Mathematics • Year 10 • Unit 3 • Lesson 5
Bearings in the Real World
Apply Lesson 5 to real Australian navigation — a yacht heading out of Pittwater, a bushwalker on the Great North Walk, a Surf Life-Saving patrol boat between Coogee and Maroubra. Each problem needs a compass-cross diagram first, then trig. Round distances to 1 d.p. and bearings to the nearest degree (in three-digit notation).
1. Word problems
For each: (i) sketch a compass cross with N up, (ii) draw the path, (iii) write the bearings used (three-digit), (iv) calculate using sin / cos / tan or back bearings.
1.1 — Yacht from Pittwater. A yacht leaves Pittwater heading on a bearing of 110° at 6 knots. After 2 hours it stops. Calculate (a) the total distance sailed, (b) the East displacement from Pittwater, (c) the South displacement (i.e. the negative North component, given as a positive number with "South"). Round to 1 d.p. 4 marks
1.2 — Bushwalker on the Great North Walk. A bushwalker walks 4.5 km from a campsite on a bearing of 065° to reach a creek. From the creek she continues on a bearing of 155° for another 3 km to a lookout. Find (a) the back bearing she'd take to return directly from the creek to the campsite, (b) the East and North displacement of the lookout from the campsite. 4 marks
1.3 — Surf Life-Saving patrol. A Coogee patrol boat travels from Coogee to a point 1.2 km due East, then turns and travels 0.8 km due South to reach Maroubra. (a) Find the straight-line distance between Coogee and Maroubra (use Pythagoras). (b) Find the true bearing the boat would take from Coogee directly to Maroubra. 4 marks
1.4 — Aircraft from Sydney to Canberra. An aircraft flies from Sydney Airport on a bearing of 220° for 240 km to reach Canberra. (a) State the back bearing the pilot would take to fly Canberra to Sydney. (b) Find the East and North displacement (using signed values, then report direction in words). 3 marks
1.5 — Lost-and-found. A hiker walks 5 km on a bearing of 045° from the trailhead and then realises she's lost her water bottle and needs to walk back. (a) What is the bearing back to the trailhead? (b) She decides instead to walk a different route: 4 km on a bearing of 270° (due West) — does this bring her closer to or further from the trailhead? Calculate her new distance from the trailhead. 3 marks
2. Explain your thinking
This question is about communication, not just numbers. Use full sentences. 4 marks
2.1 A classmate writes the bearing for "due North-East" as "45°". Explain in 4-6 sentences (i) which of the two notations from Lesson 5 they have used (compass vs true), (ii) the correct three-digit true bearing for that direction, (iii) why the three-digit convention matters in real navigation, and (iv) how to find the back bearing from "due North-East" in three-digit notation. Include the words "three-digit notation" and "clockwise from North".
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Yacht from Pittwater (110°, 6 knots, 2 hours)
(a) Distance = 6 × 2 = 12 nautical miles (12 km approximately, treating 1 knot = 1 NM/h).
(b) E = 12 × sin 110° = 12 × 0.9397 = 11.28 ≈ 11.3 km East.
(c) N = 12 × cos 110° = 12 × (−0.3420) = −4.10 km, so 4.1 km South.
Sailing south of due East at 110° gives positive E, negative N. The reported "South displacement" is the magnitude.
1.2 — Great North Walk bushwalker
(a) Back bearing from creek to campsite = 065 + 180 = 245°.
(b) Leg 1 (campsite → creek, 4.5 km on 065°):
E₁ = 4.5 × sin 65° = 4.5 × 0.9063 = 4.08 km. N₁ = 4.5 × cos 65° = 4.5 × 0.4226 = 1.90 km.
Leg 2 (creek → lookout, 3 km on 155°):
E₂ = 3 × sin 155° = 3 × 0.4226 = 1.27 km. N₂ = 3 × cos 155° = 3 × (−0.9063) = −2.72 km.
Total E = 4.08 + 1.27 = 5.35 km East. Total N = 1.90 + (−2.72) = −0.82 km, i.e. 0.82 km South.
1.3 — Coogee to Maroubra
(a) Straight-line distance = √(1.2² + 0.8²) = √(1.44 + 0.64) = √2.08 = 1.4422 ≈ 1.4 km.
(b) tan θ = E / S = 1.2 / 0.8 = 1.5, where θ is the angle from S toward East. θ = tan⁻¹(1.5) = 56.3°.
From N, the direct bearing = 180° − 56.3° = 123.7° ≈ 124°.
Alternative check: tan φ = 1.2/0.8 measured from S means the bearing in three-digit format is 180° − 56.3° = 123.7°.
1.4 — Sydney to Canberra (220°, 240 km)
(a) 220° ≥ 180°, so back bearing = 220 − 180 = 040°.
(b) E = 240 × sin 220° = 240 × (−0.6428) = −154.3 km, i.e. 154.3 km West.
N = 240 × cos 220° = 240 × (−0.7660) = −183.9 km, i.e. 183.9 km South.
1.5 — Lost-and-found
(a) Back bearing from 045° = 045 + 180 = 225°.
(b) After Leg 1: E = 5 sin 45° = 3.54 km, N = 5 cos 45° = 3.54 km.
After Leg 2 (4 km on 270° = due West): E becomes 3.54 − 4 = −0.46 km; N stays 3.54 km.
Distance from trailhead = √(0.46² + 3.54²) = √(0.21 + 12.53) = √12.74 = 3.57 ≈ 3.6 km.
Original distance was 5 km, new distance is 3.6 km, so she is closer than after the first leg (but she could be even closer by walking the back bearing).
2.1 — North-East and three-digit notation (sample response)
My classmate has used compass-bearing-style shorthand ("N 45° E"), but they have dropped the leading zero, which makes it look like a partial true bearing. In three-digit notation, all true bearings are written with three digits between 000° and 360°, so the correct true bearing for due North-East is 045° — measured clockwise from North. Real-world navigation depends on this convention because radio calls and chartwork can be misread when leading zeros are skipped: "45" could be confused with "450" (which doesn't exist on a 360° compass), or with "S 45° E" (which is 135° true). The back bearing of due North-East is 045 + 180 = 225°, which is exactly due South-West.
Marking: 1 for naming the compass-vs-true distinction, 1 for "045°" with clockwise-from-North reasoning, 1 for why three-digit notation matters, 1 for the correct back bearing.