Mathematics • Year 9 • Unit 3 • Lesson 18
Bearings: Real Journeys
Apply the component method to authentic multi-leg navigation: cargo-ship routing, a hiking detour, an air-rescue triangle, a delivery van round-trip, and finding the bearing-home from a remote campsite. Then explain why "adding leg lengths" is wrong.
1. Word problems
Each problem uses the component method (N = d cosθ, E = d sinθ) and Pythagoras for resultants. Show your working — a single final answer with no working only earns half marks.
1.1 — Cargo ship route. A cargo ship leaves Brisbane and sails 120 km on bearing 080°, then a further 90 km on bearing 130°.
(a) Find the ship's N and E displacement from Brisbane (2 d.p.).
(b) Find the straight-line distance from Brisbane (2 d.p.). 4 marks
1.2 — Hiking detour. A hiker plans a straight 5 km route on bearing 070°, but a fallen tree forces a detour: she walks 5 km on 070° (planned), then 1 km on 160° to skirt the tree, then 1 km on 340° to rejoin the original line.
(a) Find her actual end-point relative to start (N and E, 2 d.p.).
(b) Confirm she is back on the planned line — i.e. her end-point is exactly on the 070° line from start. 3 marks
1.3 — Air-rescue triangle. A rescue helicopter flies from base 30 km on bearing 045° to a search location, then 20 km on bearing 135° to a pickup point.
(a) Find the helicopter's position relative to base after the two legs (N and E, 2 d.p.).
(b) Find the straight-line distance from base to the pickup point (2 d.p.). 3 marks
1.4 — Delivery van round-trip. A delivery van leaves the depot, drives 6 km on bearing 100°, then 4 km on bearing 200°, then 5 km on bearing 300°.
(a) Find the van's net N and E displacement (2 d.p.).
(b) Find the straight-line distance back to the depot (2 d.p.). 4 marks
1.5 — Bearing home from a campsite. A group of campers walks 8 km on 060°, then 5 km on 140° to reach a remote campsite. They want the true bearing they should walk to return DIRECTLY to the start.
(a) Find the campsite's position relative to start (N and E, 2 d.p.).
(b) Find the true bearing from the start TO the campsite.
(c) Find the true bearing for the return trip (campsite back to start). 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate calculates the straight-line distance from start for a journey of "10 km on 040° then 15 km on 220°" as 25 km. They got this by adding the leg lengths: 10 + 15. In your own words, explain (i) why simply adding leg lengths is WRONG in general, (ii) the special case where the answer would actually be the DIFFERENCE (10 − 15) of the leg lengths (and why), and (iii) the correct straight-line distance for this journey. Refer to "reverse bearing" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cargo ship (120 km on 080, 90 km on 130)
Leg 1: N1 = 120 cos 80° ≈ 20.84; E1 = 120 sin 80° ≈ 118.18.
Leg 2: N2 = 90 cos 130° ≈ −57.85; E2 = 90 sin 130° ≈ 68.94.
(a) Net N = 20.84 − 57.85 ≈ −37.01 km (37.01 km south); Net E = 118.18 + 68.94 ≈ 187.12 km east.
(b) d = √(37.01² + 187.12²) = √(1369.74 + 35013.89) ≈ √36383.63 ≈ 190.74 km.
1.2 — Hiking detour
Leg 1 (5 km, 070°): N = 5 cos 70° ≈ 1.71; E = 5 sin 70° ≈ 4.70.
Leg 2 (1 km, 160°): N = 1 cos 160° ≈ −0.94; E = 1 sin 160° ≈ 0.34.
Leg 3 (1 km, 340°): N = 1 cos 340° ≈ 0.94; E = 1 sin 340° ≈ −0.34.
(a) Net N = 1.71 + (−0.94) + 0.94 = 1.71 km north; Net E = 4.70 + 0.34 + (−0.34) = 4.70 km east.
(b) These are exactly the components of the planned 5 km on 070° — the detour legs cancel because 160° and 340° are reverse bearings of equal length. ✓ Confirmed back on the planned line.
1.3 — Air-rescue triangle
Leg 1 (30 km, 045°): N = 30 cos 45 ≈ 21.21; E = 30 sin 45 ≈ 21.21.
Leg 2 (20 km, 135°): N = 20 cos 135 ≈ −14.14; E = 20 sin 135 ≈ 14.14.
(a) Net N = 21.21 − 14.14 ≈ 7.07 km north; Net E = 21.21 + 14.14 ≈ 35.36 km east.
(b) d = √(7.07² + 35.36²) ≈ √(50.0 + 1250.3) ≈ 36.06 km.
1.4 — Delivery van round-trip
Leg 1 (6, 100°): N = 6 cos 100 ≈ −1.04; E = 6 sin 100 ≈ 5.91.
Leg 2 (4, 200°): N = 4 cos 200 ≈ −3.76; E = 4 sin 200 ≈ −1.37.
Leg 3 (5, 300°): N = 5 cos 300 = 2.50; E = 5 sin 300 ≈ −4.33.
(a) Net N = −1.04 − 3.76 + 2.50 = −2.30 km (2.30 km south); Net E = 5.91 − 1.37 − 4.33 = 0.21 km east.
(b) d = √(2.30² + 0.21²) ≈ √(5.29 + 0.04) ≈ 2.31 km from depot.
1.5 — Bearing home from campsite
Leg 1 (8, 060°): N = 8 cos 60 = 4.00; E = 8 sin 60 ≈ 6.93.
Leg 2 (5, 140°): N = 5 cos 140 ≈ −3.83; E = 5 sin 140 ≈ 3.21.
(a) Net N = 4.00 − 3.83 ≈ 0.17 km north; Net E = 6.93 + 3.21 ≈ 10.14 km east.
(b) Campsite is almost due east of start. Angle from N: tan−¹(10.14 / 0.17) ≈ 89.0°. True bearing from start to campsite ≈ 089°.
(c) Return bearing = 089 + 180 = 269°.
2.1 — Why adding leg lengths is wrong (sample response)
Simply adding leg lengths is wrong in general because the legs go in different directions — the straight-line distance from start depends on how those directions COMBINE through components, not just on the total path length walked. The special case where the answer would be 10 − 15 = 5 km is when the two legs are on reverse bearings of each other (180° apart) — and that's exactly the case here: 040° and 220° differ by exactly 180°. The two legs are walked in opposite directions on the same straight line, so they partly cancel: 15 km back the way you came undoes 10 km of the original 15 km, leaving you 5 km BEHIND the start (on the bearing 220°). The correct straight-line distance from start is therefore 5 km, not 25 km.
Marking: 1 mark for explaining why simple addition is wrong (different directions); 1 for spotting "reverse bearing" (180° apart); 1 for the correct 5 km answer; 1 for noting the direction (220° or "5 km behind start").