Mathematics • Year 9 • Unit 3 • Lesson 16
Compass Bearings in the Real World
Use compass bearings to solve real navigation problems: a yacht race, bushwalking off-track, a search-and-rescue scenario, kite-flying drift, and a treasure-hunt clue. Then explain a classmate's confusion in your own words.
1. Word problems
Each problem uses the compass bearing form NθE / NθW / SθE / SθW from Lesson 16, and the components N = d cosθ / E = d sinθ (when the angle is measured from the N-S axis). Show your working — a single final answer with no working only earns half marks.
1.1 — Sydney to Bottle Buoy yacht race. A yacht leaves Sydney Harbour and sails 35 km on bearing N20°E to the first turning mark.
(a) How far north of Sydney is the yacht when it reaches the mark (2 d.p.)?
(b) How far east of Sydney (2 d.p.)?
(c) Sketch the compass rose with the leg drawn on it. 4 marks
1.2 — Bushwalking off-track. Maya leaves a marked trail and walks 1.2 km on bearing S35°E to reach an interesting rock formation.
(a) How far south of the trail is the rock (2 d.p.)?
(b) How far east of where she stepped off (2 d.p.)?
(c) On the way back she just retraces her steps. State the compass bearing of the trail FROM the rock. 4 marks
1.3 — Search and rescue. A helicopter takes off from base and flies 25 km on bearing N48°W to begin a search grid.
(a) Find the northward and westward components of the helicopter's position from base (2 d.p. each).
(b) A second helicopter takes off and needs to fly directly NORTH from base to a point that is the same northward distance from base as helicopter 1. How far north does it need to fly? 3 marks
1.4 — Kite drift. A kite is flown from a fixed peg. The wind blows the kite along bearing S15°W. After the kite settles, it is 18 m of horizontal string-length away from the peg.
(a) Find the southward drift (2 d.p.).
(b) Find the westward drift (2 d.p.).
(c) Why is the southward drift much bigger than the westward drift? Answer in one sentence using the angle 15°. 3 marks
1.5 — Treasure hunt clue. A pirate map says: "From the lone palm, walk 40 paces on bearing N25°E to the old well. From the well, walk 30 paces on bearing N25°E to the chest."
(a) Why does the chest lie on the SAME compass bearing line from the palm as the well does? Answer in one sentence.
(b) Treating one pace as 0.8 m, find the total northward distance from the palm to the chest, and the total eastward distance (2 d.p.). 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate writes the bearing of a boat heading NE as "E45°N". Their map shows the boat clearly heading half-way between N and E. In your own words, explain (i) what RULE about compass bearings they have broken, (ii) what the correct compass bearing should be, and (iii) how you would help them remember the rule in future. Refer to "first letter" somewhere in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sydney yacht (N20°E, 35 km)
(a) N-comp = 35 cos 20° ≈ 35 × 0.9397 ≈ 32.89 km north.
(b) E-comp = 35 sin 20° ≈ 35 × 0.3420 ≈ 11.97 km east.
(c) Compass rose with N up; draw a line from the centre 20° off N, swung toward E. Label "35 km" along it and "20°" at the start.
Sanity check: the small angle (20°) is from N, so the boat goes MOSTLY north — and indeed 32.89 km N > 11.97 km E. ✓
1.2 — Maya off-track (S35°E, 1.2 km)
(a) S-comp = 1.2 cos 35° ≈ 1.2 × 0.8192 ≈ 0.98 km south.
(b) E-comp = 1.2 sin 35° ≈ 1.2 × 0.5736 ≈ 0.69 km east.
(c) The trail is in the opposite direction: N35°W (from south-east of the trail, the trail lies to the north-west).
1.3 — Search-and-rescue helicopter (N48°W, 25 km)
(a) N-comp = 25 cos 48° ≈ 25 × 0.6691 ≈ 16.73 km north; W-comp = 25 sin 48° ≈ 25 × 0.7431 ≈ 18.58 km west.
(b) Second helicopter needs to fly ≈ 16.73 km north (matching helicopter 1's northward displacement).
1.4 — Kite drift (S15°W, 18 m)
(a) S-comp = 18 cos 15° ≈ 18 × 0.9659 ≈ 17.39 m south.
(b) W-comp = 18 sin 15° ≈ 18 × 0.2588 ≈ 4.66 m west.
(c) Because 15° is a small angle measured from S, the path is mostly along S — the kite drifts MOSTLY south and only slightly west. Cos 15° (≈ 0.97) is close to 1, while sin 15° (≈ 0.26) is small.
1.5 — Treasure hunt (N25°E twice)
(a) Same bearing means the same direction — palm, well, chest are collinear on a single straight line.
(b) Total distance = (40 + 30) × 0.8 = 56 m on bearing N25°E.
N-comp = 56 cos 25° ≈ 56 × 0.9063 ≈ 50.75 m north.
E-comp = 56 sin 25° ≈ 56 × 0.4226 ≈ 23.66 m east.
2.1 — Explain "E45°N" (sample response)
My classmate has broken the rule that the first letter of a compass bearing must be N or S — never E or W. The angle in a compass bearing is always measured starting from the north or south axis, then rotated toward east or west. Half-way between N and E is the same as rotating 45° from N toward E, so the correct compass bearing is N45°E. To help them remember in future, I'd tell them: "first letter = where you start, third letter = where you turn toward. You only ever START on the N-S line."
Marking: 1 mark for naming the "first letter must be N or S" rule; 1 for the correct bearing N45°E; 1 for a clear, full-sentence explanation; 1 for an actionable tip for future use.