Mathematics • Year 10 • Unit 3 • Lesson 5
Bearings & Navigation — Mixed Challenge
Pull together every idea from Lesson 5: three-digit notation, back bearings (±180°), E and N components using d × sin(bearing) and d × cos(bearing), multi-leg journeys with signed components, and computing a final straight-line bearing using inverse tan and quadrant reasoning. Spot the missing-leading-zero error, then design two paths that finish in the same place.
1. Mixed problems
Each question pulls a different idea from Lesson 5. Sketch first. Round distances to 1 d.p. and bearings to the nearest whole degree (three-digit). 2-4 marks each
1.1 Find the back bearing of (a) 075°, (b) 198°, (c) 270°, (d) 005°. 2 marks
1.2 A car drives 8 km on a bearing of 320°. Find the East and North components, using signed values then stating direction in words. 2 marks
1.3 A boat starts at port and motors 7 km East then 5 km North. Find the bearing of the boat from port, in three-digit notation, to the nearest degree. 3 marks
1.4 A drone is 600 m East and 1100 m South of its launch point. Find the bearing from the launch point to the drone, to the nearest degree. (Hint: this is in the SE quadrant — bearings 090°-180°.) 3 marks
1.5 A bushwalker walks 6 km on a bearing of 130° and then 4 km on a bearing of 250°. Find (i) the East and North displacement from her starting point, (ii) the straight-line distance from start to finish. 3 marks
1.6 Town Q is 80 km from town P on a bearing of 060°. Town R is 50 km from P on a bearing of 150°. Find (i) the East and North coordinates of both Q and R relative to P, (ii) the straight-line distance from Q to R. 4 marks
2. Find the mistake
A Year 10 student attempts this problem: "A yacht sails 10 km on a bearing of 25° from port. Find its East and North components." Their working is shown. Exactly one line contains a mistake. Spot it, explain, and redo correctly. 3 marks
Student's working:
Line 1: Bearing = 25°.
Line 2: E = 10 × sin 25° = 10 × 0.4226 = 4.226 km East.
Line 3: N = 10 × cos 25° = 10 × 0.9063 = 9.063 km North.
Line 4: Back bearing = 25 + 180 = 205°.
(a) Which line contains the convention-based mistake (not a calculation error)?
(b) Explain in one or two sentences which Lesson 5 convention has been broken.
(c) Write the corrected line(s).
Stuck? Lesson 5 § "Words You Need" requires three-digit notation for true bearings — always 025°, not 25°.3. Open-ended challenge — design two paths that end in the same place
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design two different two-leg journeys (each with bearings written in three-digit notation) that both start at the same point and finish at the same final point exactly 5 km North and 5 km East of the start. The two journeys must use different bearings on at least one leg.
For each journey:
(i) State the bearing and distance for each leg.
(ii) Show E and N components for each leg.
(iii) Verify the totals add to (E = 5, N = 5).
Bonus: Calculate the direct one-leg journey from start to finish — its bearing should come out to 045° and distance √50 ≈ 7.07 km.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Back bearings
(a) 075° → 075 + 180 = 255°.
(b) 198° → 198 − 180 = 018°.
(c) 270° → 270 − 180 = 090°.
(d) 005° → 005 + 180 = 185°.
1.2 — Car on bearing 320°, 8 km
E = 8 × sin 320° = 8 × (−0.6428) = −5.14 km, i.e. 5.1 km West.
N = 8 × cos 320° = 8 × 0.7660 = 6.13 km, i.e. 6.1 km North.
Bearing 320° is north of West — confirmed by negative E and positive N.
1.3 — Boat: 7 E then 5 N — bearing from port
The boat is at E = 7, N = 5 (NE quadrant). Angle from N (clockwise) = tan⁻¹(7 / 5) = tan⁻¹(1.4) = 54.46° ≈ 54°.
Three-digit bearing: 054°.
1.4 — Drone: 600 m E, 1100 m S
SE quadrant. Angle from S (clockwise toward East) = tan⁻¹(600 / 1100) = tan⁻¹(0.5455) = 28.61° ≈ 29°.
From N (clockwise): 180° − 29° = 151°.
1.5 — Bushwalker: 6 km @ 130°, then 4 km @ 250°
Leg 1: E₁ = 6 sin 130° = 6 × 0.7660 = 4.60 km; N₁ = 6 cos 130° = 6 × (−0.6428) = −3.86 km.
Leg 2: E₂ = 4 sin 250° = 4 × (−0.9397) = −3.76 km; N₂ = 4 cos 250° = 4 × (−0.3420) = −1.37 km.
Total E = 4.60 − 3.76 = 0.84 km East. Total N = −3.86 − 1.37 = −5.23, i.e. 5.23 km South.
Distance = √(0.84² + 5.23²) = √(0.71 + 27.35) = √28.06 = 5.3 km (to 1 d.p.).
1.6 — Two towns from P
Q (80 km @ 060°): E_Q = 80 sin 60° = 80 × 0.8660 = 69.28 km; N_Q = 80 cos 60° = 80 × 0.5 = 40.00 km.
R (50 km @ 150°): E_R = 50 sin 150° = 50 × 0.5 = 25.00 km; N_R = 50 cos 150° = 50 × (−0.8660) = −43.30 km.
Difference: ΔE = 69.28 − 25.00 = 44.28 km, ΔN = 40.00 − (−43.30) = 83.30 km.
Distance Q→R = √(44.28² + 83.30²) = √(1960.7 + 6938.9) = √8899.6 = 94.3 km.
2 — Find the mistake
(a) The mistake (a convention error, not a calculation error) is on Line 1 (and propagates through Lines 2, 3, 4).
(b) True bearings must be written in three-digit notation. "25°" should be written as 025°. The calculator gives the same numerical answer either way, but a quoted bearing of "25°" or "205°" without the leading zero (where required) violates the navigation convention from Lesson 5 and could be misread in real chartwork.
(c) Corrected: "Bearing = 025°." Likewise Line 4 should read "Back bearing = 025 + 180 = 205°" — written explicitly with leading zero on the original bearing.
The calculations themselves are correct — only the notation needs fixing.
3 — Open-ended (sample solutions)
Journey A — two cardinal legs:
Leg 1: 5 km on bearing 090° (due East). E₁ = 5, N₁ = 0.
Leg 2: 5 km on bearing 000° (due North). E₂ = 0, N₂ = 5.
Totals: E = 5 ✓, N = 5 ✓.
Journey B — slanted then corrected:
Leg 1: √50 km ≈ 7.07 km on bearing 045°. E₁ = 7.07 × sin 45° = 5.00, N₁ = 7.07 × cos 45° = 5.00.
Leg 2: 0 km (the journey is done in one leg).
(If a second leg is required: e.g. go 1 km on 090° (E₁ = 1, N₁ = 0) then walk to make up: need 4 E, 5 N. So leg 2 = √(4² + 5²) = √41 ≈ 6.40 km on bearing tan⁻¹(4/5) from N = 39° → bearing 039°. Check: E₂ = 6.40 × sin 39° = 4.03 ≈ 4 ✓, N₂ = 6.40 × cos 39° = 4.97 ≈ 5 ✓. Totals: E = 1 + 4 = 5 ✓, N = 0 + 5 = 5 ✓.)
Bonus: Direct journey = 045° for √50 = 7.07 km. ✓ (This is just Journey B simplified.)
Marking: 2 marks per journey (1 for valid bearings/distances in three-digit notation, 1 for component sums matching (5, 5)). Bonus +0 — included for understanding only. Any valid pair of journeys is acceptable.