How to Solve: Bearings and Scale Drawings

How to Solve: Bearings and Scale Drawings

Introduction

"Master bearings and scale drawings, and you’ll ace 5–10% of your GCSE Maths paper—navigating ships, designing maps, and solving real-world problems worth up to 12 marks. One wrong angle, and your answer’s off by miles. Let’s fix that."

What You Need To Know First

1. Angles on a straight line – Add up to 180°.
2. Basic trigonometry – SOHCAHTOA (for A-Level, include sine/cosine rules).
3. Scale factors – How to convert real distances to map distances.

Key Vocabulary

Formulas To Know

1. Bearing Calculation
2. Formula: Bearing = 360° - (angle from North)
3. Variables:
   • Angle from North = Measured anticlockwise from North.

4. MEMORISE THIS: Bearings are always 3 digits (e.g., 005°, not 5°).

5. Back Bearing

6. Formula: Back Bearing = Original Bearing ± 180°
7. Rule: If original bearing < 180°, add 180°. If ≥ 180°, subtract 180°.

8. MEMORISE THIS: Never exceed 360°.

9. Scale Conversion

10. Formula: Real Distance = Map Distance × Scale Factor
11. Variables:
    • Map Distance = Measured length on the drawing.
    • Scale Factor = Number after the colon (e.g., 1:10,000 → scale factor = 10,000).

12. Given on exam sheet (but know how to use it).

13. Trigonometry (A-Level)

14. Formula: Distance = (Map Distance × Scale) / cos(θ) (for slopes).
15. MEMORISE THIS: Only needed if elevation is given.

Step-by-Step Method

Part 1: Solving Bearings

Step 1: Draw a North line at the starting point. Step 2: Measure the angle clockwise from North to the line. Step 3: Write the bearing as a 3-digit number (e.g., 030°, not 30°). Step 4: For back bearings, add/subtract 180° (keep it under 360°).

Part 2: Solving Scale Drawings

Step 1: Measure the map distance with a ruler (in cm or mm). Step 2: Multiply by the scale factor to get the real distance. Step 3: If the question asks for a new drawing, divide the real distance by the scale factor. Step 4: Label all distances and bearings clearly.

Worked Example (Combined)

Question: A ship sails from point A on a bearing of 120° for 15 km. It then changes course to a bearing of 045° for 10 km to point B. A map uses a scale of 1:500,000. 1. Draw the journey on the map. 2. Find the direct distance (as the crow flies) from A to B in km.

Step 1: Draw the North line at A. - Use a protractor to mark 120° clockwise from North.

Step 2: Calculate the first leg on the map. - Scale: 1:500,000 → 1 cm = 5 km. - Real distance = 15 km → Map distance = 15 ÷ 5 = 3 cm. - Draw a 3 cm line at 120° from A.

Step 3: Draw the second leg from the new point. - New bearing = 045°. - Real distance = 10 km → Map distance = 10 ÷ 5 = 2 cm. - Draw a 2 cm line at 045° from the end of the first leg.

Step 4: Measure the direct distance A to B. - Use a ruler: Measure the straight line from A to B on the map. - Suppose it’s 4.1 cm. - Real distance = 4.1 × 5 = 20.5 km.

Step 5: Find the bearing from A to B. - Measure the angle clockwise from North to the line AB. - Suppose it’s 95° → Bearing = 095°.

Answer: 1. Map drawn as above. 2. Direct distance = 20.5 km on a bearing of 095°.

What we did and why: - Used scale conversion to shrink real distances to map size. - Drew bearings accurately to plot the ship’s path. - Measured the direct route to find the shortest distance.

Worked Examples

Example 1 – Basic (GCSE)

Question: A plane flies 200 km on a bearing of 075°. How far East and North has it travelled?

Solution: 1. Draw a right-angled triangle:
- Hypotenuse = 200 km (flight path).
- Angle = 75° from North. 2. Use SOHCAHTOA:
- East distance = 200 × sin(75°) = 193.2 km.
- North distance = 200 × cos(75°) = 51.8 km. 3. Answer: East = 193 km, North = 52 km (rounded).

What we did and why: - Split the bearing into East/North components using trig. - Rounded to 2 significant figures (common in exams).

Example 2 – Medium (GCSE/A-Level)

Question: Two ships leave a port. Ship X sails 30 km on a bearing of 030°. Ship Y sails 40 km on a bearing of 120°. Find the bearing of Ship Y from Ship X.

Solution: 1. Plot both paths:
- Ship X: 30 km at 030°.
- Ship Y: 40 km at 120°. 2. Find the angle between paths:
- Difference = 120° – 30° = 90° (right angle). 3. Use Pythagoras to find distance XY:
- XY = √(30² + 40²) = 50 km. 4. Find the bearing of Y from X:
- From X’s position, measure the angle to Y.
- Angle from North = 30° + tan⁻¹(40/30) = 83.1°.
- Bearing = 083°.

Answer: Bearing of Y from X = 083°.

What we did and why: - Used relative bearings to find the angle between paths. - Applied Pythagoras because the paths were perpendicular.

Example 3 – Exam-Style (A-Level)

Question: A hiker walks 5 km on a bearing of 200°, then 3 km on a bearing of 300°. A map uses a scale of 1:25,000. 1. Draw the route on the map. 2. Find the direct distance back to the start. 3. State the bearing the hiker must take to return directly.

Solution: 1. Map distances:
- 5 km → 5 ÷ 0.25 = 20 cm (scale: 1 cm = 0.25 km).
- 3 km → 3 ÷ 0.25 = 12 cm. 2. Plot the route:
- First leg: 20 cm at 200°.
- Second leg: 12 cm at 300°. 3. Measure direct return:
- Suppose it’s 15 cm on the map.
- Real distance = 15 × 0.25 = 3.75 km. 4. Bearing back:
- Measure angle from North to return line.
- Suppose it’s 60° anticlockwise from North → Bearing = 360° – 60° = 300°.

Answer: 1. Map drawn as above. 2. Direct distance = 3.8 km (rounded). 3. Bearing = 300°.

What we did and why: - Converted real distances to map scale first. - Measured back bearing by reversing the angle.

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s the night-before cheat sheet: 1. Bearings = 3-digit angles clockwise from North. Always draw a North line first. 2. Back bearings = Original bearing ± 180° (keep it under 360°). 3. Scale drawings = Measure map distance, multiply by scale factor for real distance. 4. Trig shortcuts – If bearings differ by 90°, use Pythagoras. 5. Double-check – Did you write 3 digits? Did you measure clockwise? Did you convert units?

Now go ace that exam—you’ve got this!