Mathematics • Year 10 • Unit 3 • Lesson 5
Compass Bearings & Navigation, Skill Drill
Build fluency with the bearings system from Lesson 5: true bearings always written as three digits (000°-360°) measured clockwise from North; back bearings found by ±180°; and the trig relationships d × sin(bearing) for the East component and d × cos(bearing) for the North component (for bearings 0°-90°). Sketch every problem on a compass cross.
1. I do, fully worked example
Read every step. Each one has a short reason on the right so you can see why, not just what.
Problem. A yacht sails 8 km on a true bearing of 060° from a harbour. Find how far East and how far North of the harbour the yacht has travelled. Round to 2 d.p.
Step 1, Sketch the compass cross.
N at the top. Draw a line from the harbour at 60° clockwise from N. Length = 8 km.
Reason: bearings start at N and rotate clockwise. The diagram fixes the orientation.
Step 2, Identify the right-angled triangle.
Angle at harbour from N = 60°. Horizontal leg = East component = E. Vertical leg = North component = N.
Reason: the right angle sits where the East and North legs meet (at the destination's "shadow" on the axes).
Step 3, East = d × sin(bearing).
E = 8 × sin 60° = 8 × 0.8660 = 6.93 km
Reason: the East component is opposite to the bearing angle measured from N, that is the sine.
Step 4, North = d × cos(bearing).
N = 8 × cos 60° = 8 × 0.5 = 4.00 km
Reason: the North component is adjacent to the bearing angle, that is the cosine.
Step 5, Verify with Pythagoras.
√(6.93² + 4.00²) = √(48.02 + 16.00) = √64.02 ≈ 8.00 ✓
Answer: the yacht is 6.93 km East and 4.00 km North of the harbour.
2. We do, fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank. 5 marks
Problem. Anika walks from her front door on a bearing of 075° for 200 m. Then she stops. Find the back bearing she would walk to return home, and her East and North displacement from her front door. Round to 1 d.p.
Step 1, Sketch the compass cross.
N at top. Walk from front door at ______° clockwise from N for ______ m.
Step 2, Back bearing rule.
If forward bearing is less than 180°: back bearing = forward + ______. So back bearing = 075 + ______ = ______°.
Step 3, East component.
E = 200 × sin ______° = 200 × ______ = ______ m
Step 4, North component.
N = 200 × cos ______° = 200 × ______ = ______ m
Step 5, Verify with Pythagoras.
√(E² + N²) = √(______ + ______) = ______ ≈ 200 ✓
3. You do, independent practice
Show working. Round distances to 1 d.p. and bearings to the nearest whole degree (always written as three digits, e.g. 045°).
Foundation, bearings notation and back bearings
3.1 Convert these directions to true bearings (three-digit notation): (a) due North, (b) due East, (c) due South, (d) due West. 1 mark
3.2 Find the back bearing for a forward bearing of 030°. 1 mark
3.3 Find the back bearing for a forward bearing of 215°. 1 mark
3.4 A ship sails 12 km on a bearing of 090°. How far East and how far North does it travel? 1 mark
Standard, finding E and N components
3.5 A bushwalker walks 6 km on a bearing of 040°. Find the East and North components of her displacement. 2 marks
3.6 A drone flies 1500 m on a bearing of 020°. Find the East and North components, to the nearest metre. 2 marks
Extension, push your thinking
3.7 A yacht sails 15 km on a bearing of 030°, then turns and sails another 10 km on a bearing of 120°. Find (i) the total East displacement from the start, (ii) the total North displacement, (iii) the straight-line distance from the starting point to the final point using Pythagoras. 4 marks
3.8 An aircraft flies 200 km from town P on a bearing of 050° to reach town Q. (i) What true bearing does the pilot need to fly to return from Q to P (the back bearing)? (ii) How far East of P is Q? 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (Anika 200 m, bearing 075°)
Step 2: forward 075° is less than 180°, so back bearing = 075 + 180 = 255°.
Step 3: E = 200 × sin 75° = 200 × 0.9659 = 193.2 m East.
Step 4: N = 200 × cos 75° = 200 × 0.2588 = 51.8 m North.
Step 5: √(193.2² + 51.8²) = √(37326 + 2683) = √40009 ≈ 200.02 ≈ 200. ✓
3.1, Cardinal bearings
(a) N = 000°, (b) E = 090°, (c) S = 180°, (d) W = 270°. Three-digit notation always.
3.2, Back bearing of 030°
030° is less than 180°, so back bearing = 030 + 180 = 210°.
3.3, Back bearing of 215°
215° is ≥ 180°, so back bearing = 215 − 180 = 035°.
3.4, Ship on bearing 090°
Bearing 090° = due East. So E = 12 km, N = 0 km.
Check: 12 × sin 90° = 12 × 1 = 12 ✓; 12 × cos 90° = 12 × 0 = 0 ✓.
3.5, Bushwalker, 6 km, 040°
E = 6 × sin 40° = 6 × 0.6428 = 3.86 ≈ 3.9 km East.
N = 6 × cos 40° = 6 × 0.7660 = 4.60 ≈ 4.6 km North.
3.6, Drone, 1500 m, 020°
E = 1500 × sin 20° = 1500 × 0.3420 = 513.03 ≈ 513 m East.
N = 1500 × cos 20° = 1500 × 0.9397 = 1409.54 ≈ 1410 m North.
3.7, Yacht: 15 km @ 030° then 10 km @ 120°
Leg 1: E₁ = 15 × sin 30° = 15 × 0.5 = 7.5 km; N₁ = 15 × cos 30° = 15 × 0.8660 = 12.99 km.
Leg 2: E₂ = 10 × sin 120° = 10 × 0.8660 = 8.66 km; N₂ = 10 × cos 120° = 10 × (−0.5) = −5.00 km (i.e. South).
Total E = 7.5 + 8.66 = 16.16 km East. Total N = 12.99 − 5.00 = 7.99 km North.
Straight-line distance = √(16.16² + 7.99²) = √(261.1 + 63.8) = √324.9 ≈ 18.0 km.
3.8, Aircraft 200 km on 050°
(i) Back bearing = 050 + 180 = 230°.
(ii) E = 200 × sin 50° = 200 × 0.7660 = 153.21 ≈ 153.2 km East.