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.