Contents · 0% complete

Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn7357Paper 1 · PureTrigonometrySine rule, cosine rule and area

Trigonometry · Subtopic 2 of 13

Sine Rule, Cosine Rule and Area

For any triangle: the sine rule links each side to its opposite angle; the cosine rule handles two sides and the included angle, or three sides; and the area is $\tfrac12ab\sin C$ . Watch for the ambiguous case when using the sine rule to find an angle.

Edexcel YMA01 WMA11Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2AQA 7357 Paper 3≈ 40 minRevision 5 · updated 2026-10-27

Practise this topicTimed test Ask Salam TutorSwitch board

Section 1 of 14

Learning objectives

Choose between the sine rule and the cosine rule.
Find sides and angles in any triangle.
Recognise and solve the ambiguous case.
Use area $=\tfrac12ab\sin C$ , forwards and backwards.

Pause and think

You know $AB=10$ , $BC=7$ and angle $A=35^\circ$ . Draw it: put $A$ and the $35^\circ$ line down, then swing a $7$ cm arc from $B$ . How many places can $C$ go?

Section 2 of 14

Core idea

Start hereMeasuring across a lake

You can't stretch a tape across a lake. But stand at a point, measure the distances to two trees on opposite shores and the angle between them, and the gap between the trees follows. Non-right-angled triangles are solved with the sine and cosine rules.

The big idea

$\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}$ ; $a^2=b^2+c^2-2bc\cos A$ ; area $=\frac12ab\sin C$ . Side $a$ is opposite angle $A$ .

1. The rules

Naming a triangle: side a is opposite angle A

Small letters for sides, capitals for angles; each side faces the angle with the same letter.

Area and sine rule

$h=b\sin C$
Height from $A$ onto side $a$ , using the right-angled triangle containing $C$ .
$\text{Area}=\tfrac12ah=\tfrac12ab\sin C$
Half base times height.
$\tfrac12bc\sin A=\tfrac12ac\sin B=\tfrac12ab\sin C$
Same area, any pair of sides.
$\frac{a}{\sin A}=\frac b{\sin B}=\frac c{\sin C}$
Divide through by $\frac12abc$ and invert.

Cosine rule

$C=(0,0),\ B=(a,0),\ A=(b\cos C,\ b\sin C)$
Place $C$ at the origin.
$c^2=(b\cos C-a)^2+(b\sin C)^2$
Distance $AB$ .
$=b^2(\cos^2C+\sin^2C)-2ab\cos C+a^2$
Expand.
$c^2=a^2+b^2-2ab\cos C$
$\cos^2+\sin^2=1$ .

With

C=90^\circ

it is Pythagoras.

Which rule?

Drawing diagram…

Find a "pair" (side + opposite angle) for the sine rule; otherwise use the cosine rule.

Worked example $a=7$ , $b=8$ , $C=60^\circ$

Find $c$ and the area.

Goal: $c$ ; area.
Strategy: Two sides and the included angle: cosine rule.

$c^2=49+64-2(7)(8)(0.5)=57$
$\cos60^\circ=\frac12$ .
$c=7.55$
Square root.
$\text{Area}=\tfrac12(7)(8)\sin60^\circ=14\sqrt3=24.2$
Area formula.

Answer

c=7.55

, area

24.2

✓ Check

60^\circ

triangle with sides 7, 8: third side between them ✓.

Worked exampleSides 5, 7, 8

Find all angles.

Goal: Three angles.
Strategy: Cosine rule rearranged: $\cos A=\frac{b^2+c^2-a^2}{2bc}$ .

$\cos\theta_8=\frac{25+49-64}{70}=\frac17\Rightarrow81.8^\circ$
Opposite 8.
$\cos\theta_7=\frac{25+64-49}{80}=\frac12\Rightarrow60^\circ$
Opposite 7.
$180-81.8-60=38.2^\circ$
Angle sum.

Answer

81.8^\circ

60^\circ

38.2^\circ

✓ CheckLargest angle opposite largest side ✓.

2. The ambiguous case

Two possible triangles with the same two sides and angle

The side of length 3.4 can reach the base in two places, giving an acute angle at

B_1

and an obtuse one at

B_2

Worked example $A=40^\circ$ , $a=5$ , $b=7$

Find $B$ .

Goal: All possible $B$ .
Strategy: Sine rule; check the obtuse alternative.

$\sin B=\frac{7\sin40^\circ}5=0.900$
Sine rule.
$B=64.2^\circ\text{ or }115.8^\circ$
$\sin(180^\circ-\theta)=\sin\theta$ .
$40+115.8=155.8<180$
Room for a third angle, so both valid.

Answer

B=64.2^\circ

115.8^\circ

: two triangles.

✓ Check

b>a

and

a>b\sin A=4.5

: the classic two-triangle condition ✓.

Edge caseOnly one triangle If the obtuse option plus the known angle reaches

180^\circ

, discard it.

Pause and think

Area of a triangle with sides 6, 6 and included angle $30^\circ$ ?

Summary

Sine rule with a pair; cosine rule otherwise.
Area $\frac12ab\sin C$ .
Check for two triangles.

Included angle: Angle between two given sides.
Ambiguous case: Two triangles from the same data.

HistoryAstronomers' triangles. The cosine rule appears in geometric form in Euclid's Elements (Book II), centuries before trigonometry existed. The 15th-century Persian astronomer Jamshid al-Kashi stated it in a form close to today's, and in France it is still called le théorème d'Al-Kashi.

Fun factHeron's shortcut. Heron of Alexandria found the area of a triangle from its three sides alone:

\sqrt{s(s-a)(s-b)(s-c)}

, where

s

is half the perimeter. It is equivalent to

\tfrac12ab\sin C

with the cosine rule built in.

Section 3 of 14

Key formulae

Sine rule

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

Cosine rule

a^2=b^2+c^2-2bc\cos A

Area

\text{Area}=\tfrac12ab\sin C

TrickPair test. Look for a known side with its opposite angle. Got a pair? Sine rule. No pair? Cosine rule. One question settles it every time.

Section 4 of 14

Worked examples

Example 1 · Exact cosine rule0 / 2 steps

In triangle

ABC

b=3

c=5

and

A=60^\circ

. Find the exact value of

a

Example 2 · Ambiguous case0 / 2 steps

AB=10

BC=7

A=35^\circ

. Find the possible values of

C

Section 5 of 14

Explore the graph

Two sides fixed, angle varies

Loading the interactive graph…

Try this

Sides $5$ and $7$ meet at the origin with angle $c$ . The values are the area $\tfrac12(5)(7)\sin c$ and the third side from the cosine rule.
Which angle gives the largest area? Why?
Compare $c=30$ and $c=150$ . The areas match but the third sides do not. Why?

$\sin c=\sin(180^\circ-c)$ , but $\cos c$ changes sign.

In the real worldSurveying without crossing. Surveyors measure the width of a river by taking a baseline on one bank and measuring angles to a tree on the far bank. The sine rule then gives the distance without anyone getting wet.

Section 6 of 14

Common mistakes

IncorrectWorking out

(b^2+c^2-2bc)\cos A

CorrectFind

2bc\cos A

first

Order of operations.

IncorrectIgnoring the obtuse answer

CorrectCheck

180^\circ-\theta

Both may give valid triangles.

IncorrectArea with an angle not between the sides

CorrectUse the included angle

C

must be between

a

and

b

IncorrectCalculator in radians

CorrectDegrees mode

Check the mode before every triangle question.

Sine rule

I found you an angle!

The other angle

$180^\circ$ minus that one also works

the ambiguous case, my old enemy

Section 7 of 14

Exam tip

Exam tipKeep the unrounded value. Store each intermediate side or angle in your calculator and use the stored value in the next step. Rounding to 3 s.f. midway can push the final answer outside the accepted range.

Section 8 of 14

Guided practice

SA-IAL-P1-TR-G013 marksL1 FoundationCalculatorSine & cosine rules

In triangle

ABC

a=7

cm,

b=5

cm and angle

C=60^\circ

. Find the length

c

, to 3 significant figures.

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-IAL-P1-TR1-0012 marksL1 FoundationNon-calculatorSine & cosine rules

In triangle

ABC

b=3

cm,

c=5

cm and angle

A=60^\circ

. Find the exact length of

a

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-P1-TR-0013 marksL1 FoundationCalculatorSine & cosine rules

In triangle

ABC

, angle

A=40^\circ

, angle

B=65^\circ

and

a=8

cm. Find

b

, to 3 significant figures.

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-P1-TR1-0023 marksL2 StandardCalculatorSine & cosine rules

In triangle

ABC

a=6

cm,

b=8

cm and angle

A=30^\circ

. Angle

B

is acute. Find

B

, in degrees to 1 decimal place.

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-P1-TR-0023 marksL2 StandardCalculatorSine & cosine rules

A triangle has sides

5

cm,

6

cm and

8

cm. Find the largest angle, in degrees to 1 decimal place.

Salam Academy Original Practice · Proposed mark scheme

5 more practice questions with hints, worked solutions and instant marking
Included in Student Premium.

See Premium

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

1 more challenge question with hints, worked solutions and instant marking
Included in Student Premium.

See Premium

Section 11 of 14

Summary

Sine rule: needs a side–angle pair.
Cosine rule: two sides and the included angle, or three sides.
Ambiguous case: test $180^\circ-\theta$ .
Area $=\tfrac12ab\sin C$ .

Section 12 of 14

Quick quiz

1 more quiz question with hints, worked solutions and instant marking
Included in Student Premium.

See Premium

Section 13 of 14

Same topic, different board

Specification: 7357 A-level Mathematics
Where: Section E, Trigonometry
Expected: The same, including the ambiguous case.
Calculator: Permitted; some answers must still be exact.
Typical style: AQA mixes short multiple-choice questions with longer structured ones.
Open in: 7357 · Paper 1 7357 · Paper 2 7357 · Paper 3

Section 14 of 14

Past paper questions

3 questions (29 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

WMA11/01 · June 2025Question 69 marks
Sector and triangle in radians: an angle, then area and perimeter.
Question paper, page 14 Mark scheme
WMA11/01 · January 2025Question 813 marks
Angles, areas and arc lengths in radians for a badge design, and its perimeter.
Question paper, page 26 Mark scheme
WMA11/01 · June 2024Question 57 marks
Sector area, trigonometry in radians and a composite area.
Question paper, page 10 Mark scheme

All YMA01 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7357, First teaching 2017 (official specification).