OCR GCSE Mathematics: Sine rule and cosine rule (J560)

This study guide covers sections 10.05d (sine rule) and 10.05e (cosine rule) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). Both statements sit only in the Higher tier column of the specification, so everything below is Higher tier only. The same column lists the area formula ½ab sin C in 10.03a, and 6.02d lists all three formulae among those Higher tier learners should recall and use, so the area formula is included here for problem solving.

Use it with the OCR GCSE Mathematics hub and the printable checklist. The condensed version is the revision notes, and the practice set tests everything below. To find your weak spots first, try the free 10-minute diagnostics.

What this unit covers

Ref	What you must be able to do	Tier
10.05d	Know and apply the sine rule, a/sin A = b/sin B = c/sin C, to find lengths and angles	Higher tier only
10.05e	Know and apply the cosine rule, a² = b² + c² - 2bc cos A, to find lengths and angles	Higher tier only
10.03a	Area of a triangle = ½ab sin C (used here to solve problems)	Higher tier only

The rules work in any triangle, not only right-angled ones. Right-angled triangles, Pythagoras and exact trigonometric values are separate units, so this page assumes you can already use sin, cos and tan on a right-angled triangle.

Labelling the triangle

Name the angles A, B and C. Name the side opposite each angle with the matching lower-case letter: a is opposite A, b is opposite B, c is opposite C.

        A
       / \
    c /   \ b
     /     \
    B-------C
         a

Every error in this topic starts with a mislabelled triangle. Before you write any formula, mark each given angle and the side facing it. A side and the angle opposite it are always a pair.

The sine rule

The sine rule says that in every triangle, each side divided by the sine of its opposite angle gives the same number:

a / sin A = b / sin B = c / sin C

You never use all three fractions at once. Pick the two that contain your unknown and the pair of values you know.

Use the sine rule when you know a side and its opposite angle, plus one more angle or side.

Finding a length

Put the unknown side on the top of the left-hand fraction so that you do not have to rearrange afterwards.

Worked example 1. In triangle ABC, A = 48°, B = 71° and a = 9.3 cm. Find b.

b / sin 71° = 9.3 / sin 48°
b = 9.3 × sin 71° / sin 48°
b = 11.83...

So b = 11.8 cm (3 s.f.). The angle C = 61° was not needed here, because a and A form the known pair and B is the angle opposite the side we want.

If the angle opposite the unknown side is not given, find it first with the angle sum: C = 180° - A - B.

Finding an angle

Turn the rule upside down so the unknown angle is on top: sin A / a = sin B / b.

Worked example 2. In triangle ABC, a = 14.2 cm, b = 11.5 cm and A = 64°. Find B.

sin B / 11.5 = sin 64° / 14.2
sin B = 11.5 × sin 64° / 14.2
sin B = 0.72789...
B = sin⁻¹(0.72789...) = 46.71...°

So B = 46.7° (1 d.p.).

When there are two possible angles

Sine is positive for both an acute angle and the obtuse angle that adds with it to 180°. If sin B = 0.8333, then B could be 56.4° or 123.6°. Check whether the obtuse angle fits in the triangle by adding it to the angle you already know. It must leave a positive third angle.

Worked example 3. Triangle ABC has A = 30°, a = 6 cm and b = 10 cm. Find the possible values of B.

sin B = 10 × sin 30° / 6 = 5/6 = 0.8333...
B = 56.4° or B = 180° - 56.4° = 123.6°

Test each: 30° + 56.4° = 86.4° (less than 180°, fits) and 30° + 123.6° = 153.6° (less than 180°, fits). Two different triangles are possible, so B = 56.4° or 123.6°.

In Example 2, the given angle A faced the longer side (a > b). Then B must be smaller than A, so the obtuse answer cannot fit. When the given angle faces the shorter side, as here (a < b), expect to check both. Only give two answers when the question allows both, for example when it says “find the possible values”. Otherwise use the diagram or wording to decide.

The cosine rule

The cosine rule links all three sides with one angle:

a² = b² + c² - 2bc cos A

The angle A is the one between sides b and c, and a is the side opposite A. You may relabel to suit the triangle, for example c² = a² + b² - 2ab cos C.

Use the cosine rule when you know two sides and the angle between them, or all three sides.

Finding a length

Worked example 4. In triangle ABC, b = 7.4 cm, c = 10.6 cm and A = 112°. Find a.

a² = 7.4² + 10.6² - 2 × 7.4 × 10.6 × cos 112°
a² = 54.76 + 112.36 - 156.88 × (-0.3746...)
a² = 167.12 + 58.77...
a² = 225.89...
a = 15.03...

So a = 15.0 cm (3 s.f.). Here cos 112° is negative, so the last term adds. Many errors come from losing that sign. Work out 2bc cos A as one number, with its sign, before you subtract it.

Do the multiplication before the subtraction. Do not subtract 2bc from b² + c² first and then multiply by cos A.

Finding an angle

Rearrange the rule to make cos A the subject:

cos A = (b² + c² - a²) / (2bc)

The angle you want is opposite the side a, which is the side on its own at the start of the numerator.

Worked example 5. A triangle has sides 6.5 cm, 9.0 cm and 12.5 cm. Find its largest angle.

The largest angle is opposite the longest side, so call 12.5 cm the side c and find C.

cos C = (6.5² + 9.0² - 12.5²) / (2 × 6.5 × 9.0)
cos C = (42.25 + 81 - 156.25) / 117
cos C = -33 / 117 = -0.28205...
C = cos⁻¹(-0.28205...) = 106.38...°

So the largest angle is 106.4° (1 d.p.). A negative cosine means the angle is obtuse. That is a useful check.

Starting with the largest angle is a good habit. Cosine gives only one angle between 0° and 180°, so there is no second answer to test. If you then need another angle, you can use the sine rule safely, because the other two angles must both be acute.

Choosing the rule

You know	You want	Use
Two angles and any side	another side	Sine rule
A side and its opposite angle, and one more side	the angle opposite that side	Sine rule
Two sides and the angle between them	the third side	Cosine rule
All three sides	any angle	Cosine rule

If the information does not match any row, you probably need the angle sum first. Two angles always give the third.

Notice the pattern. The sine rule needs a matching pair (a side and the angle opposite it). The cosine rule is for the cases where no such pair exists.

Area of a triangle: ½ab sin C

The area formula uses two sides and the angle between them:

Area = ½ × a × b × sin C

Worked example 6. Two sides of a triangle are 12.4 cm and 9.7 cm and the angle between them is 38°. Find the area.

Area = ½ × 12.4 × 9.7 × sin 38°
     = 60.14 × 0.61566...
     = 37.02...

The area is 37.0 cm² (3 s.f.).

You can also run the formula backwards. If you know the area and two sides, work out sin C and then find C. Take care here: sin C gives an acute angle and an obtuse angle, so look in the question for which one is meant.

Combined problems

Longer questions chain the rules together. A sensible plan is to draw one clear triangle, mark everything you know, choose the rule from the table, and then ask what the answer gives you for the next step.

Worked example 7. A boat leaves A and sails 8.5 km on a bearing of 072° to B. It then sails 6.2 km on a bearing of 140° to C. Find the distance AC and the bearing of C from A.

Bearings are measured clockwise from north. The bearing of A from B is 072° + 180° = 252°. The angle ABC is the difference between the bearing back to A and the bearing on to C:

angle ABC = 252° - 140° = 112°

Two sides and the included angle are known, so use the cosine rule:

AC² = 8.5² + 6.2² - 2 × 8.5 × 6.2 × cos 112°
AC² = 72.25 + 38.44 + 39.48...
AC² = 150.17...
AC = 12.25...

So AC = 12.3 km (3 s.f.).

Now find angle BAC. The side BC and the angle ABC are a known pair, so use the sine rule with the unknown angle on top:

sin(BAC) / 6.2 = sin 112° / 12.25...
sin(BAC) = 6.2 × sin 112° / 12.25...
sin(BAC) = 0.46909...
angle BAC = 27.98...°

Angle BAC is acute because BC is shorter than AC. The bearing of C from A is 072° + 27.98…° = 99.98…°, so it is 100.0° (1 d.p.).

Use your unrounded value of AC in the second step. Rounding early can change the final angle in the first decimal place.

Common errors

• Wrong pairing. Using a side with the sine of an angle that is not opposite it. Draw the triangle and tick each pair.
• Rule mismatch. Trying the sine rule when you know two sides and the angle between them. There is no matching pair, so use the cosine rule.
• Subtracting first. In a² = b² + c² - 2bc cos A, multiplying (b² + c² - 2bc) by cos A is wrong. The cos A belongs to the 2bc term only.
• Sign slip with obtuse angles. Forgetting that cos is negative for angles over 90°, so -2bc cos A becomes a positive addition.
• Missing second solution. Giving only 56.4° when the question allows 123.6° too (or giving both when only one fits).
• Wrong angle in the area formula. The angle in ½ab sin C must lie between the two sides you use.
• Early rounding. Keep full unrounded values through a multi-step answer and round once at the end.
• Wrong angle unit. Give angles in degrees, as the questions here do, not in radians.

Where to go next

Work through the practice set, then use the revision notes as a quick checklist before a mock. The free diagnostics will show you which topics need attention next.

Official syllabus

OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), for first assessment in 2017, published by OCR. Sections 10.05d, 10.05e, 10.03a and 6.02d. All content on this page is Higher tier only.