By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
(For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script)
"Imagine you’re a pilot navigating a plane—one wrong bearing, and you’re 100 miles off course. Bearings problems test the exact same skill: precision. Master this, and you’ll nail every navigation question on your exam—guaranteed."
Before diving into bearings, ensure you understand: 1. Angles on a straight line (add to 180°). 2. Compass directions (North, East, South, West) and how they divide the plane. 3. Basic trigonometry (SOHCAHTOA for right-angled triangles).
If any of these are shaky, pause and review them first.
Follow these 6 steps for EVERY bearings problem:
Question: A ship sails from point P on a bearing of 120° for 8 km to point Q. Find the bearing from Q back to P.
Step 1: Draw a North line at P. Step 2: Label the bearing 120° from P to Q. Step 3: Draw the line PQ at 120°. Step 4: Mark PQ = 8 km. Step 5: Find the back bearing: - Original bearing = 120° - Back bearing = 120° + 180° = 300° Step 6: Verify: 300° is clockwise from North at Q.
Answer: The bearing from Q to P is 300°.
Question: A plane flies from town A to town B on a bearing of 075°. What is the bearing from B back to A?
Solution: 1. Original bearing (A→B) = 075°. 2. Back bearing (B→A) = 075° + 180° = 255°. 3. Check: 255° is between 180° and 270° (Southwest direction).
Answer: 255°
What we did and why: - Used the back bearing formula because the question asked for the reverse direction. - Added 180° to the original bearing to find the opposite direction.
Question: A hiker walks 5 km from point X on a bearing of 210° to point Y. How far west of X is Y?
Solution: 1. Draw North line at X. 2. Bearing 210° = 180° + 30° (Southwest direction). 3. Split into right-angled triangle: - Hypotenuse (XY) = 5 km. - Angle between XY and South line = 30°. 4. Use cosine for west distance: - West distance = 5 × cos(30°) = 5 × 0.866 = 4.33 km.
Answer: 4.33 km west
What we did and why: - Recognised 210° is in the southwest quadrant. - Used trigonometry (cosine) to find the horizontal (west) component.
Question: Two ships leave a port at the same time. Ship A sails on a bearing of 040° at 15 km/h. Ship B sails on a bearing of 130° at 20 km/h. After 2 hours, how far apart are the ships?
Solution: 1. Calculate distances: - Ship A: 15 km/h × 2 h = 30 km. - Ship B: 20 km/h × 2 h = 40 km. 2. Find angle between paths: - Bearing A = 040°, Bearing B = 130°. - Angle between = 130° – 040° = 90° (right angle!). 3. Use Pythagoras’ theorem: - Distance apart = √(30² + 40²) = √(900 + 1600) = √2500 = 50 km.
Answer: 50 km
What we did and why: - Calculated distances first using speed × time. - Found the angle between bearings (90° = right angle). - Used Pythagoras because the paths formed a right-angled triangle.
"Alright, let’s lock this in. Bearings are just angles measured clockwise from North, always written as three digits. To solve any problem: 1. Draw North lines at the starting point. 2. Label the bearing given. 3. Use back bearings (original ± 180°) for reverse directions. 4. Split into triangles and use SOHCAHTOA or Pythagoras if needed. 5. Double-check your angles are measured clockwise from North.
Common traps? Forgetting the 3-digit format, mixing up clockwise, and missing right angles between bearings. Now go crush those exam questions—you’ve got this!
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