Mathematics Standard • Year 11 • Module 2 • Lesson 17
Bearings and Navigation — Past-Paper Style
HSC Mathematics Standard 2-style writing for bearings: structured short answers plus one multi-mark extended response with explicit marking criteria.
1. Short-answer questions
1.1 A ship sails 28 km on a bearing of 145°T. Calculate how far south of its starting point the ship now lies, correct to 2 decimal places. 3 marks Band 3
1.2 From port P, a yacht sails 36 km on a bearing of 060°T to buoy B. Then it sails 27 km on a bearing of 150°T to buoy C. (a) Show that angle PBC is 90°. (b) Find the straight-line distance PC. 3 marks Band 3–4
1.3 A plane flies on a bearing of 240°T from airport A for 320 km to airport B.
(a) Convert 240°T to a compass bearing.
(b) State the bearing on which the plane must fly to return directly from B to A. Justify your method in one sentence. 4 marks Band 4
2. Extended response
2.1 A search-and-rescue boat S leaves Newcastle harbour H and travels 25 km on a bearing of 050°T to reach a reported incident at point I. Bad weather then forces it to travel 20 km on a bearing of 140°T to a sheltered cove at point C.
(a) Show that the leg HI and the leg IC are perpendicular.
(b) Calculate the straight-line distance HC, correct to 2 decimal places.
(c) Calculate the bearing from H to C, correct to the nearest degree.
(d) State, with brief justification, the bearing the boat must travel on to return directly from C to H. 7 marks Band 5–6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — explicitly shows 140° − 050° = 90°, hence HI ⊥ IC.
Part (b) — 2 marks
• 1 mark — correct Pythagoras setup: HC² = 25² + 20².
• 1 mark — correct evaluation HC = √1025 ≈ 32.02 km, with units.
Part (c) — 2 marks
• 1 mark — correct tan calculation: tan(IHC) = 20/25, angle IHC ≈ 38.66°.
• 1 mark — adds to leg-1 bearing: 050° + 38.66° ≈ 089°T (to nearest degree).
Part (d) — 2 marks
• 1 mark — correctly applies back-bearing rule to part (c) answer.
• 1 mark — final answer 269°T with one-sentence justification (e.g. "089° < 180° so add 180°").
Your response:
Stuck on (c)? In the right-angled triangle at I, angle IHC is the angle at H between HI and HC. tan(IHC) = opposite/adjacent = IC/HI = 20/25.How did this worksheet feel?
What I'll revisit before next class:
1.1 — South component on 145°T, 28 km (3 marks)
Sample response.
145°T is in the SE quadrant; angle from S = 180° − 145° = 35°.
South component = 28 × cos 35° = 28 × 0.81915… ≈ 22.94 km.
Marking notes. 1 mark — identifies SE quadrant and correct angle from S (35°). 1 mark — correct substitution into 28 cos 35°. 1 mark — correct numerical answer with units (km). A bare "22.94" without identifying the angle scores 2/3.
1.2 — Perpendicular legs + Pythagoras (3 marks)
(a) Sample response. The two consecutive bearings from B differ by 150° − 060° = 90°, so the leg PB and the leg BC meet at 90° at B. Hence angle PBC = 90°.
(b) Sample response. PC² = 36² + 27² = 1296 + 729 = 2025; PC = √2025 = 45 km.
Marking notes. (a) 1 mark — shows the 90° difference and concludes perpendicular. (b) 1 mark — correct Pythagoras setup. 1 mark — correct final value with units.
1.3 — Compass conversion and back bearing (4 marks)
(a) Sample response. 240°T is in the SW quadrant (180°–270°); compass bearing = S(240 − 180)°W = S60°W.
(b) Sample response. Bearing A→B = 240°. Since 240° ≥ 180°, back bearing B→A = 240° − 180° = 060°T. (Justification: the back bearing rule subtracts 180° when the original bearing is 180° or greater.)
Marking notes. (a) 1 mark — correct quadrant identification. 1 mark — correct compass form S60°W. (b) 1 mark — correct numerical answer 060°T. 1 mark — clear one-sentence justification using the back-bearing rule. A bare "060°" with no method or rule reference scores 1/2.
2.1 — Extended response, sample Band-6 (7 marks): annotated
Sample Band-6 response.
(a) Show legs perpendicular.
Difference in bearings at I = 140° − 050° = 90°. The leg HI (heading 050°T from H) and leg IC (heading 140°T from I) therefore form a right angle at I. Hence angle HIC = 90°. [1 mark — perpendicular shown via 90° difference.]
(b) HC distance.
Since angle HIC = 90°, Pythagoras applies:
HC² = HI² + IC² = 25² + 20² = 625 + 400 = 1025. [1 mark — Pythagoras setup.]
HC = √1025 ≈ 32.02 km (2 d.p.). [1 mark — evaluation with units.]
(c) Bearing from H to C.
In right triangle HIC, the angle at H (call it θ = angle IHC) satisfies tan θ = IC / HI = 20 / 25 = 0.8.
θ = tan⁻¹(0.8) ≈ 38.66°. [1 mark — correct trig setup and angle.]
Bearing H→C = bearing H→I + θ = 050° + 38.66° ≈ 88.66° → 089°T (to nearest degree). [1 mark — adds to leg-1 bearing and rounds.]
(d) Bearing C → H.
Applying the back-bearing rule to 089°T: 089° < 180°, so back bearing = 089° + 180° = 269°T. [1 mark — correct rule applied. 1 mark — final answer with justification.]
Conclusion: HC ≈ 32.02 km on bearing 089°T from H; return bearing from C is 269°T.
Total: 7/7.
Band descriptors for marker.
Band 3: Recognises right angle and uses Pythagoras correctly for HC; struggles with the bearing calculation or omits part (d). ≈ 3 marks.
Band 4: Correct HC; attempts the bearing using tan but adds it to the wrong reference (e.g. adds to 140° instead of 050°), or omits the rounding step. ≈ 4–5 marks.
Band 5: Correct HC, correct bearing H→C, but does not justify the back-bearing rule in (d) or makes a rule error (subtracts instead of adds). ≈ 6 marks.
Band 6: Complete and correct throughout, with units on distance, three-digit bearings, and a clear sentence justifying the back-bearing rule. 7/7.