Sine Rule, Cosine Rule and Area of Triangles

Mathematics revision notes · Cambridge (CIE) · 0580

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1. Labelling and choosing a rule

The labelling convention

These three rules work in any triangle, not just right-angled ones, which is what makes them different from Pythagoras and SOHCAHTOA
All three are printed in the List of formulas, so none has to be memorised, but each must be labelled correctly before it can be used
Capital letters mark the angles and the matching small letter marks the side opposite that angle
- Side a lies opposite angle A, side b opposite angle B, and side c opposite angle C
Relabel the triangle in the question to match, rather than trying to fit the formula to whatever letters the diagram already uses

A triangle whose corners are labelled with capitals A, B and C and whose sides carry the matching small letters, each small letter placed on the side directly opposite its capital: a opposite A, b opposite B and c opposite C — Source: Area of a triangle by Save My Exams

Getting a side paired with the wrong angle is the commonest cause of a wrong answer here

Which rule

Choose the rule from what you already know and what you are being asked for

Rule	Use it when
Sine rule	a side and the angle facing it are both known, and you want the partner of another such pair
Cosine rule	two sides and their included angle are known, and the third side is wanted
Cosine rule	all three sides are known, and any angle is wanted
Area formula	two sides and their included angle are known, and the area is wanted

The included angle means the one wedged between the two sides, at the corner where they meet
Quick test: spot a side sitting opposite a known angle and the sine rule applies; without such a pair, it is the cosine rule
A question worth many marks usually needs two rules in succession, often the cosine rule then the sine rule
Angles in a triangle add to 180°, which frequently supplies the missing angle without any rule at all
Papers never name the rules in the question, so recognising which applies is part of what is being tested

A decision tree for picking the rule: if the triangle is right-angled use SOHCAHTOA; if not, ask whether the question involves a side opposite a known angle, and if it does use the sine rule; if it does not, use the area formula when area is involved and the cosine rule otherwise — Source: Applications of trigonometry by Save My Exams

Worked example

Deciding which rule fits

For each triangle, say which rule you would use. (a) Angle A = 40°, side a = 9 cm, angle B = 65°. (b) Sides b = 7 cm and c = 10 cm with angle A = 55° between them. (c) All three sides known, angle B wanted.

Solution:

The deciding question is always the same: is a side known together with the angle opposite it?
(a) Side a and angle A are a matching pair, and angle B is known, so the sine rule finds side b
(b) The angle given lies between the two known sides, so no side has its opposite angle known
That is the cosine rule, finding side a from b, c and angle A
(c) Three sides and no angles, so again no matching pair: the cosine rule rearranged for cos B
If (b) had asked for the area rather than a side, two sides and the angle between them is exactly ½ab sin C
The papers never print the words "sine rule" or "cosine rule", so this choice is always yours to make

2. The sine rule

Finding a side

The printed form is a ÷ sin A = b ÷ sin B = c ÷ sin C
Only two of the three parts are needed, so pick the pair that contains your unknown and the pair you know completely
Substitute, then multiply across to make the unknown side the subject

Worked example

A side by the sine rule

In triangle ABC, angle A = 52°, angle B = 71° and side a = 9.4 cm. Find side b, correct to 3 significant figures.

Triangle ABC with B at the bottom left, C at the bottom right and A above. The angle at A is marked 52 degrees, the angle at B is marked 71 degrees, and the side BC facing A is labelled 9.4 cm. The side b, facing B, is unlabelled. The figure is marked NOT TO SCALE.

Solution:

Side a and angle A are an opposite pair, and side b faces angle B, so use the sine rule
b ÷ sin 71° = 9.4 ÷ sin 52°
Multiply both sides by sin 71°: b = 9.4 × sin 71° ÷ sin 52°
b = 11.27887…
So b = 11.3 cm to 3 significant figures
Check: b faces the larger angle, so it should be the longer side, and it is

Finding an angle

Inverting all three fractions moves the angles into the numerators, giving sin A ÷ a = sin B ÷ b = sin C ÷ c
That flipped version is not printed, but it follows from the printed one, so either form can be written down
Substitute, make the sine of the unknown angle the subject, then apply the inverse sine
Give angles correct to one decimal place, as the syllabus requires

The ambiguous case

Inverse sine on a calculator only ever returns the acute answer, but an obtuse angle can have the same sine
Whenever the sine rule is used to find an angle, there is a second candidate: 180° minus the acute answer
The syllabus names this the ambiguous case and includes it explicitly, along with obtuse angles generally
The obtuse candidate is possible only if it and the angle you already know add up to less than 180°; when both are possible, the diagram or the wording decides

Why two triangles fit the same three facts: with the angle theta and the side a fixed, a side of length b can reach the base in two places, giving one triangle with an acute angle alpha and another with the obtuse angle 180 − alpha at that corner — Source: Sine rule by Save My Exams

This never happens with the cosine rule, which returns the obtuse angle directly

Worked example

An angle, with the ambiguous case

In triangle ABC, side a = 7.6 cm, side b = 5.2 cm and angle B = 34°. Find angle A.

Solution:

Side b and angle B are an opposite pair, and angle A faces side a, so the sine rule applies, inverted to put the angles in the numerators
sin A ÷ 7.6 = sin 34° ÷ 5.2
Multiply both sides by 7.6: sin A = 7.6 × sin 34° ÷ 5.2
sin A = 0.8172…
The calculator gives A = sin⁻¹(0.8172…) = 54.8136…, so A = 54.8° to 1 decimal place
The obtuse alternative is 180 − 54.8136… = 125.2°, and both are consistent with the given information: the third angle C would be 91.2° or 20.8°
Either can be correct here, so a real question settles it, with its diagram or by saying that the angle is obtuse

3. The cosine rule

Finding a side

The printed form is a² = b² + c² − 2bc cos A
Angle A must be the one between sides b and c, and side a must be the one facing it
Substitute, work out the right-hand side, then square root
Keep the subtraction intact: the whole 2bc cos A is subtracted, and where A is obtuse its cosine is negative, so the term is added in effect

Triangle ABC marked not to scale with all three sides given, AB 4.2 km, BC 3.8 km and AC 7.1 km, and the angle theta at B between the two shorter sides, which is the shape the cosine rule solves for an angle — Source: Cosine rule by Save My Exams

Worked example

A side by the cosine rule

In triangle ABC, b = 6.3 cm, c = 9.1 cm and their included angle is A = 47°. Find side a, correct to 3 significant figures.

Triangle ABC with A at the bottom left. The side AB along the bottom is labelled 9.1 cm and the side AC rising to the right is labelled 6.3 cm. The angle at A, between those two labelled sides, is marked 47 degrees. The side BC is unlabelled. The figure is marked NOT TO SCALE.

Solution:

Two sides with their included angle, and the third side wanted, so the cosine rule applies
a² = 6.3² + 9.1² − 2 × 6.3 × 9.1 × cos 47°
a² = 39.69 + 82.81 − 114.66 × cos 47°
a² = 122.5 − 78.1979… = 44.3020…
a = √44.3020… = 6.6559…
So a = 6.66 cm to 3 significant figures

Finding an angle

With all three sides known, rearrange to cos A = (b² + c² − a²) ÷ 2bc
That rearrangement is not printed, so either derive it or substitute into the printed form and solve
Apply the inverse cosine at the end
A negative value for cos A simply means the angle is obtuse, and the calculator returns it directly with no ambiguity

Worked example

An obtuse angle from three sides

A triangle has sides 6 cm, 8 cm and 12 cm. Find the angle opposite the 12 cm side, correct to 1 decimal place.

A triangle drawn with its longest side horizontal along the bottom and labelled 12 cm, the left-hand side labelled 6 cm and the right-hand side labelled 8 cm. The angle facing the 12 cm side is visibly obtuse and is not marked with a value.

Solution:

All three sides are known, so use the cosine rule for an angle, with a = 12 as the side facing the unknown angle
cos A = (6² + 8² − 12²) ÷ (2 × 6 × 8)
cos A = (36 + 64 − 144) ÷ 96 = −44 ÷ 96 = −0.4583…
The cosine is negative, so the angle is obtuse
A = cos⁻¹(−0.4583…) = 117.2796…
So A = 117.3° to 1 decimal place

4. Area of a triangle

Two sides and their included angle

The printed formula is Area = ½ab sin C
Angle C must sit between the two sides a and b, and any pair of sides with their included angle will do
Where that angle is not given, find it first with the sine or cosine rule
If C happens to be 90°, sin 90° = 1 and the formula collapses to the familiar half base times height
Convert all the lengths to the same unit before substituting, and give the area in the unit the question asks for

Triangle ABC marked not to scale with AC given as 1.1 m, AB as 32 cm and the angle at A between them as 74 degrees, so the two sides and their included angle are known but the units must be matched before the area formula is used — Source: Area of a triangle by Save My Exams

Worked example

Area from two sides and the included angle

A triangle has sides of 8.5 cm and 6.2 cm with an angle of 63° between them. Find its area, correct to 3 significant figures.

A triangle with its base along the bottom labelled 8.5 cm and its left side labelled 6.2 cm, with the angle of 63 degrees between them marked. The third side and the height are not shown. The figure is marked NOT TO SCALE.

Solution:

The angle lies between the two given sides, so the area formula applies directly
Area = ½ × 8.5 × 6.2 × sin 63°
Area = 26.35 × sin 63°
Area = 23.4780…
The area is 23.5 cm² to 3 significant figures
On the non-calculator paper the angle has an exact sine instead: with 30° in place of 63°, sin 30° = ½ and the area is ½ × 8.5 × 6.2 × ½ = 13.175 cm²

Exam tip

Carry your earlier answers into the next part

Long questions on these rules often chain their parts: a length or angle you calculate in one part is needed in a later one, for another rule or for an area. The method marks that use it are given for your own answer, even a wrong one, so substitute it and keep going rather than leave the part blank.

Worked example

The area of a quadrilateral in two triangles

In quadrilateral PQRS the diagonal PR is drawn. PQ = 7.8 cm, QR = 5.4 cm, PS = 11.6 cm, angle PQR = 118° and angle RPS = 49°. Find the length of PR and the area of PQRS, each correct to 3 significant figures.

Quadrilateral PQRS with P at the bottom left, Q at the bottom right, R on the right and S at the top, and the diagonal PR drawn. PQ is labelled 7.8 cm, QR 5.4 cm and PS 11.6 cm; the angle PQR is marked 118 degrees and the angle between PR and PS at P is marked 49 degrees. PR and RS are not labelled. The figure is marked NOT TO SCALE.

Solution:

The diagonal splits PQRS into two triangles, PQR and PRS
In triangle PQR the angle Q lies between PQ and QR, so the cosine rule gives PR
PR² = 7.8² + 5.4² − 2 × 7.8 × 5.4 × cos 118°
PR² = 60.84 + 29.16 − 84.24 × cos 118° = 90 + 39.5482… = 129.5482…
The angle is obtuse, so cos 118° is negative and subtracting 84.24 × cos 118° adds 39.5482…
PR = √129.5482… = 11.3819…, so PR = 11.4 cm to 3 significant figures
Triangle PQR: area = ½ × 7.8 × 5.4 × sin 118° = 18.5948… cm²
Triangle PRS: angle RPS lies between PR and PS, so area = ½ × 11.3819… × 11.6 × sin 49° = 49.8222… cm²
Area of PQRS = 18.5948… + 49.8222… = 68.4171…, so the area is 68.4 cm² to 3 significant figures
Use the unrounded PR in the second triangle: with 11.4 the total comes to 68.496…, which rounds to 68.5 instead