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.