Mathematics Standard • Year 11 • Module 2 • Lesson 17
Bearings and Navigation — Skill Drill
Drill the core bearings skills: converting between true and compass formats, finding back bearings, and resolving a journey into north–south and east–west components.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 True bearings are always measured ____________ from ____________, written with ____________ digits.
Q1.2 Complete the back-bearing rule:
If the original bearing is less than 180°, the back bearing = bearing + ____.
If the original bearing is 180° or more, the back bearing = bearing − ____.
Q1.3 In the journey diagram, how many separate North lines should be drawn if a walker visits three points A, B, then C? ____________
2. Worked example — single-leg N–S / E–W components
Every line of working has a reason on the right.
Problem. A plane flies on a bearing of 035°T for 120 km. Find how far north and how far east of its starting point the plane is, correct to 2 decimal places.
Step 1 — Identify the quadrant and angle from North.
035°T is in the NE quadrant; angle from N = 35°.
Reason: 0°–90°T means north-east of start, measured 35° clockwise from N.
Step 2 — North component uses cos.
N = 120 × cos 35° = 120 × 0.81915… ≈ 98.30 km
Reason: in the right-angled triangle, the side along North is adjacent to the angle from N — use cos.
Step 3 — East component uses sin.
E = 120 × sin 35° = 120 × 0.57358… ≈ 68.83 km
Reason: the side perpendicular to the N axis is opposite the angle — use sin.
Conclusion. The plane is 98.30 km north and 68.83 km east of its start.
3. Faded example — fill in the missing steps
A ship sails on a bearing of 220°T for 60 km. Find how far south and how far west of its starting point the ship is, correct to 2 decimal places. 4 marks
Step 1 — Identify the quadrant. 220°T is in the ____ quadrant.
Step 2 — Angle from South line: 220° − 180° = ____°
Step 3 — South component: S = 60 × cos(____°) = ____________ km
Step 4 — West component: W = 60 × sin(____°) = ____________ km
Conclusion. The ship is ____________ km south and ____________ km west of its start.
4. Graduated practice — bearings and components
Show your working in the space below each part. Round to 2 decimal places unless told otherwise.
Foundation — single-step recall (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Convert N20°E to a true bearing. | |
| 4.2 1 | Convert 250°T to a compass bearing. | |
| 4.3 1 | Find the back bearing of 075°T. | |
| 4.4 1 | Find the back bearing of 210°T. |
Standard — typical HSC difficulty (6 questions)
Sketch each scenario with a North line at every point. Label the angle you use for trig.
4.5 A yacht sails 90 km on bearing 050°T. Find how far north of its start the yacht is, to 2 d.p. 2 marks
4.6 A bushwalker walks 12 km on bearing 165°T. Find how far south of the start they end up, to 2 d.p. 2 marks
4.7 From A, walk 6 km due North to B, then 8 km due East to C. Find the straight-line distance AC. 2 marks
4.8 Using the AC from Q4.7, find the bearing from A to C, to the nearest degree. 2 marks
4.9 Convert each true bearing to a compass bearing: (a) 095°T (b) 305°T 2 marks
4.10 A boat travels on bearing 280°T for 40 km. Find how far west of the start it is, to 2 d.p. 2 marks
Extension — two-leg journey (2 questions)
4.11 A helicopter leaves base H, flies 40 km on bearing 030°T to P, then 60 km on bearing 120°T to Q. (a) Show angle PHQ-or-PBQ-equivalent: prove the two legs PH and PQ are perpendicular. (b) Find HQ to 2 d.p. 3 marks
4.12 From H, fly 50 km on 080°T to A, then 70 km on 170°T to B. Find the bearing from H back to B, to the nearest degree. 3 marks
5. Self-check the easy 3
Tick once you've verified each one.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Definition of true bearing
True bearings are always measured clockwise from North, written with three digits (e.g. 045°T, 270°T).
Q1.2 — Back-bearing rule
If bearing < 180°: back bearing = bearing + 180°. If bearing ≥ 180°: back bearing = bearing − 180°.
Q1.3 — Number of North lines
Three North lines (one at A, one at B, one at C) — each bearing is measured from North at the local point.
Q3 — Faded example (SW journey)
Step 1: SW quadrant. Step 2: 220 − 180 = 40° from South. Step 3: S = 60 × cos 40° ≈ 45.96 km. Step 4: W = 60 × sin 40° ≈ 38.57 km. Conclusion: 45.96 km south, 38.57 km west.
Q4.1 — Convert N20°E to true bearing
NE quadrant: bearing = 20° directly → 020°T.
Q4.2 — Convert 250°T to compass
SW quadrant (180°–270°): compass = S(250 − 180)°W = S70°W.
Q4.3 — Back bearing of 075°T
075° < 180°, so add 180°: 075° + 180° = 255°T.
Q4.4 — Back bearing of 210°T
210° ≥ 180°, so subtract 180°: 210° − 180° = 030°T.
Q4.5 — North component on 050°T, 90 km
NE quadrant, angle from N = 50°. N = 90 × cos 50° ≈ 57.85 km.
Q4.6 — South component on 165°T, 12 km
SE quadrant, angle from S = 180 − 165 = 15°. S = 12 × cos 15° ≈ 11.59 km.
Q4.7 — AC from N then E legs (Pythagoras)
AC² = 6² + 8² = 36 + 64 = 100; AC = 10 km.
Q4.8 — Bearing from A to C
θ = angle from N at A; tan θ = 8/6 = 1.333…; θ = tan⁻¹(1.333…) ≈ 53°. C is NE of A → bearing = 053°T.
Q4.9 — Compass bearings
(a) 095°T is SE: S(180 − 95)°E = S85°E. (b) 305°T is NW: N(360 − 305)°W = N55°W.
Q4.10 — West component on 280°T, 40 km
NW quadrant, angle from N = 360 − 280 = 80°. W = 40 × sin 80° ≈ 39.39 km.
Q4.11 — Two perpendicular legs (40 km then 60 km)
(a) 120° − 030° = 90° → angle PHQ at the turn point P is 90°. (Or: the two legs differ by exactly 90° in bearing, so they are perpendicular.) Hence HPQ is a right-angled triangle with right angle at P.
(b) HQ² = 40² + 60² = 1600 + 3600 = 5200; HQ = √5200 ≈ 72.11 km.
Q4.12 — Bearing from H back to B (50 km on 080°T then 70 km on 170°T)
Legs differ by 170° − 080° = 90°, so right-angle at A.
Components from H to B (E = positive, N = positive):
Leg 1 (080°T, 50 km): N = 50 cos 80° ≈ 8.68; E = 50 sin 80° ≈ 49.24.
Leg 2 (170°T, 70 km, SE quadrant, angle from S = 10°): southward component = 70 cos 10° ≈ 68.94, eastward = 70 sin 10° ≈ 12.16.
Net from H: N = 8.68 − 68.94 = −60.26 (so 60.26 S); E = 49.24 + 12.16 = 61.40.
Bearing H→B: in SE quadrant, angle from S = tan⁻¹(61.40 / 60.26) ≈ 45.5°.
Bearing from H = 180° − 45.5° ≈ 135°T (to the nearest degree). The question asks H→B, not B→H, so answer = 135°T.