Bearings and Navigation Problems

Draw a North line at every point — not just the first one. Every bearing is measured clockwise from North at that specific location. Get the diagram right and the trig reduces to a standard right-angled triangle problem.

55–60 min MS-M2 — HIGH 3 MC 3 SA Lesson 17 of 22 Free
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Think First

A ship leaves port and sails northeast for 20 km, then turns and sails southeast for 20 km. Where does it end up relative to the port? How far is it from port? What direction would it need to sail to return directly?

Type your initial response below — you will revisit this at the end of the lesson.

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Come back to this at the end of the lesson.

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Bearing Relationships — This Lesson

$\text{True bearing}$: 000°T–360°T
Measured clockwise from North — always 3 digits: 045°T, 270°T 000°/360° = North; 090° = East; 180° = South; 270° = West
$\text{Back bearing} = \theta \pm 180°$
Reverse direction — add 180° if $\theta < 180°$; subtract 180° if $\theta \geq 180°$ Check: result must be in 000°–360°
$\text{N-S component} = d\cos\alpha$
Journey components — $\alpha$ = angle from N-S line (from North or South); $d$ = distance E-W component $= d\sin\alpha$
$D = \sqrt{(\Delta x)^2 + (\Delta y)^2}$
Return distance — Pythagoras on total East-West and North-South displacements Return bearing: $\tan^{-1}(\Delta x / \Delta y)$, then convert to true bearing based on quadrant
TRUE BEARINGS — CLOCKWISE FROM N N E S W 055°T 090° 180° 270° ↻ clockwise BACK BEARING RULE If bearing < 180°: add 180° e.g. 055° + 180° = 235°T If bearing ≥ 180°: subtract 180° e.g. 235° − 180° = 055°T N 055° 235° Opposite directions Always draw a North line at every point

🧠 Know

  • True bearing: 3-digit clockwise angle from North
  • Compass bearing format: N/S [angle] E/W
  • Back bearing = add or subtract 180°
  • Every point in a diagram needs its own North line

💡 Understand

  • Why North lines at every point are parallel — and why this matters
  • Why the bearing angle ≠ the triangle angle in most problems
  • How N-S and E-W components combine to give resultant distance and bearing

✅ Can Do

  • Convert between true and compass bearings
  • Draw accurate bearing diagrams with North lines at every point
  • Find North-South and East-West components of a journey
  • Find the return distance and bearing after a two-leg journey
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Key Terms

True bearing A direction expressed as a three-digit angle measured clockwise from North — always written with three digits (e.g. 045°T, 270°T)
Compass bearing A direction expressed relative to North or South with a deviation toward East or West — format: N[angle]E, S[angle]W, etc.
North line A vertical line pointing upward drawn at every reference point in a bearing diagram — the baseline from which all bearings are measured clockwise
Back bearing The bearing from B to A when the bearing from A to B is known — found by adding or subtracting 180°

Misconceptions to Fix

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Wrong: Converting units only requires multiplying by 10.

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Right: Metric conversions use powers of 10, but area conversions use powers of 100 and volume uses powers of 1000.

True Bearings and Compass Bearings

Key Point

Always check your units before substituting into formulas. Converting to consistent units is a common source of errors in assessment tasks.

Key Terms
FormulaA rule showing the relationship between variables using symbols.
SubstitutionReplacing variables with their known values in an equation.
Unit ConversionChanging a measurement from one unit to another.
CapacityThe amount of liquid a container can hold, measured in litres or millilitres.
PerimeterThe total distance around the outside of a shape.
AreaThe amount of space inside a two-dimensional shape.

Two Ways to Express Direction

Both systems express direction from a reference point. True bearings use a three-digit clockwise angle from North. Compass bearings use a deviation from North or South toward East or West.

Converting Between Bearing Systems
000°–090°T (NE) ⟺ N[bearing]E
090°–180°T (SE) ⟺ S[180°−bearing]E
180°–270°T (SW) ⟺ S[bearing−180°]W
270°–360°T (NW) ⟺ N[360°−bearing]W
Worked Example 1 Converting Bearings

Problem

(a) Convert 155°T to a compass bearing.   (b) Convert N65°W to a true bearing.

Solution

1 (a) 155° is in the SE quadrant (090°–180°). Compass bearing $= S(180°-155°)E = \mathbf{S25°E}$ Identify quadrant; subtract from 180° for SE

Drawing Bearing Diagrams

North Lines at Every Point

The single most important habit in bearing problems is drawing a North line at every point in the journey — not just the starting point. Each bearing is measured from North at that location, not from the previous leg.

  1. Mark the starting point and draw a vertical North line through it
  2. Measure the bearing angle clockwise from North; draw the direction of travel
  3. Mark the distance along that line to the next point
  4. At the new point, draw a new North line (parallel to the first)
  5. Repeat for each leg; label all known quantities
Common error: Using the full true bearing angle (e.g. 130°) directly in sin or cos. The trig angle is the angle within the right-angled triangle — usually measured from North or South, not the full bearing. Read it from your diagram.
Worked Example 2 Single Leg Components

Problem

A ship sails on a bearing of 130°T for 85 km. Find how far east and how far south of its starting position the ship is, correct to 2 decimal places.

Solution

1 130°T is SE quadrant. Angle from South line $= 180° - 130° = 50°$. Angle from East line $= 130° - 90° = 40°$. Draw direction is 40° past East (or 50° from South toward East)

Two-Leg Journey Problems

Worked Example 3 N then E — Distance and Bearing

Problem

A bushwalker starts at A, walks 12 km due North to B, then 9 km due East to C. Find (a) the straight-line distance AC, and (b) the true bearing from A to C, correct to the nearest degree.

Solution

1 $AC^2 = AB^2 + BC^2 = 12^2 + 9^2 = 144 + 81 = 225$; $AC = 15$ km Right angle at B (North leg and East leg are perpendicular); Pythagoras
Worked Example 4 Two Legs — Non-Cardinal Bearings

Problem

A yacht leaves port P and sails 20 km on bearing 040°T to buoy B, then 15 km on bearing 130°T to point C. Find (a) the straight-line distance PC, and (b) the bearing from C back to P, to the nearest degree.

Solution

1 Bearings differ by $130° - 040° = 90°$, so legs PB and BC are perpendicular. When two consecutive bearings differ by exactly 90°, the legs form a right angle — Pythagoras applies directly
Practice

Practice Questions

Draw the diagram first with North lines at every point. Label all distances and angles before calculating.

Section A — Converting Bearings

  1. Convert to compass bearings: (a) 070°T   (b) 160°T   (c) 240°T   (d) 320°T
  2. Convert to true bearings: (a) N35°E   (b) S60°E   (c) S15°W   (d) N80°W
  3. Find the back bearing: (a) 045°T   (b) 200°T   (c) 310°T   (d) 090°T

Section B — Single Leg Components

  1. A plane flies on bearing 060°T for 200 km. Find how far north and how far east of its start it is (to 2 d.p.).
  2. A ship sails S40°W for 150 km. Find how far south and how far west of its start it is (to 2 d.p.).

Section C — Two-Leg Problems

  1. From A, walk 8 km due North to B, then 6 km due East to C. (a) Find the distance AC. (b) Find the bearing from A to C to the nearest degree. (c) Find the bearing from C back to A.
  2. A helicopter leaves base H, flies 50 km on bearing 025°T to point P, then 70 km on bearing 115°T to point Q. (a) Show the two legs are perpendicular. (b) Find HQ. (c) Find the bearing from Q back to H, to the nearest degree.
  3. From town X, a road runs 15 km on bearing 310°T to town Y. From Y, another road runs 20 km on bearing 040°T to town Z. Find the straight-line distance XZ and the bearing from X to Z (to the nearest degree).

Q1

(a) N70°E   (b) S20°E   (c) S60°W   (d) N40°W

Q2

(a) 035°T   (b) 120°T   (c) 195°T   (d) 280°T

Q3

(a) 225°T   (b) 020°T   (c) 130°T   (d) 270°T

Q4

Angle from N $= 60°$; N $= 200\cos60° = \mathbf{100 \text{ km}}$; E $= 200\sin60° \approx \mathbf{173.21 \text{ km}}$

Q5

S40°W $= 220°$T; angle from South $= 40°$; S $= 150\cos40° \approx \mathbf{114.91 \text{ km}}$; W $= 150\sin40° \approx \mathbf{96.42 \text{ km}}$

Q6

(a) $\sqrt{8^2+6^2} = \mathbf{10 \text{ km}}$; (b) $\tan\theta = 6/8$; $\theta \approx 37°$; bearing $= \mathbf{037°T}$; (c) Back bearing $= 217°T$

Q7

(a) $115° - 025° = 90°$ ✓; (b) $HQ = \sqrt{50^2+70^2} = \sqrt{7400} \approx \mathbf{86.02 \text{ km}}$; (c) N components: $50\cos25°=45.32$ (N) + $70\cos65°= -29.58$ (N→actually: 115°T is SE, so N-comp is $-70\cos65°$... let's compute: N total $= 50\cos25° - 70\sin25° = 45.32 - 29.58 = 15.74$... hmm let me recalculate properly. For 025°T: N=50cos25°=45.32, E=50sin25°=21.13. For 115°T: N=−70sin25°=−29.58 (southward), E=70cos25°=63.44. Total N=45.32−29.58=15.74; E=21.13+63.44=84.57. $\theta = \tan^{-1}(84.57/15.74) \approx 80°$; bearing H to Q $= 080°$T; back bearing Q to H $= \mathbf{260°T}$

Q8

XY bearing 310°T (NW): N$=15\cos50°=9.64$ km (N), W$=15\sin50°=11.49$ km (W). YZ bearing 040°T: N$=20\cos40°=15.32$ km (N), E$=20\sin40°=12.86$ km (E). Total from X: N$=9.64+15.32=24.96$, E/W$=12.86-11.49=1.37$ km E. $XZ=\sqrt{24.96^2+1.37^2}\approx \mathbf{25.0 \text{ km}}$; bearing $= \tan^{-1}(1.37/24.96) \approx 3°$ from North $\to \mathbf{003°T}$

Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Multiple Choice

1 The true bearing 245°T expressed as a compass bearing is:

A   N65°W
B   S65°W
C   S65°E
D   N65°E

? Regarding this topic, 1 The true bearing 245°T expressed as a compass bearing is:

A   &nbsp; N65°W
B   &nbsp; S65°W
C   &nbsp; S65°E
D   &nbsp; N65°E
B - Correct!
B — 245° is in the SW quadrant; compass bearing $= S(245°-180°)W = S65°W$.

2 A ship sails from port on a bearing of 035°T for 40 km. How far north of port is the ship, correct to 2 decimal places?

A   22.94 km
B   32.77 km
C   35.02 km
D   49.02 km

? Regarding this topic, 2 A ship sails from port on a bearing of 035°T for 40 km. How far north of port is the ship, correct to 2 decimal places?

A   &nbsp; 22.94 km
B   &nbsp; 32.77 km
C   &nbsp; 35.02 km
D   &nbsp; 49.02 km
B - Correct!
B — North component $= 40\cos35° = 40 \times 0.81915... \approx 32.77$ km.

3 A bushwalker walks 10 km due East then 10 km due South. The bearing from the walker's final position back to the start is:

A   045°T
B   135°T
C   270°T
D   315°T

? Regarding this topic, 3 A bushwalker walks 10 km due East then 10 km due South. The bearing from the walker's final position back to the start is:

A   &nbsp; 045°T
B   &nbsp; 135°T
C   &nbsp; 270°T
D   &nbsp; 315°T
D - Correct!
D — Walker is 10 km E and 10 km S of start; direction back is NW. Isosceles triangle → 45° from North toward West; bearing $= 360° - 45° = 315°T$.

Short Answer

01

SA 4 2 marks A plane flies from airport A on a bearing of 112°T for 380 km to airport B.

(a) Convert 112°T to a compass bearing.  (1 mark)

(b) State the bearing from B back to A.  (1 mark)

Work in your book
Saved

(a)

SE quadrant; $S(180°-112°)E = \mathbf{S68°E}$

(b)

$112° < 180°$, so add 180°: $112° + 180° = \mathbf{292°T}$

02

SA 5 3 marks A boat leaves a marina and travels 18 km on a bearing of 055°T to reach a buoy. It then travels due South until it is directly east of the marina.

(a) Draw a fully labelled diagram of this situation.  (1 mark)

(b) How far south does the boat travel from the buoy to its final position? Give your answer to 2 decimal places.  (1 mark)

(c) How far east of the marina is the boat's final position? Give your answer to 2 decimal places.  (1 mark)

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(a)

Marina M at bottom-left; North line at M; line 18 km at 55° clockwise from North to buoy B; vertical line south from B to final point C; right angle at C; horizontal dotted line from M due East to C

(b)

North component of MB $= 18\cos55° \approx 10.32$ km; the boat travels south this same distance $= \mathbf{10.32 \text{ km}}$

(c)

East component $= 18\sin55° \approx \mathbf{14.74 \text{ km}}$

03

SA 6 4 marks Two bushwalkers start from base camp B. Walker 1 travels 24 km on a bearing of 050°T to reach point P. Walker 2 travels 24 km on a bearing of 140°T to reach point Q.

(a) Show that angle $PBQ = 90°$.  (1 mark)

(b) Find the distance $PQ$.  (1 mark)

(c) Find the bearing from $P$ to $Q$, correct to the nearest degree.  (2 marks)

Work in your book
Saved

(a)

Angle $PBQ = 140° - 050° = 90°$ — the two bearings from B differ by 90°, so BP and BQ are perpendicular.

(b)

$PQ^2 = 24^2 + 24^2 = 1152$; $PQ = \sqrt{1152} = 24\sqrt{2} \approx \mathbf{33.94 \text{ km}}$

(c)

Using coordinates (B at origin, N = +y, E = +x):
P $= (24\sin50°,\ 24\cos50°) = (18.385,\ 15.429)$
Q $= (24\sin140°,\ 24\cos140°) = (15.427,\ -18.385)$
Vector P→Q: $\Delta x = -2.958$ (W), $\Delta y = -33.814$ (S)
Direction is almost due South with tiny westward component; $\phi = \tan^{-1}(2.958/33.814) \approx 5°$ west of South
Bearing $= 180° + 5° = \mathbf{185°T}$

Interactive

Bearing Dial — True Bearing & Back Bearing

055°T
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Boss Battle

Boss Battle — Bearings Final!

The ultimate Module 2 challenge — use all your measurement knowledge to defeat the boss. Pool: lessons 1–17.