Sine rule, cosine rule and triangle area | AQA GCSE Maths Notes

Visual model

Label opposite pairs before using sine rule

A

B

C

b

c

a

a\text{ is opposite }A

Check if the triangle is right-angled.

Use cosine rule for two sides and the included angle.

Use sine rule for matching opposite pairs.

Gold-standard guide

26 mins

What you will learn

Solve non-right-angled triangles.

Use a clear step-by-step method for sine rule, cosine rule and triangle area.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier geometry and measures skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Use the cosine rule when you have SAS (two sides and included angle) or SSS

Step 1

Identify SAS — use the cosine rule

We have b, c and the included angle A

Step 2

Substitute values

a² = 9² + 7² − 2(9)(7) × cos(65°) = 81 + 49 − 126 × 0.4226 ≈ 130 − 53.24 = 76.76

Step 3

Take the square root

a = $\sqrt{76.76}$ ≈ 8.76 cm

Watch out

Students use the sine rule when the cosine rule is needed (or vice versa)

Sine rule

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

Cosine rule

$a^{2} = b^{2} + c^{2} - 2bc cos(A).$

Triangle area

$area = \frac{1}{2} ab sin(C).$

Worked example

In triangle ABC, side b = 9 cm, side c = 7 cm and angle A = 65°. Find side a.

Identify SAS — use the cosine rule: We have b, c and the included angle A. Cosine rule: a² = b² + c² − 2bc × cos(A).

Substitute values: a² = 9² + 7² − 2(9)(7) × cos(65°) = 81 + 49 − 126 × 0.4226 ≈ 130 − 53.24 = 76.76.

Take the square root: a = $\sqrt{76.76}$ ≈ 8.76 cm.

Final answer

a ≈ 8.76 cm

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

In triangle ABC, side b = 9 cm, side c = 7 cm and angle A = 65°. Find side a.

3 marks4 minssine-rule-cosine-rule-and-triangle-area-worked

Show solution

Worked solution

1.Identify SAS — use the cosine rule: We have b, c and the included angle A. Cosine rule: a² = b² + c² − 2bc × cos(A).
2.Substitute values: a² = 9² + 7² − 2(9)(7) × cos(65°) = 81 + 49 − 126 × 0.4226 ≈ 130 − 53.24 = 76.76.
3.Take the square root: a = $\sqrt{76.76}$ ≈ 8.76 cm.

Final answer

a ≈ 8.76 cm

Mark points

M1: identify sas — use the cosine rule
M1: substitute values
M1: take the square root
A1: a ≈ 8.76 cm

Watch out

Students use the sine rule when the cosine rule is needed (or vice versa).Key check: if the angle between the two given sides is the one in the formula (angle A opposite side a), use the cosine rule.Otherwise try the sine rule.

Diagnosticrecall

Find angle B in triangle ABC where a = 5, b = 7, c = 8.

1 mark2 minssine-rule-cosine-rule-and-triangle-area-q1

Show solution

Worked solution

1.Spot the skill: Use the cosine rule when you have SAS (two sides and included angle) or SSS.
2.Use the identify sas — use the cosine rule stage first, then substitute values.
3.Keep the final answer visible: cos B = (25 + 64 − 49)/(2×5×8) → B ≈ 57.9°.

Final answer

cos B = (25 + 64 − 49)/(2×5×8) → B ≈ 57.9°

Mark points

M1: use the correct use the cosine rule when you have sas (two sides and included angle) or sss.use the sine rule when you have aas or ssa. area = $\frac{1}{2}$ × a × b × sin(c).choose the rule based on what information is given.
A1: cos B = (25 + 64 − 49)/(2×5×8) → B ≈ 57.9°

Watch out

Easyprocedure

Use the sine rule: a/sin A = b/sin B. If a = 10, A = 40°, B = 70°, find b.

2 marks3 minssine-rule-cosine-rule-and-triangle-area-q2

Show solution

Worked solution

1.Spot the skill: Use the cosine rule when you have SAS (two sides and included angle) or SSS.
2.Use the substitute values stage first, then take the square root.
3.Keep the final answer visible: b ≈ 14.9 cm.

Final answer

b ≈ 14.9 cm

Mark points

M1: use the correct use the cosine rule when you have sas (two sides and included angle) or sss.use the sine rule when you have aas or ssa. area = $\frac{1}{2}$ × a × b × sin(c).choose the rule based on what information is given.
A1: b ≈ 14.9 cm

Watch out

Mediumreasoning

Find the area of triangle with sides 6 cm and 8 cm and included angle 50°.

3 marks4 minssine-rule-cosine-rule-and-triangle-area-q3

Show solution

Worked solution

1.Spot the skill: Use the cosine rule when you have SAS (two sides and included angle) or SSS.
2.Use the take the square root stage first, then identify sas — use the cosine rule.
3.Keep the final answer visible: ≈ 18.4 cm².

Final answer

≈ 18.4 cm²

Mark points

M1: use the correct use the cosine rule when you have sas (two sides and included angle) or sss.use the sine rule when you have aas or ssa. area = $\frac{1}{2}$ × a × b × sin(c).choose the rule based on what information is given.
A1: ≈ 18.4 cm²

Watch out

Hardproblem solving

In triangle PQR, PQ = 11, QR = 9, angle Q = 100°. Find PR.

3 marks5 minssine-rule-cosine-rule-and-triangle-area-q4

Show solution

Worked solution

1.Spot the skill: Use the cosine rule when you have SAS (two sides and included angle) or SSS.
2.Use the identify sas — use the cosine rule stage first, then substitute values.
3.Keep the final answer visible: ≈ 16.1 cm.

Final answer

≈ 16.1 cm

Mark points

M1: use the correct use the cosine rule when you have sas (two sides and included angle) or sss.use the sine rule when you have aas or ssa. area = $\frac{1}{2}$ × a × b × sin(c).choose the rule based on what information is given.
A1: ≈ 16.1 cm

Watch out

Exam-stylemulti-step

Find all angles of triangle with sides 4, 5 and 6 cm.

4 marks6 minssine-rule-cosine-rule-and-triangle-area-q5

Show solution

Worked solution

1.Spot the skill: Use the cosine rule when you have SAS (two sides and included angle) or SSS.
2.Use the substitute values stage first, then take the square root.
3.Keep the final answer visible: ≈ 41.4°, 55.8° and 82.8°.

Final answer

≈ 41.4°, 55.8° and 82.8°

Mark points

M1: use the correct use the cosine rule when you have sas (two sides and included angle) or sss.use the sine rule when you have aas or ssa. area = $\frac{1}{2}$ × a × b × sin(c).choose the rule based on what information is given.
A1: ≈ 41.4°, 55.8° and 82.8°

Watch out

Grade 9 stretchproblem solving

Two sides of a triangle are 7 cm and 11 cm with included angle 60 degrees. Find the third side to 3 significant figures.

60^\circ

7\text{ cm}

11\text{ cm}

c

4 marks7 minstrig-g9

Show solution

Worked solution

1.Use the cosine rule.
2.Substitute the included angle.
3.Square root the result.

Final answer

9.64 cm

Mark points

M1: c² = 7² + 11² - 2(7)(11)cos(60)
A1: 9.64 cm

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint

16 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Sine rule, cosine rule and triangle area - 2 marksFind angle B in triangle ABC where a = 5, b = 7, c = 8.Mark answer

Answer

cos B = (25 + 64 − 49)/(2×5×8) → B ≈ 57.9°

2Angles, lines and polygons - 2 marksThe exterior angle of a regular polygon is 24°. How many sides?Mark answer

Answer

3Properties of shapes - 2 marksName all 2D shapes with equal diagonals that bisect each other.Mark answer

Answer

Rectangle, square

4Perimeter, area and volume - 3 marksFind the perimeter of a rectangle with length 13 cm and width 8 cm.Mark answer

Answer

42 cm

Mastery check

I can explain the method for sine rule, cosine rule and triangle area.
I can show clear working without skipping key steps.
I can avoid this mistake: Students use the sine rule when the cosine rule is needed (or vice versa).Key check: if the angle between the two given sides is the one in the formula (angle A opposite side a), use the cosine rule.Otherwise try the sine rule.

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