Mathematics • Year 9 • Unit 3 • Lesson 18

Bearing Problems — Mixed Challenge

Combine every tool from Lessons 16-18: compass and true bearings, components (cos/sin of true bearing), Pythagoras for resultant distance, inverse trig for resultant bearing, quadrant adjustments, and reverse bearings.

Master · Mixed Challenge

1. Mixed problems — pick the right tool

Each question combines two or more skills. Sketch the situation BEFORE you write any numbers. Show your working. 3 marks each

1.1 A boat sails 15 km on bearing 060°, then 10 km on bearing 320°. Find the straight-line distance from start (2 d.p.) and the resultant true bearing.

1.2 A plane flies 200 km on bearing 090°, then 200 km on bearing 180°. (a) Find the straight-line distance from start. (b) Find the resultant true bearing. (c) Express the resultant bearing in compass form.

1.3 A walker travels on bearing N50°E (compass form) for 8 km, then on true bearing 230° for 8 km. (a) Convert N50°E to a true bearing. (b) What is the relationship between the two true bearings? (c) Find the walker's final position.

1.4 A drone takes off and flies 12 km on bearing 075°, then 9 km on bearing 165°. Find (a) net N and E components (2 d.p.), (b) the straight-line distance from base, (c) the true bearing from base to the drone (2 d.p.).

1.5 A yacht races on a course of three legs: 14 km on 030°, then 10 km on 120°, then 14 km on 210°. Find the net displacement from start. Then state the true bearing from the start to the end-point (2 d.p.).

1.6 A surveyor walks 100 m on bearing 050°, then 100 m on bearing 230°. Predict the result before computing anything (one sentence), then verify with components.

Stuck on 1.6? 050 and 230 are reverse bearings. Same distance + opposite directions = back to start.

2. Find the mistake

A student has tried a two-leg problem: 30 km on bearing 040°, then 25 km on bearing 160°. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks

Student's working — 30 km on 040° then 25 km on 160°:

Line 1: Leg 1: N₁ = 30 cos 40° ≈ 22.98; E₁ = 30 sin 40° ≈ 19.28.

Line 2: Leg 2: N₂ = 25 cos 160° ≈ 23.49; E₂ = 25 sin 160° ≈ 8.55.

Line 3: Net N = 22.98 + 23.49 ≈ 46.47.

Line 4: Net E = 19.28 + 8.55 ≈ 27.83.

Line 5: Distance = √(46.47² + 27.83²) ≈ 54.16 km.

(a) Which line contains the mistake?

(b) Explain in one or two sentences why that line is wrong.

(c) Write out the corrected working in full, including the corrected final distance (2 d.p.).

Stuck? 160° is in the SE quadrant (between 090 and 180). What should the SIGN of cos 160° be? (Hint: cos of any angle between 90 and 270 is negative.)

3. Open-ended challenge — return to start

This question has more than one valid answer — many designs work. 4 marks

3.1 Design a three-leg journey that returns EXACTLY to the start. Your design must satisfy:

(i) Each leg has a distinct true bearing (no two legs the same direction).
(ii) No two of your bearings are simply reverse bearings of each other (so you cannot just "do a leg, undo it, go home").
(iii) Each leg is at least 5 km long.
(iv) The total net displacement (after all three legs) is exactly zero — within 0.05 km in both N and E.

For your design:
(a) State your three bearings and distances.
(b) Compute the N and E components of each leg (2 d.p.).
(c) Show that the sums of N and E components are both (essentially) zero.

Bonus: Notice the geometric shape your three-leg path traces — what is it?

Stuck? An equilateral triangle works: three legs of equal length, on bearings 120° apart, e.g. 000°, 120°, 240°.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — 15 km on 060° then 10 km on 320°

Leg 1: N₁ = 15 cos 60 = 7.50; E₁ = 15 sin 60 ≈ 12.99.
Leg 2: N₂ = 10 cos 320 ≈ 7.66; E₂ = 10 sin 320 ≈ −6.43.
Net N ≈ 15.16; Net E ≈ 6.56.
Distance = √(15.16² + 6.56²) ≈ 16.52 km.
Bearing: NE quadrant, tan−¹(6.56/15.16) ≈ 23.4° from N. True bearing ≈ 023°.

1.2 — 200 km east then 200 km south

Leg 1 (200, 090°): N = 0; E = 200.
Leg 2 (200, 180°): N = −200; E = 0.
Net N = −200; Net E = 200.
(a) Distance = √(200² + 200²) = 200√2 ≈ 282.84 km.
(b) End-point is in the SE quadrant; angle from S = tan−¹(200/200) = 45°; true bearing = 180 − 45 = 135°.
(c) Compass: S45°E.

1.3 — N50°E for 8 km, then 230° true for 8 km

(a) N50°E in NE quadrant: true = 50° = 050°.
(b) 050 and 230 differ by 180 → reverse bearings.
(c) Equal-distance reverse-bearing legs cancel completely → the walker is back at the start (0 km, 0 km).

1.4 — 12 km on 075° then 9 km on 165°

Leg 1: N₁ = 12 cos 75 ≈ 3.11; E₁ = 12 sin 75 ≈ 11.59.
Leg 2: N₂ = 9 cos 165 ≈ −8.69; E₂ = 9 sin 165 ≈ 2.33.
(a) Net N = 3.11 − 8.69 ≈ −5.58 km (5.58 km south); Net E = 11.59 + 2.33 ≈ 13.92 km east.
(b) Distance = √(5.58² + 13.92²) ≈ 15.00 km.
(c) End-point in SE quadrant. Angle from S = tan−¹(13.92/5.58) ≈ 68.2°. True bearing = 180 − 68.2 ≈ 111.8°.

1.5 — Three-leg yacht race (14 on 030, 10 on 120, 14 on 210)

Leg 1: N = 14 cos 30 ≈ 12.12; E = 14 sin 30 = 7.00.
Leg 2: N = 10 cos 120 = −5.00; E = 10 sin 120 ≈ 8.66.
Leg 3: N = 14 cos 210 ≈ −12.12; E = 14 sin 210 = −7.00.
Net N = 12.12 − 5.00 − 12.12 = −5.00 km (5 km south); Net E = 7.00 + 8.66 − 7.00 = 8.66 km east.
End-point in SE quadrant. Angle from S = tan−¹(8.66/5.00) ≈ 60.0°. True bearing = 180 − 60.0 = 120°.
Notice: legs 1 and 3 are reverse bearings of equal length, so they cancel completely. Only leg 2 (10 km on 120°) is left — and its bearing IS 120°. ✓

1.6 — 100 m on 050° then 100 m on 230°

Prediction: 050 and 230 are reverse bearings (180° apart). Equal distances on reverse bearings cancel — the surveyor returns to the start.
Verification: N₁ = 100 cos 50 ≈ 64.28; N₂ = 100 cos 230 ≈ −64.28. Sum = 0. ✓ E₁ = 100 sin 50 ≈ 76.60; E₂ = 100 sin 230 ≈ −76.60. Sum = 0. ✓ Net displacement = 0 m.

2 — Find the mistake (30 km on 040 then 25 km on 160)

(a) The mistake is on Line 2.
(b) 160° is in the SE quadrant (between 090 and 180), so cos 160° is negative, not positive. The student forgot the sign — N₂ should be a NEGATIVE number (south-ward), not +23.49.
(c) Corrected working:
Line 2: N₂ = 25 cos 160° ≈ −23.49; E₂ = 25 sin 160° ≈ 8.55.
Line 3: Net N = 22.98 + (−23.49) ≈ −0.51.
Line 4: Net E = 19.28 + 8.55 ≈ 27.83.
Line 5: Distance = √(0.51² + 27.83²) ≈ √(0.26 + 774.51) ≈ 27.83 km.
The correct end-point is barely south of the start and almost 28 km east — quite different from the student's 54 km answer.

3 — Three-leg return to start (sample solution)

Design (equilateral triangle): three 10 km legs on bearings 000°, 120°, 240°.

Components:
Leg 1 (10 km, 000°): N = 10.00, E = 0.00.
Leg 2 (10 km, 120°): N = 10 cos 120 = −5.00; E = 10 sin 120 ≈ 8.66.
Leg 3 (10 km, 240°): N = 10 cos 240 = −5.00; E = 10 sin 240 ≈ −8.66.

Sums: Net N = 10 − 5 − 5 = 0. ✓ Net E = 0 + 8.66 − 8.66 = 0. ✓

Each pair of bearings differs by 120°, not 180°, so condition (ii) is met. All legs are 10 km ≥ 5 km. ✓

Bonus: the path traces an equilateral triangle. Three equal sides on bearings 120° apart always closes back to the start.

Marking: 1 mark for a valid set of three bearings/distances meeting all four conditions; 1 for correctly computed components; 1 for showing both sums = 0 (within tolerance); 1 for identifying the equilateral triangle (or for a different valid closed shape such as scalene triangle 040, 160, 280 with appropriate distances).