IGCSEMathsCambridge (CIE)ExtendedRevision NotesPythagoras & TrigonometrySine, Cosine Rule & Area of TrianglesThe Sine Rule

The Sine Rule (Cambridge (CIE) IGCSE Maths: Extended): Revision Note

Exam code: 0580 & 0980

Written by: Amber

Reviewed by: Dan Finlay

Updated on 30 April 2026

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Sine rule

What is the sine rule?

The sine rule is used in non right-angled triangles
- It allows us to find missing side lengths or angles
It states that for any triangle with angles A, B and C

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Where
- $a$ is the side opposite angle A
- $b$ is the side opposite angle B
- $c$ is the side opposite angle C

Non Right-Angled Triangle labelled with angles A, B and C and opposite corresponding sides a, b and c.

How do I use the sine rule to find missing lengths?

Use the sine rule
- when you have opposite pairs of sides and angles in the question
  - a and A, or b and B, or c and C
Start by labelling your triangle with the angles and sides
- Angles have upper case letters
- Sides opposite the angles have the equivalent lower case letter
To find a missing length, substitute numbers into the formula
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
- You only need to have two parts equal to each other (not all three)
  - Then solve to find the side you need

How do I use the sine rule to find missing angles?

To find a missing angle, it is easier to rearrange the formula first by flipping each part
$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$
- The angles are now in the numerators of the fractions
- Substitute the values you have into the formula and solve
  - You will need to use inverse sine in your calculation, $\sin^{- 1} (. . .)$

Examiner Tips and Tricks

You are given the formula $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ in your exam.

You are not given the formula $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ . However, you can derive it by flipping each fraction in the formula you are given.

What is the ambiguous case of the sine rule?

Given information about a triangle, there may be two different ways to draw it
In the diagram below, the lengths of two sides are given, a and b
- A base angle is also given, $θ$ , but no angle near b is given
- It turns out that there are two possible ways to arrange b to complete the triangle!
  - Both triangles have the correct values of a, b and $θ$
The other base angle could either be obtuse or acute
- The sine rule only gives the acute answer on your calculator
  - You need to check the diagram to see if the angle you need is actually obtuse
  - If it is, use this rule: obtuse angle = 180 - acute angle

aa-sl-3-3-2-ambiguous-sine-rule-diagram-1

Worked Example

The following diagram shows triangle ABC.

$AB = 8.1 cm$ , $BC = 12.3 cm$ and angle $BCA = 27 °$ .

Triangle ABC with AB = 8.1 cm, BC = 12.3 cm, AC = y cm, angle BAC = xº and angle BCA = 27º.

(a) Calculate the value of $x$ .

Answer:

Label the sides of the diagram

Triangle ABC with sides opposite angles labelled with corresponding lowercase letters.

x is an angle so use the sine rule with the angles on top

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$
In practice, you only need to equate two of these three parts

$\begin{array}{rcl} \frac{\sin x}{12.3} & = & \frac{\sin 27}{8.1} \\ \sin x & = & \frac{12.3 \sin 27}{8.1} \\ x & = & \sin^{- 1} (\frac{12.3 \sin 27}{8.1}) \\ x & = & 43.58207 . . . \end{array}$

$43.6 °$ (to 1 d.p)

(b) Calculate the value of $y$ .

Answer:

To find y you need to know the angle opposite (angle ABC)
You know 27 and x from above, so subtract these from 180

$\begin{array}{rcl} Angle A B C & = & 180 - 27 - 43.58207 . . . \\ = & 109.41792 . . . \end{array}$

y is a length so use the sin rule with the sides on the top
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

$\begin{array}{rcl} \frac{y}{\sin (109.41792 . . .)} & = & \frac{8.1}{\sin 27} \\ y & = & \frac{8.1 \sin (109.41792 . . .)}{\sin 27} \\ y & = & 16.82691 . . . \end{array}$

$y = 16.8 cm$ (to 3 s.f.)

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Questions students ask about The Sine Rule

Real questions from students revising this topic on Save My Exams

How do you know whether to use a/sin A or sin A/a?

Put what you are looking for on the top: use a / sin A to find a side, and sin A / a to find an angle.

Both versions are the same equation with every fraction flipped, so either is correct
Keeping the unknown on top saves you a messy rearrangement
You are given a / sin A = b / sin B = c / sin C in the exam; flip each fraction to get the angle version
You only ever need two of the three parts

Example from this page:

(a) Finding the angle x, so angles on top:

sin x / 12.3 = sin 27° / 8.1
sin x = 12.3 sin 27° / 8.1
x = sin⁻¹(12.3 sin 27° / 8.1) = 43.6° (1 d.p.)

(b) Finding the length y, so sides on top:

Angle ABC = 180° − 27° − 43.58…° = 109.41…°
y / sin 109.41…° = 8.1 / sin 27°
y = 8.1 sin 109.41…° / sin 27° = 16.8 cm (3 s.f.)

How do you find an obtuse angle using the sine rule?

Your calculator's sin⁻¹ only ever gives the acute angle, so if the angle you need is obtuse, use obtuse angle = 180° − acute angle.

This is the ambiguous case: sin θ = sin(180° − θ), so an acute and an obtuse angle share the same sine
Check the diagram — if the angle clearly looks bigger than 90°, it is the obtuse one
Or rule it out with the sides: a triangle can only have one obtuse angle and it must be opposite the longest side, so if your angle is not opposite the longest side, take the acute value

Example using this page's triangle: the page finds angle ABC from the angle sum, but you could find it with the sine rule instead.

sin B / 16.826… = sin 27° / 8.1
sin⁻¹ gives B = 70.58…°
B is opposite the longest side (16.826… cm, against 12.3 cm and 8.1 cm), so it is the only angle that could be obtuse, and the diagram shows it is
So B = 180° − 70.58…° = 109.4° (1 d.p.)

That matches the page's angle-sum answer, 180° − 27° − 43.58…° = 109.41…°.