Mathematics • Year 9 • Unit 3 • Lesson 16
Compass Bearings — Mixed Challenge
Pull together everything from Lesson 16: reading and drawing a bearing, decomposing into N-S and E-W components, multi-leg journeys, spotting a classic mistake, and an open-ended bearing puzzle.
1. Mixed problems — pick the right approach
Each question uses a different combination of compass-bearing skills. Decide what the question is really asking BEFORE you start writing. Show your working. 3 marks each
1.1 A drone flies 20 km on bearing N72°E. Find its position relative to base (km north and km east, 2 d.p.).
1.2 A boat heads on bearing S15°E for 18 km, then changes to bearing S15°W for 22 km. Without computing components, explain why the boat ends up DIRECTLY south of its start. What is the total southward distance (2 d.p.)?
1.3 A runner trains on bearing N45°E for 4 km, then on bearing N45°W for 4 km. Find (a) the total northward distance from the start and (b) the net east-west position (2 d.p.).
1.4 Express each of the following as a compass bearing of the form NθE, NθW, SθE or SθW: (a) "due east"; (b) "exactly half-way between W and S"; (c) "due north".
1.5 A ship sails 60 km on bearing S55°W. Find the southward and westward components. Then find the value of $\theta$ such that the eastward component of "60 km on bearing NθE" would equal the WESTWARD component of the original leg.
1.6 A glider released from a balloon flies 5 km on bearing S30°E. A second glider released at the same point flies 5 km on bearing N30°W. (a) What is the relationship between these two bearings? (b) How far apart are the two gliders at the end of their flights? (2 d.p.)
2. Find the mistake
Another student has tried to find the northward and eastward components of a 50 km leg on bearing N65°E. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — 50 km on bearing N65°E:
Line 1: Angle is 65°, measured from the N axis.
Line 2: Right triangle: hypotenuse 50, N-component along N, E-component along E.
Line 3: N-component = 50 sin 65° ≈ 45.32 km.
Line 4: E-component = 50 cos 65° ≈ 21.13 km.
Line 5: So the boat is about 45.32 km N and 21.13 km E of start.
(a) Which line(s) contain the mistake?
(b) Explain in one or two sentences why those lines are wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? When the angle is measured FROM the N axis, the N side is ADJACENT to that angle. Which trig ratio uses adjacent + hypotenuse?3. Open-ended challenge — design your own treasure hunt
This question has many valid answers — there are several different bearing combinations that work. 4 marks
3.1 Design a two-leg treasure hunt that satisfies ALL of the following:
(i) Leg 1 has bearing Nθ1E and length 100 m.
(ii) Leg 2 has bearing Sθ2E and length 50 m.
(iii) The chest ends up exactly due east of the starting palm tree (i.e. zero net northward displacement).
For your design:
(a) State your chosen values for θ1 and θ2.
(b) Show working that the total northward displacement is zero.
(c) Find how far east the chest is from the palm (2 d.p.).
Bonus: Try at least two different valid pairs (θ1, θ2) and notice what they have in common.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 20 km on N72°E
N-comp = 20 cos 72° ≈ 20 × 0.3090 ≈ 6.18 km north; E-comp = 20 sin 72° ≈ 20 × 0.9511 ≈ 19.02 km east. The angle is large (72°), so the path runs MOSTLY east — confirmed by 19.02 > 6.18.
1.2 — S15°E for 18 km then S15°W for 22 km
The east-component of leg 1 is 18 sin 15° eastward; the west-component of leg 2 is 22 sin 15° westward. Those are not equal in general, so the boat does NOT end directly south unless the two distances are equal.
Correction: the statement "boat ends up directly south" only holds when 18 sin 15° = 22 sin 15° — which is false here. The student should compute the net E-W: net E = 18 sin 15° − 22 sin 15° ≈ (18 − 22)(0.2588) ≈ −1.04 km, i.e. 1.04 km WEST of start. Total southward = 18 cos 15° + 22 cos 15° = 40 cos 15° ≈ 38.64 km south. So the boat is about 38.64 km S and 1.04 km W — not directly south.
1.3 — 4 km on N45°E then 4 km on N45°W
Leg 1: N = 4 cos 45° ≈ 2.83 km, E = 4 sin 45° ≈ 2.83 km.
Leg 2: N = 4 cos 45° ≈ 2.83 km, W = 4 sin 45° ≈ 2.83 km.
(a) Total northward = 2.83 + 2.83 = 5.66 km north (or 4√2 km).
(b) Net E-W: 2.83 east − 2.83 west = 0 km — directly north of start.
1.4 — Special directions
(a) Due east is 90° from N toward E: N90°E (boundary case).
(b) Half-way between W and S = 45° off S toward W: S45°W.
(c) Due north is 0° from N: N0°E (or simply N — a boundary case).
1.5 — 60 km on S55°W; matching W-comp
S-comp = 60 cos 55° ≈ 34.41 km south; W-comp = 60 sin 55° ≈ 49.15 km west.
For 60 km on NθE to have E-comp = 49.15: 60 sin θ = 49.15, so sin θ = 0.8192 ⇒ θ ≈ 55°.
Makes sense — the same 55° off the N-S axis gives the same perpendicular (sin) component.
1.6 — Two gliders on S30°E and N30°W (5 km each)
(a) S30°E and N30°W point in exactly opposite directions (180° apart) — they are reverse bearings of one another.
(b) Same length on opposite directions = the two gliders land 5 + 5 = 10.00 km apart along the S30°E / N30°W line.
2 — Find the mistake (N65°E, 50 km)
(a) The mistake is on Lines 3 and 4 (a single conceptual error swapping sin/cos in both).
(b) When the angle is measured FROM the N axis, N is adjacent to the angle (so N uses cos) and E is opposite (so E uses sin). The student has used sin for N and cos for E — exactly swapped.
(c) Corrected working:
N-component = 50 cos 65° ≈ 50 × 0.4226 ≈ 21.13 km north.
E-component = 50 sin 65° ≈ 50 × 0.9063 ≈ 45.32 km east.
So the boat is about 21.13 km N and 45.32 km E of start. Sanity check: the angle (65°) is closer to 90° than to 0°, so the path runs MOSTLY east — confirmed.
3 — Open-ended treasure hunt (sample solution)
We need zero net northward displacement: leg 1's northward (100 cos θ1) must equal leg 2's southward (50 cos θ2). So 100 cos θ1 = 50 cos θ2, or cos θ2 = 2 cos θ1.
Design 1: θ1 = 60°, then cos θ2 = 2 × 0.5 = 1.0, so θ2 = 0°.
Leg 1 (N60°E, 100 m): N = 100 cos 60° = 50, E = 100 sin 60° ≈ 86.60.
Leg 2 (S0°E = due south, 50 m): N = −50, E = 0.
Net N = 50 − 50 = 0 ✓. Net E ≈ 86.60 m east.
Design 2: θ1 = 70°, then cos θ2 = 2 cos 70° ≈ 2 × 0.342 = 0.684, so θ2 ≈ 46.8°.
Leg 1 (N70°E, 100 m): N ≈ 34.2, E ≈ 93.97.
Leg 2 (S46.8°E, 50 m): N ≈ −34.2, E ≈ 36.46.
Net N ≈ 0 ✓. Net E ≈ 130.43 m east.
Common thread: all valid pairs satisfy cos θ2 = 2 cos θ1.
Marking: 1 mark for valid (θ1, θ2) pair; 1 for working showing N-displacement = 0; 1 for correct net E (with units); 1 for noting the general relationship between θ1 and θ2 (or producing a second valid design).