The Sine Rule
x

Revision notes for Edexcel GCSE Maths The Sine Rule. Open the guide for explanations and worked examples. Written against the Edexcel GCSE Maths (1MA1) specification, so the content matches what's examinable rather than general Maths background.

The Sine Rule

What you'll learn

  • How to match an angle with the side opposite it.
  • When to use the sine rule instead of right-angled trigonometry.
  • How to find missing sides, missing angles, perimeters and areas.
  • How to handle the “obtuse angle” twist in exam questions.

1. The key idea: opposite pairs

The sine rule connects angles to the sides directly opposite them.

In triangle ABC:

Triangle ABC showing each side paired with the angle directly opposite it.

  • Side AB is opposite angle C.
  • Side AC is opposite angle B.
  • Side BC is opposite angle A.
Definition

Opposite side

The opposite side to an angle is the side across the triangle from that angle, not one of the sides touching the angle.

You will often need the angle sum of a triangle first.

Key Idea

Angle sum

The angles in any triangle add to 180°.

Example

Spotting the opposite pair

A triangle PQR has PQ = 9 cm, angle P = 47° and angle R = 68°. Find angle Q and name the complete opposite pair.

Triangle PQR with PQ = 9 cm opposite angle R = 68°, and angle Q left to be found.

  1. Use the angle sum:

    Q=180∘−47∘−68∘=65∘Q = 180^\circ - 47^\circ - 68^\circ = 65^\circQ=180∘−47∘−68∘=65∘

  2. Side PQ is opposite angle R.

  3. So the complete opposite pair is 9 cm with angle 68°.

Tip

Degrees mode

Make sure your calculator is in degrees mode. At GCSE, triangle angles are measured in degrees unless stated otherwise.

2. The sine rule formula

Use the sine rule when you know one complete opposite pair and at least one more side or angle from another pair.

Definition

The sine rule

If sides aaa, bbb and ccc are opposite angles AAA, BBB and CCC, then:

  • To find a missing side, use:

    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}sinAa​=sinBb​=sinCc​

  • To find a missing angle, use:

    sin⁡Aa=sin⁡Bb=sin⁡Cc\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}asinA​=bsinB​=csinC​
Common Mistake

Pairing the wrong side

The side must be opposite the angle, not next to it. Before substituting, draw or trace the line straight across from the angle to its opposite side.

3. Finding a missing side

When the unknown is a length, put the side lengths on top.

Example

Finding a missing side

In triangle DEF, side DE = 14 cm. Angle F = 101° and angle E = 37°. Work out the length of DF to 1 decimal place.

Triangle DEF showing the known opposite pair DE and angle F, and the unknown side DF opposite angle E.

  1. Identify the complete opposite pair: DE = 14 cm is opposite angle F = 101°.

  2. The side we want, DF, is opposite angle E = 37°. Let DF be xxx.

  3. Use the side form of the sine rule:

    xsin⁡37∘=14sin⁡101∘\frac{x}{\sin 37^\circ} = \frac{14}{\sin 101^\circ}sin37∘x​=sin101∘14​

  4. Rearrange by multiplying by sin⁡37∘\sin 37^\circsin37∘:

    x=14sin⁡37∘sin⁡101∘x = \frac{14\sin 37^\circ}{\sin 101^\circ}x=sin101∘14sin37∘​

  5. Calculate and round: x≈8.6x \approx 8.6x≈8.6 cm to 1 decimal place.

4. Finding a missing angle

When the unknown is an angle, put the sines on top. You will then use inverse sine.

Definition

Inverse sine

The inverse sine, written sin⁡−1\sin^{-1}sin−1, is the calculator function that finds an angle from its sine value.

Example

Finding a missing angle

In triangle GHI, GH = 9.5 m, GI = 7.0 m and angle I = 72°. Work out angle H to 3 significant figures.

Triangle GHI showing GH = 9.5 m opposite angle I, and GI = 7.0 m opposite the unknown angle H.

  1. The complete opposite pair is GH = 9.5 m and angle I = 72°.

  2. Side GI = 7.0 m is opposite angle H.

  3. Use the angle form of the sine rule:

    sin⁡H7.0=sin⁡72∘9.5\frac{\sin H}{7.0} = \frac{\sin 72^\circ}{9.5}7.0sinH​=9.5sin72∘​

  4. Rearrange:

    sin⁡H=7.0sin⁡72∘9.5\sin H = \frac{7.0\sin 72^\circ}{9.5}sinH=9.57.0sin72∘​

  5. Use inverse sine:

    H=sin⁡−1(7.0sin⁡72∘9.5)≈44.5∘H = \sin^{-1}\left(\frac{7.0\sin 72^\circ}{9.5}\right) \approx 44.5^\circH=sin−1(9.57.0sin72∘​)≈44.5∘

  6. The angle is 44.5° to 3 significant figures.

Common Mistake

Misreading inverse sine

sin⁡−1\sin^{-1}sin−1 does not mean “divide by sine”. It means “find the angle whose sine is this number”.

5. Perimeter questions

A perimeter question usually means you must find the missing side lengths first, then add all the sides.

Example

Using the sine rule to find a perimeter

Triangle ABC has AC = 10.4 m, angle B = 112° and angle C = 39°. Work out the perimeter to 3 significant figures.

Triangle ABC for a perimeter problem, with one side known and the other two sides to be found using the sine rule.

  1. Find the third angle:

    A=180∘−112∘−39∘=29∘A = 180^\circ - 112^\circ - 39^\circ = 29^\circA=180∘−112∘−39∘=29∘

  2. The known pair is AC = 10.4 m opposite angle B = 112°.

  3. Find AB, which is opposite angle C:

    AB=10.4sin⁡39∘sin⁡112∘≈7.06AB = \frac{10.4\sin 39^\circ}{\sin 112^\circ} \approx 7.06AB=sin112∘10.4sin39∘​≈7.06

  4. Find BC, which is opposite angle A:

    BC=10.4sin⁡29∘sin⁡112∘≈5.44BC = \frac{10.4\sin 29^\circ}{\sin 112^\circ} \approx 5.44BC=sin112∘10.4sin29∘​≈5.44

  5. Add the three sides: P≈10.4+7.06+5.44=22.9P \approx 10.4 + 7.06 + 5.44 = 22.9P≈10.4+7.06+5.44=22.9 m.

Tip

Do not round too early

Keep full calculator values until the final answer. Rounding side lengths halfway through can change the final perimeter or area.

6. Area questions

For a non-right-angled triangle, the area formula is:

Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin CArea=21​absinC

Here, aaa and bbb are two sides, and CCC is the included angle between them.

The included angle C is the angle between the two sides used in the area formula.

Definition

Included angle

The included angle is the angle between the two sides you are using.

Example

Using the sine rule before finding area

Triangle LMN has LM = 12 m, angle L = 74° and angle N = 38°. Work out the area to 1 decimal place.

Triangle LMN showing the known side and angles, with LN needed so that LM and LN enclose angle L for the area formula.

  1. Find the third angle:

    M=180∘−74∘−38∘=68∘M = 180^\circ - 74^\circ - 38^\circ = 68^\circM=180∘−74∘−38∘=68∘

  2. The known pair is LM = 12 m opposite angle N = 38°.

  3. Find LN, which is opposite angle M:

    LN=12sin⁡68∘sin⁡38∘≈18.07LN = \frac{12\sin 68^\circ}{\sin 38^\circ} \approx 18.07LN=sin38∘12sin68∘​≈18.07

  4. Use LM and LN with the included angle L:

    Area=12×12×18.07×sin⁡74∘\text{Area} = \frac{1}{2} \times 12 \times 18.07 \times \sin 74^\circArea=21​×12×18.07×sin74∘

  5. The area is 104.2 m² to 1 decimal place.

7. The obtuse angle case

Sometimes inverse sine gives you the acute angle, but the triangle may need the obtuse angle instead.

Common Mistake

The ambiguous case

For positive sine values, sin⁡θ=sin⁡(180∘−θ)\sin \theta = \sin(180^\circ - \theta)sinθ=sin(180∘−θ). If a question says an angle is obtuse, use the larger angle.

Example

Choosing the obtuse angle

Triangle ABC has AC = 18 cm, AB = 12 cm and angle C = 32°. Angle B is obtuse. Work out angle B to 3 significant figures.

Triangle ABC in the ambiguous sine rule case, with angle B shown as obtuse rather than the acute inverse-sine value.

  1. AC is opposite angle B, and AB is opposite angle C.

  2. Use the sine rule:

    sin⁡B18=sin⁡32∘12\frac{\sin B}{18} = \frac{\sin 32^\circ}{12}18sinB​=12sin32∘​

  3. Rearrange and use inverse sine:

    B=sin⁡−1(18sin⁡32∘12)≈52.6∘B = \sin^{-1}\left(\frac{18\sin 32^\circ}{12}\right) \approx 52.6^\circB=sin−1(1218sin32∘​)≈52.6∘

  4. This is the acute angle, but angle B is obtuse:

    B=180∘−52.6∘=127.4∘B = 180^\circ - 52.6^\circ = 127.4^\circB=180∘−52.6∘=127.4∘

  5. So angle B is 127° to 3 significant figures.

Exam technique

In the exam

  1. Mark the side opposite each angle before writing the formula.

  2. Check you have one complete opposite pair.

  3. Use side-over-sine for missing sides, and sine-over-side for missing angles.

  4. If using inverse sine, check whether an obtuse answer is possible or stated.

  5. Round only at the end, to the accuracy asked for.

Self review

Check yourself

  • Can you explain why side AB is opposite angle C in triangle ABC?

  • If you know two angles and one side, what should you do before using the sine rule?

  • Why might an inverse sine calculation give an acute angle when the final answer is obtuse?

Recap questions

Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.

You've reached the end

Test yourself on this topic, or move on to the next guide.

Practice questionsTake a quick quiz on this topicFlashcardsSelf-test with active recall
The Cosine RuleUp next

How was this guide?

The Sine Rule Revision Guide

  1. GCSE
  2. /Maths
  3. /The Sine Rule