The cosine rule Quiz (Grade 7)

Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8/9

Topic:

Score: 0 / 10

The cosine rule — Notes & Worked Examples Show / Hide

Explanation:

Use the Cosine Rule when you do not have a complete opposite side-angle pair: either you know 2 sides and the included angle (SAS) and want the third side using a² = b² + c² − 2bc cos A, or you know all 3 sides (SSS) and want an angle using cos A = b² + c² − a²2bc.

Make sure the side labeled a is always the side directly opposite the angle A in the formula. Remember to take the square root at the end when finding a side length.

Example 1: Find side a when b = 7 cm, c = 10 cm, and included angle A = 58° (3 s.f.).

Solution:

a² = 7² + 10² − (2 × 7 × 10 × cos 58°)

a² = 149 − 140 cos 58° = 74.8112...

a = √74.8112... = 8.6493... cm

Answer: a = 8.65 cm (3 s.f.)

Example 2: A triangle has sides 6 cm, 8 cm, and 11 cm. Find the largest angle (1 d.p.).

Solution:

Largest angle A is opposite longest side (a = 11):

cos A = 6² + 8² − 11²2 × 6 × 8 = −2196

A = cos⁻¹(−0.21875) = 102.635...°

Answer: 102.6° (1 d.p.)

One Page Guide (PDF) The cosine rule Worksheets

Question 1

[2 marks]

Use the cosine rule to calculate the length of side BC (marked x) correct to 1 decimal place.

Worked Solution & Answer

x² = 6² + 9² - 2(6)(9)\cos(60°) = 63 ⇒ x = 7.9 cm B1

Answer: 7.9 cm cm

Question 2 AQA Jun 2025 2H Q18

[4 marks]

Lines AB, BC, CD and AD form two triangles.

Work out the length of AD.

Answer cm

Worked Solution & Answer

In bottom isosceles triangle: base angles are both 32°, so vertex angle = 180° – 2(32°) = 116° M1
Vertically opposite angle in top triangle = 116° M1
By Cosine Rule: AD² = 9.6² + 2.8² – 2(9.6)(2.8) cos(116°) = 100 + 23.567 = 123.567 M1
AD = √123.567 ≈ 11.1 cm. A1

Answer: Answer 11.1 cm

Question Paper (PDF) • Mark Scheme (PDF)

Question 3

[4 marks]

Here is triangle ABC.

Work out the area of triangle ABC.
Give your answer correct to 3 significant figures.

cm²

Worked Solution & Answer

cos A = (14.6² + 7.9² – 18.2²) / (2 × 14.6 × 7.9) = –55.67 / 230.68 = –0.24133... ⇒ A = 103.965° M3
Area = 1/2 × 14.6 × 7.9 × sin(103.965°) = 55.967... = 56.0 cm² A1

Answer: 56.0 cm²

Question Paper (PDF) • Mark Scheme (PDF)

Question 4

[4 marks]

PQR and QRS are triangles.

Calculate the length of QS.
Give your answer correct to 3 significant figures.
You must show all your working.

Worked Solution & Answer

By cosine rule in ΔPQR: QR² = 11² + 9.4² – 2(11)(9.4)cos 27° = 209.36 – 206.8 cos 27° = 25.099... ⇒ QR = 5.0099... cm M3
By sine rule in ΔQRS: QS / sin 88° = QR / sin 41° ⇒ QS = (5.0099 × sin 88°) / sin 41° = 7.631... → 7.63 cm A1

Answer: 7.63 cm

Question Paper (PDF) • Mark Scheme (PDF)

Question 5 AQA Nov 2021 2H Q23

[4 marks]

A ship sails from P to Q and then from Q to R.

Q is 12 miles from P, on a bearing of 080°
R is 28 miles from Q, on a bearing of 155°
Work out the direct distance from P to R.

Answer miles

Worked Solution & Answer

Angle at Q between QP and QR: M1
Back bearing of P from Q is 80° + 180° = 260°
Bearing of R from Q is 155° M1
Angle PQR = 260° - 155° = 105°
By the cosine rule on triangle PQR: M1
PR² = 12² + 28² - 2(12)(28)\cos(105°)
PR² = 144 + 784 - 672\cos(105°) = 928 - 672(-0.2588) ≈ 1101.9
PR = √(1101.9) ≈ 33.2 miles. A1

Answer: Answer 33.2 miles

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 AQA Nov 2022 2H Q24

[5 marks]

B is 60 miles from A on a bearing of 170°

A ship sails from A on a bearing of 247°
It travels at a constant speed of 23 mph for 112 hours.
Is the ship now closer to B than it was when it left A?
You **must** show your working.

Worked Solution & Answer

(Average speed = distance ÷ time)
Distance AC = 23 × 1.5 = 34.5 miles M2
Angle BAC = 247° – 170° = 77° M1
By Cosine Rule: BC² = 60² + 34.5² – 2 × 60 × 34.5 × cos 77° = 3600 + 1190.25 – 931.30 = 3858.95 M1
BC = √3858.95 = 62.1 miles > 60 miles, so No (it is further away). C1

Answer: Distance travelled = 23 × 1.5 = 34.5 miles. Angle at A = 247° – 170° = 77°.
Distance to B = √(60² + 34.5² – 2 × 60 × 34.5 × cos 77°) = 62.1 miles > 60 miles, so No.

Question Paper (PDF) • Mark Scheme (PDF)

Question 7

[5 marks]

ABCD is a quadrilateral.

Find the size of angle ABC.
Give your answer correct to the nearest degree.

AD = 6 cm, CD = 9 cm, BC = 12 cm
Angle BCD = 60°, Angle DAB = 125°

Worked Solution & Answer

In triangle BCD: BD² = 12² + 9² – 2(12)(9)cos(60°) = 144 + 81 – 108 = 117 ⇒ BD = √117 ≈ 10.817 cm M2
Sine rule in BCD: sin(CBD)/9 = sin(60°)/√117 ⇒ sin(CBD) ≈ 0.7206 ⇒ angle CBD ≈ 46.10° M1
Sine rule in ABD: sin(ABD)/6 = sin(125°)/√117 ⇒ sin(ABD) ≈ 0.4543 ⇒ angle ABD ≈ 27.02° M1
Angle ABC = 46.10° + 27.02° = 73.12° ⇒ 73°. A1

Answer: 73 °

Question Paper (PDF) • Mark Scheme (PDF)

Question 8 OCR Nov 2023 4H Q19

[5 marks]

The diagram shows the positions of three towns A, B and C.

The bearing of town B from town A is 053°.
The bearing of town C from town B is 108°.
AB = 35 km and BC = 61 km.
Calculate AC.
You must show your working.

AC = km

Worked Solution & Answer

Interior angle at B: (180° − 108°) + 53° = 72° + 53° = 125° M2
By the cosine rule: M1
AC² = 35² + 61² − 2 × 35 × 61 × cos 125° = 1225 + 3721 − 4270 cos 125° = 4946 + 2449.17 = 7395.17 M1
AC = √7395.17 = 85.995... ≈ 86 km. A1

Answer: AC = 86 km

Question Paper (PDF) • Mark Scheme (PDF)

Question 9

[5 marks]

The diagram shows a hexagon ABCDEF.

ABEF and CBED are congruent parallelograms where AB = BC = x cm.
P is the point on AF and Q is the point on CD such that BP = BQ = 10 cm.
Given that angle ABC = 30°,
prove that \cos PBQ = 1 - (2 - √(3))200x²

Worked Solution & Answer

In △ABC: AC² = x² + x² – 2(x)(x)cos 30° = 2x² – 2x²(√3 / 2) = (2 – √3)x² M2
Since AF is parallel to CD, PQ = AC, so PQ² = (2 – √3)x² M2
In △PBQ: cos PBQ = (10² + 10² – PQ²) / (2 × 10 × 10) = (200 – (2 – √3)x²) / 200 = 1 – ((2 – √3)/200)x². C1

Answer: Proof by Cosine Rule in △ ABC and △ PBQ

Question Paper (PDF) • Mark Scheme (PDF)

Question 10 AQA Nov 2024 1H Q13

[5 marks]

(a) Estimate (√401 + 1.9³) / cos 58.7° by rounding to 1 s.f. [3 marks]
13 (b) Identify two errors Sam has made in his cosine rule working. [2 marks]

(a) By rounding each number to one significant figure,
[INDENT] estimate the value of √(401) + 1.9³\cos 58.7°
You **must** show your working. [3 marks]

Answer

(b) Here is a triangle. [2 marks]

Worked Solution & Answer

(a)

(√400 + 2³) / cos 60° = (20 + 8) / 0.5 = 28 / 0.5 = 56 M2 A1

Answer (a): Answer 56

(b)

Error 1: Cosine rule uses –2bc cos A, not +2bc cos A M1
Error 2: 3² + 7² = 9 + 49 = 58 ≠ 10² (and 100 + 42 cos x ≠ 142 cos x) A1

Answer (b): Error 1: It should be – 2 × 3 × 7 × cos x (minus sign in cosine rule). Error 2: 3² + 7² = 58, not 10² (or 100 + 42 cos x ≠ 142 cos x).

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 43 marks)

The cosine rule Worksheets Next: Congruent triangles →

Grade 7 Quizzes

19 Topics

7A - Surds	7B - Bounds
7C - Proportion (harder)	7D - Algebraic fractions
7E - The quadratic formula	7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs	7H - Rearranging (harder)
7J - Functions (harder)	7K - Iteration
7L - Area of any triangle	7M - The sine rule
7N - The cosine rule	7P - Congruent triangles
7Q - 3D pythagoras / trig.	7R - Histograms
7S - Conditional probability	7T - Exact trig. values (harder)
7U - Area of a segment