sqrt{2}}{\sin 30^\circ}\) — M1
  • \(b = \dfrac{3\sqrt{2} \times \frac{\sqrt{2}}{2}}{\frac{1}{2}}\) — M1
  • \(\dfrac{3}{\frac{1}{2}} = 6\), with the working shown to the end — Q1

Model answer

\(\dfrac{b}{\sin 45^\circ} = \dfrac{3\sqrt{2}}{\sin 30^\circ}\), so \(b = \dfrac{3\sqrt{2} \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = \dfrac{3}{\frac{1}{2}} = 6\).

7. Multiple choice 1 mark Easier

In triangle \(ABC\), which side is opposite angle \(A\)?

  1. A \(b\)
  2. B \(c\)
  3. C \(a\) Correct
  4. D \(AB\)

Why: Each side is opposite the angle with the same letter.

8. Multiple choice 1 mark Easier

Which is the sine rule for finding a side?

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

Why: The sine rule says that side over the sine of its opposite angle is constant.

9. Multiple choice 1 mark Easier

What do you need to use the sine rule?

  1. A A side and its opposite angle, plus one more side or angle Correct
  2. B Three sides
  3. C Two sides and the angle between them
  4. D A right angle

Why: The sine rule needs a matching pair.

10. Multiple choice 1 mark Easier

What is \(\sin 45^\circ\)?

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

Why: This is one of the exact values.

11. Multiple choice 1 mark Core

In triangle \(ABC\), \(A = 30^\circ\), \(B = 90^\circ\) and \(a = 5\). What is \(b\)?

  1. A 5
  2. B \(\dfrac{5}{2}\)
  3. C 10 Correct
  4. D \(5\sqrt{3}\)

Why: \(\dfrac{b}{\sin 90^\circ} = \dfrac{5}{\sin 30^\circ}\), so \(b = \dfrac{5}{\frac{1}{2}} = 10\).

12. Multiple choice 1 mark Core

In triangle \(ABC\), \(A = 30^\circ\), \(B = 45^\circ\) and \(a = 4\). What is \(b\)?

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

Why: \(b = \dfrac{4\sin 45^\circ}{\sin 30^\circ} = \dfrac{4 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 4\sqrt{2}\).

13. Multiple choice 1 mark Core

In triangle \(ABC\), \(a = 6\), \(b = 6\sqrt{2}\) and \(A = 30^\circ\). What is \(\sin B\)?

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

Why: \(\sin B = \dfrac{b\sin A}{a} = \dfrac{6\sqrt{2} \times \frac{1}{2}}{6} = \dfrac{\sqrt{2}}{2}\).

14. Multiple choice 1 mark Core

In triangle \(ABC\), \(A = 60^\circ\) and \(B = 45^\circ\). What is angle \(C\)?

  1. A \(105^\circ\)
  2. B \(15^\circ\)
  3. C \(45^\circ\)
  4. D \(75^\circ\) Correct

Why: \(180 - 60 - 45 = 75\).

15. Multiple choice 1 mark Stretch

In triangle \(ABC\), \(A = 60^\circ\), \(B = 45^\circ\) and \(a = 3\sqrt{3}\). What is \(b\)?

  1. A \(3\sqrt{3}\)
  2. B \(\dfrac{3\sqrt{2}}{2}\)
  3. C \(3\sqrt{2}\) Correct
  4. D \(6\)

Why: \(b = \dfrac{3\sqrt{3} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = 3\sqrt{2}\).