Mathematics • Year 9 • Unit 3 • Lesson 18
Solving Bearing Problems
Build fluency with the component method for solving bearing problems: split each leg into N-S and E-W parts using N = d cosθ, E = d sinθ; sum the components; then apply Pythagoras and inverse trig to find the resultant.
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 yacht sails 40 km on bearing 060°, then 25 km on bearing 150°. Find the straight-line distance from start (2 d.p.) and the true bearing of the end-point from the start.
Step 1 — Components of leg 1 (40 km on 060°).
N1 = 40 cos 60° = 40 × 0.5 = 20.00 km
E1 = 40 sin 60° ≈ 40 × 0.866 ≈ 34.64 km
Reason: with true bearings, N = d cosθ and E = d sinθ. Bearing 060 is in NE — both components positive.
Step 2 — Components of leg 2 (25 km on 150°).
N2 = 25 cos 150° ≈ −21.65 km (southward)
E2 = 25 sin 150° = 25 × 0.5 = 12.50 km
Reason: bearing 150 is in SE (between 090 and 180). Cos 150° is negative — calculator does the sign for you.
Step 3 — Sum the components.
Net N = 20.00 + (−21.65) = −1.65 km (1.65 km SOUTH)
Net E = 34.64 + 12.50 = 47.14 km (east)
Reason: signs handled automatically; net N is slightly south, net E is well east.
Step 4 — Resultant distance using Pythagoras.
d = √(1.65² + 47.14²) ≈ √(2.72 + 2222.18) ≈ √2224.9 ≈ 47.17 km
Reason: net N and net E are perpendicular — Pythagoras gives the straight-line distance.
Step 5 — Resultant bearing.
End-point is 1.65 km S and 47.14 km E of start → SE quadrant.
Angle from S axis: tan−¹(47.14 / 1.65) ≈ 88°.
True bearing (SE formula): 180 − 88 = 092°.
Reason: in SE, true = 180 − (angle off S); here the path runs nearly due east.
Answer: distance ≈ 47.17 km; bearing ≈ 092°.
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 boat sails 30 km on bearing 040°, then 20 km on bearing 110°. Find the straight-line distance from start (2 d.p.) and the net N-S and E-W displacement.
Step 1 — Leg 1 (30 km on 040°):
N1 = 30 cos __________ ° ≈ __________ km
E1 = 30 sin __________ ° ≈ __________ km
Step 2 — Leg 2 (20 km on 110°):
N2 = 20 cos __________ ° ≈ __________ km (sign: __________ )
E2 = 20 sin __________ ° ≈ __________ km
Step 3 — Sum:
Net N = __________ + __________ = __________ km
Net E = __________ + __________ = __________ km
Step 4 — Pythagoras:
d = √( __________ ² + __________ ²) ≈ __________ km
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (single component). The middle two are standard (single leg position). The last two are extension (two-leg journey).
Foundation — single component
3.1 A car travels 10 km on true bearing 030°. Find the northward component (2 d.p.). 1 mark
3.2 A boat sails 12 km on true bearing 075°. Find the eastward component (2 d.p.). 1 mark
3.3 A drone flies 8 km on bearing 270°. State the N and E components (no calculator needed). 1 mark
3.4 A walker travels 5 km on bearing 200°. Compute cos 200° and sin 200° on your calculator. Which one is negative, and what does that mean physically? 1 mark
Standard — single leg position
3.5 A plane flies 60 km on bearing 135°. Find the N and E components, stating the direction (north/south, east/west) for each (2 d.p.). 2 marks
3.6 A ship sails 18 km on bearing 310°. Find the N and E components, with directions (2 d.p.). 2 marks
Extension — two-leg journey
3.7 A yacht sails 10 km on bearing 050° then 6 km on bearing 230°. Find the net displacement from start (a single distance + direction, 2 d.p.). 3 marks
3.8 A delivery truck travels 8 km on bearing 045° then 8 km on bearing 135°. Find (a) the net N-S and E-W displacement and (b) the straight-line distance from start (2 d.p.). 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (30 km on 040° then 20 km on 110°)
Step 1: N1 = 30 cos 40° ≈ 22.98 km; E1 = 30 sin 40° ≈ 19.28 km.
Step 2: N2 = 20 cos 110° ≈ −6.84 km (sign: negative → south); E2 = 20 sin 110° ≈ 18.79 km.
Step 3: Net N = 22.98 + (−6.84) = 16.14 km north; Net E = 19.28 + 18.79 = 38.07 km east.
Step 4: d = √(16.14² + 38.07²) = √(260.50 + 1449.32) ≈ √1709.8 ≈ 41.35 km.
3.1 — 10 km on 030° (N-comp)
N = 10 cos 30° ≈ 10 × 0.866 ≈ 8.66 km north.
3.2 — 12 km on 075° (E-comp)
E = 12 sin 75° ≈ 12 × 0.9659 ≈ 11.59 km east.
3.3 — 8 km on 270°
Cos 270° = 0; sin 270° = −1. So N = 8 × 0 = 0 km and E = 8 × (−1) = −8 km (i.e. 8 km west). 270° = due west.
3.4 — 5 km on 200°
Cos 200° ≈ −0.9397; sin 200° ≈ −0.3420. Both are negative because 200° is in the SW quadrant — physically, the walker moves SOUTH (negative N) and WEST (negative E).
3.5 — 60 km on 135°
N = 60 cos 135° ≈ −42.43 km → 42.43 km south.
E = 60 sin 135° ≈ 42.43 km → 42.43 km east.
135° is exactly half-way between E and S, so the components are equal in magnitude.
3.6 — 18 km on 310°
N = 18 cos 310° ≈ 11.57 km → 11.57 km north.
E = 18 sin 310° ≈ −13.79 km → 13.79 km west.
310° is in NW quadrant — positive N, negative E.
3.7 — 10 km on 050° then 6 km on 230°
050 and 230 differ by 180 — they are reverse bearings. So leg 2 cancels part of leg 1.
Net displacement = 10 − 6 = 4 km on bearing 050°.
Answer: 4.00 km on bearing 050° (or by components: Net N = 10 cos 50 + 6 cos 230 = 6.43 − 3.86 = 2.57; Net E = 10 sin 50 + 6 sin 230 = 7.66 − 4.60 = 3.06; d = √(2.57² + 3.06²) ≈ 4.00 km ✓).
3.8 — 8 km on 045° then 8 km on 135°
Leg 1: N1 = 8 cos 45 ≈ 5.66, E1 = 8 sin 45 ≈ 5.66.
Leg 2: N2 = 8 cos 135 ≈ −5.66, E2 = 8 sin 135 ≈ 5.66.
(a) Net N = 5.66 − 5.66 = 0 km; Net E = 5.66 + 5.66 = 11.31 km east.
(b) d = √(0² + 11.31²) = 11.31 km (= 8√2, due east).
The two perpendicular equal legs combine to send the truck exactly east.