Mathematics • Year 9 • Unit 3 • Lesson 16
Compass Bearings
Build fluency with the NθE / NθW / SθE / SθW bearing form. Start at the N-S axis, rotate by an acute angle toward E or W, then break a journey into N-S and E-W components.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. A ship sails 50 km on bearing N30°E. Draw the bearing and find (a) the northward component and (b) the eastward component, to 2 d.p.
Step 1 — Decode the bearing.
First letter N → start facing north. Number 30° → rotate that much. Third letter E → rotate toward east.
Reason: a compass bearing always starts at N or S, never at E or W.
Step 2 — Sketch the compass rose.
Draw N up, E right, S down, W left. From the centre, draw a line 30° clockwise off the N axis. The line lands in the NE quadrant.
Reason: N up always — that's the convention. The third letter tells you which way to swing.
Step 3 — Set up the right triangle.
The 50 km path is the hypotenuse. The angle 30° sits at the start, measured from the N axis. N-component runs along N (adjacent); E-component runs along E (opposite).
Reason: angle measured FROM N means N is adjacent to the angle, E is opposite.
Step 4 — Apply the trig ratios.
N-component = 50 cos 30° ≈ 50 × 0.866 ≈ 43.30 km
E-component = 50 sin 30° = 50 × 0.5 = 25.00 km
Reason: adjacent uses cos; opposite uses sin (SOH-CAH-TOA).
Answer: (a) ≈ 43.30 km north, (b) 25.00 km east.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. A walker travels 80 km on bearing S40°W. Find (a) the southward and (b) the westward component, to 2 d.p.
Step 1 — Decode: first letter is __________ → start facing south. Number is __________ . Third letter is __________ → rotate toward __________ .
Step 2 — Quadrant: the path lands in the __________ quadrant (between S and W).
Step 3 — Right triangle: hypotenuse = __________ km. Angle of 40° sits at the start, measured from the __________ axis.
Step 4 — Trig ratios:
S-component = 80 cos __________ ° ≈ 80 × __________ ≈ __________ km
W-component = 80 sin __________ ° ≈ 80 × __________ ≈ __________ km
Answer: __________ km south and __________ km west of start.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (read/draw a bearing). The middle two are standard (compute one component). The last two are extension (full position or multi-leg).
Foundation — read and draw
3.1 A boat heads N25°E. State (i) which axis the angle is measured from, and (ii) which quadrant the boat is heading into. 1 mark
3.2 Write a sentence describing the direction S70°E in plain English (use words like "from", "rotate", "toward"). 1 mark
3.3 Sketch a compass rose (N up) and draw the bearing N60°W on it. Label the 60° angle. 2 marks
3.4 Express "directly south-east" (exactly half-way between S and E) as a compass bearing. 1 mark
Standard — one component
3.5 A drone flies 12 km on bearing N55°E. Find the eastward component to 2 d.p. 2 marks
3.6 A yacht sails 24 km on bearing S65°W. Find the southward component to 2 d.p. 2 marks
Extension — full position
3.7 A hiker walks 6 km on bearing N50°W. (a) Find the northward and (b) westward displacement from the start, to 2 d.p. 3 marks
3.8 A glider released from a balloon flies 4 km on bearing S20°E. Then a second glider released at the same point flies 3 km on bearing N20°E. Find how far apart (north-to-south) the two gliders end up. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded S40°W, 80 km)
Step 1: first letter S; number 40°; third letter W → rotate toward west.
Step 2: lands in the SW quadrant.
Step 3: hypotenuse = 80 km; angle measured from the S (south) axis.
Step 4: S-comp = 80 cos 40° ≈ 80 × 0.766 ≈ 61.28 km; W-comp = 80 sin 40° ≈ 80 × 0.643 ≈ 51.42 km.
Answer: ≈ 61.28 km south and 51.42 km west of start.
3.1 — N25°E
(i) Angle measured from the N (north) axis. (ii) The boat heads into the NE quadrant.
3.2 — S70°E in plain English
"From the centre, face south, then rotate 70° toward east." (Lands in the SE quadrant, much closer to E than to S.)
3.3 — Sketch N60°W
Compass rose with N up. From the centre, draw a line that goes 60° off N, swung to the west (so it lands in the NW quadrant, closer to W than to N). Label the 60° angle between the line and the N axis.
3.4 — South-east
Exactly half-way between S and E is 45° from each. Bearing form: S45°E.
3.5 — 12 km on N55°E (east-comp)
Angle from N, so E is opposite. E-comp = 12 sin 55° ≈ 12 × 0.8192 ≈ 9.83 km east.
3.6 — 24 km on S65°W (south-comp)
Angle from S, so S is adjacent. S-comp = 24 cos 65° ≈ 24 × 0.4226 ≈ 10.14 km south.
3.7 — 6 km on N50°W
(a) N-comp = 6 cos 50° ≈ 6 × 0.6428 ≈ 3.86 km north.
(b) W-comp = 6 sin 50° ≈ 6 × 0.7660 ≈ 4.60 km west.
3.8 — Two gliders, N-S separation
Glider 1 (4 km on S20°E): S-comp = 4 cos 20° ≈ 3.76 km south of start.
Glider 2 (3 km on N20°E): N-comp = 3 cos 20° ≈ 2.82 km north of start.
N-S separation = 3.76 + 2.82 ≈ 6.58 km apart (north-to-south).
Why add: one glider is south of the start, the other is north — their N-S gap is the sum, not the difference.