Coordinates and trigonometry · Ages 14–15

Bearings worksheet

Part of: Trigonometry

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Coordinates and trigonometry · Ages 14–15

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Write each answer in the space. For word problems, show your working and write a sentence answer.

1.B is N73°W of A. What is B's true bearing from A, in degrees?
Answer:
2.B is S6°W of A. What is B's true bearing from A, in degrees?
Answer:
3.B is on a true bearing of 206° from A. What is A's true bearing from B, in degrees?
Answer:
4.A ship sails 3 km from O on a bearing of 076° to A, then 4 km on 166° to B. How far is B from O, in km?
Answer:
5.A is 32 km due north of O, and B 12 km due east of O. To the nearest 0.1°, what is B's true bearing from A?
Answer:
6.A plane flies 30 km from O on a true bearing of 165°. Correct to 1 decimal place, how far east has it moved, in km?
Answer:
7.A is 6 km due north of O, and B 17 km due east of O. To the nearest 0.1°, what is B's true bearing from A?
Answer:
8.A plane flies 35 km from O on a true bearing of 044°. Correct to 1 decimal place, how far east has it moved, in km?
Answer:

9.B is on a true bearing of 080° from A. As a compass bearing, what is B's bearing from A?
- A. N80°E
- B. S80°E
- C. N80°W
- D. S80°W
Choose the right answer
Answer:
10.B is on a true bearing of 135° from A. As a compass bearing, what is B's bearing from A?
- A. N45°E
- B. S45°E
- C. S45°W
- D. N45°W
Choose the right answer
Answer:
11.B is on a true bearing of 134° from A, so A is on a true bearing of 226° from B.
True or false?
Answer:
12.B is on a true bearing of 143° from A, so A is on a true bearing of 217° from B.
True or false?
Answer:

13.A hiker walks 28 km from camp on 047°, then 96 km on 137°. Find how far from camp he is: put the steps in order.
- A. Distance² = 28² + 96²
- B. 137° − 47° = 90°
- C. Distance = 100 km
Write the letters in the right order
Working:
Answer:
14.A ship sails 8 km from port O on 017° to A, then 15 km on 107° to B. B is km from the port; to the nearest 0.1°, B's true bearing from O is °.
Fill in each blank
Working:
Answer:
15.A plane flies 270 km from an airport on a bearing of 023°. Correct to 1 decimal place, how far north of the airport is it, in km?
Working:
Answer:
16.From a lookout, a helicopter is on a true bearing of 285°. From the helicopter, what is the lookout's true bearing, in degrees?
Working:
Answer:
17.A plane flies 640 km from an airport on a bearing of 049°. Correct to 1 decimal place, how far north of the airport is it, in km?
Working:
Answer:
18.A hiker walks 72 km from camp on 050°, then 21 km on 140°. Find how far from camp he is: put the steps in order.
- A. Distance = 75 km
- B. 140° − 50° = 90°
- C. Distance² = 72² + 21²
Write the letters in the right order
Working:
Answer:

Show the answers and working

Bearings worksheet: Answers and working

1. 287°
73° from north towards west is anticlockwise; clockwise it is 360° − 73° = 287°.
2. 186°
Clockwise from north to south is 180°, then 6° on towards west: 180° + 6° = 186°.
3. 26°
The way back differs by 180°: 206° ± 180° = 26° (kept between 0° and 360°).
4. 5 km
The two legs' bearings differ by 166° − 76° = 90°, a right angle at A: OB = 3² + 4² = 5 km.
5. 159.4°
Looking south from A, B is tan⁻¹(1232) ≈ 20.6° towards east: S20.6°E of A, that is 180° − 20.6° ≈ 159.4° (rounded at the end).
6. 7.8 km
Split the flight into east and north parts: east = 30 × sin 165° ≈ 7.8 km.
7. 109.4°
Looking south from A, B is tan⁻¹(176) ≈ 70.6° towards east: S70.6°E of A, that is 180° − 70.6° ≈ 109.4° (rounded at the end).
8. 24.3 km
Split the flight into east and north parts: east = 35 × sin 44° ≈ 24.3 km.

9. A. N80°E
See which quarter 080° is in, then the angle from north or south, 80°.
10. B. S45°E
See which quarter 135° is in, then the angle from north or south, 45°.
11. False
The way back differs by 180°: 134° + 180° = 314°; 360° − 134° mirrors the direction instead.
12. False
The way back differs by 180°: 143° + 180° = 323°; 360° − 143° mirrors the direction instead.

Marking: 2 for the working, 1 for the answer, 1 for the sentence.

13. B → A → C
First the bearings differ by 90° (a right angle at the turn), then Pythagoras: 28² + 96² = 100 km.
14. 17, 78.9
The legs meet at a right angle at A: OB = 8² + 15² = 17; ∠AOB = tan⁻¹(158), so the bearing = 17° + ∠AOB ≈ 78.9°.
15. 248.5 km
The northward part is adjacent to the 23° angle with north: 270 × cos 23° ≈ 248.5 km.
16. 105°
The way back differs by 180°: 285° − 180° = 105°.
17. 419.9 km
The northward part is adjacent to the 49° angle with north: 640 × cos 49° ≈ 419.9 km.
18. B → C → A
First the bearings differ by 90° (a right angle at the turn), then Pythagoras: 72² + 21² = 75 km.

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