Mathematics Standard • Year 11 • Module 2 • Lesson 17
Bearings and Navigation — Problem Set
Apply true and compass bearings to real navigation scenarios — yachts, walkers, planes and surveyors — using N-line diagrams and trig to find distances and directions.
Problem 1 — Yacht with a perpendicular turn
A yacht leaves port P and sails 24 km on a bearing of 040°T to a marker buoy B. From B, the yacht then sails 18 km on a bearing of 130°T to anchor at point C.
Set up: What are we solving for?
(i) Show that the leg PB and the leg BC are perpendicular. 1 mark
(ii) Find the straight-line distance PC, correct to 2 decimal places. 2 marks
(iii) Find the bearing from P to C, correct to the nearest degree. 2 marks
Stuck? Two bearings that differ by exactly 90° guarantee Pythagoras applies on the two legs.Problem 2 — Cargo plane heading
A cargo plane leaves Sydney airport on a bearing of 295°T and flies for 480 km in a straight line before landing at a regional airfield.
Set up: What are we solving for?
(i) Identify the quadrant of 295°T and state the trig angle from the closest cardinal line. 1 mark
(ii) Calculate how far north of Sydney the airfield is, to the nearest km. 2 marks
(iii) Calculate how far west of Sydney the airfield is, to the nearest km. 2 marks
Stuck? 295°T is in the NW quadrant; angle from N is 360 − 295 = 65°. North = d cos 65°, West = d sin 65°.Problem 3 — Bushwalking out and back
Three bushwalkers leave base camp B. They walk 14 km on a bearing of 075°T to checkpoint K. They then need to return directly to B along the shortest path.
Set up: What are we solving for?
(i) State the bearing they used from B to K. 1 mark
(ii) Calculate the bearing they must walk from K to return directly to B. 2 marks
(iii) A second group leaves B and walks 14 km on bearing 255°T. Explain, with reference to back-bearings, whether group 1 and group 2 are walking along the same straight line. 2 marks
Stuck? Compare 255°T to the back bearing of 075°T.Problem 4 — Surveying a paddock
A surveyor starts at peg A, walks 45 m due North to peg B, then 28 m due East to peg C. She then needs to return directly from C back to A.
Set up: What are we solving for?
(i) Find the distance AC, correct to 2 decimal places. 2 marks
(ii) Find the bearing from A to C, correct to the nearest degree. 2 marks
(iii) State the bearing the surveyor must walk from C back to A. Show how you derived it from the answer to (ii). 2 marks
Stuck? Apply the back-bearing rule to your answer for the A→C bearing.Problem 5 — Search-and-rescue helicopter
A search-and-rescue helicopter leaves base H and flies 60 km on a bearing of 020°T to point P. At P it changes course and flies 80 km on a bearing of 110°T to its final search position S.
Set up: What are we solving for?
(i) Show that the angle HPS at point P is 90°. 1 mark
(ii) Calculate the straight-line distance HS, correct to 2 decimal places. 2 marks
(iii) Calculate the bearing from H to S, correct to the nearest degree, then state the bearing the helicopter must fly from S back to H. 3 marks
Stuck? In triangle HPS, angle at P = 90°. The angle PHS can be found from tan(PHS) = PS / HP. Then bearing H→S = 020° + angle PHS.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Yacht P → B → C
Set up. We are confirming a right angle at B, then using Pythagoras for PC, then finding the bearing P→C using components from P.
(i) Bearings differ by 130° − 040° = 90°, so legs PB and BC are perpendicular.
(ii) PC² = 24² + 18² = 576 + 324 = 900; PC = √900 = 30.00 km.
(iii) In right triangle at B, angle BPC has tan(BPC) = BC/PB = 18/24 = 0.75; angle ≈ 36.87°. Bearing P→C = bearing P→B (040°) + 36.87° ≈ 76.87° → 077°T (to nearest degree). Quick check: bearing must lie between 040° and 130°.
Problem 2 — Cargo plane on 295°T, 480 km
Set up. We are splitting the single 480 km leg into its N and W components.
(i) 295°T is in the NW quadrant (270°–360°); angle from N (toward W) = 360 − 295 = 65°.
(ii) North = 480 × cos 65° ≈ 480 × 0.4226 ≈ 203 km north (to nearest km).
(iii) West = 480 × sin 65° ≈ 480 × 0.9063 ≈ 435 km west (to nearest km).
Problem 3 — Out-and-back on 075°T
Set up. We need the back bearing of 075°T, then compare it to 255°T to see if both groups walk the same line.
(i) B→K bearing = 075°T.
(ii) 075° < 180°, so back bearing = 075° + 180° = 255°T. Walkers return on 255°T.
(iii) Group 2 walks 255°T, which is exactly the back bearing of group 1's 075°T. Yes — both groups walk along the same straight line, but in opposite directions.
Problem 4 — Surveyor A → B → C
Set up. A right-angled triangle (N leg + E leg) with the right angle at B; find AC and the bearing A→C; then apply the back-bearing rule.
(i) AC² = 45² + 28² = 2025 + 784 = 2809; AC = √2809 = 53.00 m.
(ii) θ at A from North = tan⁻¹(28/45) = tan⁻¹(0.6222…) ≈ 31.89° → 032°T (to nearest degree).
(iii) Back bearing of 032°T: 032° < 180°, so add 180° → 212°T. The surveyor must walk on bearing 212°T from C to return to A.
Problem 5 — Helicopter H → P → S
Set up. Two legs differing by exactly 90° → right triangle at P; use Pythagoras for HS and right-triangle trig for the bearing H→S; then apply back-bearing rule.
(i) 110° − 020° = 90° → right angle at P. ✓
(ii) HS² = 60² + 80² = 3600 + 6400 = 10 000; HS = 100.00 km.
(iii) tan(PHS) = PS/HP = 80/60 = 1.333…; angle PHS ≈ 53.13°. Bearing H→S = 020° + 53.13° ≈ 73.13° → 073°T (to nearest degree).
Back bearing S→H: 073° < 180°, add 180° → 253°T. (Common slip: subtracting 180° instead of adding when the original is < 180°.)