Bearings Quiz (Grade 4)

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Score: 0 / 10

Bearings — Notes & Worked Examples Show / Hide

Explanation:

Every bearing must obey three rules: (1) measure from North, (2) measure clockwise, and (3) write the answer as 3 digits (e.g. 45° must be written as 045°). Look carefully at the word “from”: for the bearing of B from A, stand at A and measure clockwise from the North line at A.

Because North lines are parallel, the interior angles between two North lines are co-interior angles and add up to 180°. Use this (together with angles around a point = 360°) to calculate back bearings without a protractor.

Example 1: The bearing of ship B from port A is 065°. Work out the bearing of port A from ship B.

Co-interior at B:

180° − 65° = 115°

360° − 115° = 245°

Example 2: Town Q is on a bearing of 295° from Town P. Work out the bearing of Town P from Town Q.

At P: 360° − 295° = 65°

At Q: 180° − 65° = 115°

Answer: 115°

One Page Guide (PDF) Bearings Worksheets

Question 1 Specimen Set 2 3F Q18

[1 mark]

The diagram shows the positions of two churches, A and B.

Amber says,
"The bearing of church B from church A is 025°"
Amber is wrong.
Explain why.

Worked Solution & Answer

360° – 25° = 335° clockwise from North C1

Answer: Bearings are measured clockwise from North, so the bearing is 335° (not 025°)

Question Paper (PDF) • Mark Scheme (PDF)

Question 2 AQA Nov 2023 2H Q4

[1 mark]

A, B and C are three points.

Work out the bearing of C from A.

Answer

Worked Solution & Answer

360° – (80° + 50°) = 360° – 130° = 230°. B1

Answer: Answer 230

Question Paper (PDF) • Mark Scheme (PDF)

Question 3 IGCSE Jun 2025 2F Q16

[2 marks]

The diagram shows the positions of two villages, A and B

The bearing of B from A is 073°
Work out the bearing of A from B

Worked Solution & Answer

180 + 73 = 253° M1 A1

Answer: 253 °

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Question 4 AQA Nov 2023 2F Q21

[2 marks]

The 3-figure bearing of B from A is 021°

Work out the 3-figure bearing of C from A.

Answer °

Worked Solution & Answer

Bearing from A to C = 021° + 58° = 079°. M1 A1

Answer: Answer 079 °

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Question 5 AQA Nov 2019 2F Q28 / 2H Q9

[3 marks]

J and K are ships. P is a port.
J is due South of P.
Angle JPK = 56°
JP = KP

Work out the bearing of J from K.

Answer °

Worked Solution & Answer

Triangle JPK is isosceles with JP = KP M1
Base angles: ∠ PJK = ∠ PKJ = 180° - 56°2 = 124°2 = 62°
Since J is due South of P, line PJ lies along the North-South line
The North line at K is parallel to PJ M1
By alternate angles, the angle between the North line at K and line KP (measured anticlockwise) is 56°
Angle ∠ PKJ = 62° extends further anticlockwise to line KJ
Anticlockwise angle from North at K to KJ = 56° + 62° = 118°
Therefore, the clockwise bearing of J from K = 360° - 118° = 242°. A1

Answer: Answer 242 °

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Question 6

[3 marks]

The accurately drawn diagram shows two points, C and D, on a map.

(a) Find the bearing of D from C. [1 mark]

(b) Work out the real distance from C to D. [2 marks]

Worked Solution & Answer

(a)

Measure clockwise from North line at C: 115° B1

Answer (a): 115 °

(b)

Measured distance CD on printed paper is 7.5 cm (range 7.3 to 7.7 cm) M1
Real distance = 7.5 × 4 = 30 km A1

Answer (b): 30 km

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Question 7 AQA Jun 2022 3H Q12

[3 marks]

(a) B is 6 km due South of A and 6 km due West of C.
Work out the bearing of A from C.
12 (b) Here is a scale drawing. A ship is going to sail from D to E.
Mia works out that the ship needs to sail on a bearing of 068°
Why must Mia be wrong?

(a) B is 6 km due South of A and 6 km due West of C.
Work out the bearing of A from C. [2 marks]

Answer °

(b) Here is a scale drawing. A ship is going to sail from D to E.
Mia works out that the ship needs to sail on a bearing of 068°
Why must Mia be wrong? [1 mark]

Worked Solution & Answer

(a)

Triangle ABC has right angle at B, with AB = BC = 6 km M1
Therefore △ ABC is an isosceles right-angled triangle, so ∠ BCA = 45°
From C, due West is on bearing 270°
A is 45° north of west from C
Bearing of A from C = 270° + 45° = 315° (or 360° - 45° = 315°). A1

Answer (a): Answer 315 °

(b)

A bearing of 068° points towards the north-east
From D to E, the direction is towards the south-west, which must have a bearing between 180° and 270° (specifically 068° + 180° = 248°). C1

Answer (b): 068° is North-East, but sailing from D to E is South-West (bearing must be between 180° and 270°)

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Question 8 OCR Nov 2017 6H Q9

[4 marks]

The diagram shows the positions of two towns, Amton and Bisham.

The bearing of Bisham from Amton is b°.
The bearing of Amton from Bisham is 6b°.
Calculate the 3-figure bearing of Amton from Bisham.

Worked Solution & Answer

The back-bearing of Amton from Bisham is b + 180° M2
So 6b = b + 180 ⇒ 5b = 180 ⇒ b = 36 M1
Bearing of Amton from Bisham = 6 × 36° = 216°. A1

Answer: 216 °

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Question 9 Eduqas Jun 2019 1F Q12

[4 marks]

Work out
(a) The diagram shows the position of two points, A and B. The scale is 1cm represents 1km.
Measure and write down the bearing of B from A.
(b) Point C is 5km from A and 6km from B.
The bearing of C from B is an acute angle.
Complete the diagram to find the position of point C.

(a) The diagram shows the position of two points, A and B. The scale is 1cm represents 1km.
Measure and write down the bearing of B from A. [1 mark]

(b) Point C is 5km from A and 6km from B.
The bearing of C from B is an acute angle.
Complete the diagram to find the position of point C. [3 marks]

Worked Solution & Answer

(a)

Measuring clockwise angle from North at A to line AB gives 220° (or 218° to 222°). B1

Answer (a): 220° °

(b)

Draw arc of radius 5 cm centered at A and arc of radius 6 cm centered at B M2
Point C is at the intersection giving an acute bearing from B. A1

Answer (b): Point C constructed

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Question 10 OCR Jun 2022 2F Q16

[5 marks]

A town square has a fountain (F) at the centre.
There is also a bell tower (B) and a statue (S).
The bearing of the statue from the fountain is x°.

(a) The bearing of the bell tower from the fountain is 140° more than the bearing of the statue from the fountain.
Write down, in terms of x, the bearing of the bell tower from the fountain. [1 mark]

(b) The bearing of the bell tower from the fountain is also three times the bearing of the statue from the fountain.
Work out the bearing of the bell tower from the fountain. [4 marks]

Worked Solution & Answer

(a)

Bearing of B from F = (x + 140)°. B1

Answer (a): x + 140 °

(b)

3x = x + 140 ⇒ 2x = 140 ⇒ x = 70 M3
Bearing of bell tower = 3 × 70 = 210°. A1

Answer (b): 210 °

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Score: 0 / 10 (0 / 28 marks)

Bearings Worksheets Next: Cylinders →

Grade 4 Quizzes

24 Topics

4A - Compound interest	4B - Indices (index laws)
4C - Prime factors HCF and LCM	4D - Real life graphs
4E - Inequalities on a number line	4F - Solving inequalities
4G - Forming and solving equations	4H - nth term of a linear sequence
4J - Expanding and factorising	4K - Pythagoras theorem
4L - Angles in parallel lines	4M - Angles in polygons
4N - Surface area	4P - Volume of a cuboid
4Q - Volume of a prism	4R - Bearings
4S - Cylinders	4T - Plans and elevations
4U - Scatter graphs	4V - Probability
4W - Averages from tables	4X - Reverse mean
4Y - Area of a trapezium	4Z - Loci and construction