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 minMS-M2 — HIGH3 MC3 SALesson 17 of 22Free
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Choose how you work: type answers on screen, or work in your book.
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.
Write your initial response in your book. You will revisit it at the end of the lesson.
Write your initial thinking in your book
Saved
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°T000°/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); $d$ = distanceE-W component $= d\sin\alpha$
$D = \sqrt{(\Delta x)^2 + (\Delta y)^2}$
Return distance — Pythagoras on total East-West and North-South displacementsReturn bearing: $\tan^{-1}(\Delta x / \Delta y)$, then convert to true bearing based on quadrant
🧠 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 bearingA direction expressed as a three-digit angle measured clockwise from North — always written with three digits (e.g. 045°T, 270°T)
Compass bearingA direction expressed relative to North or South with a deviation toward East or West — format: N[angle]E, S[angle]W, etc.
North lineA vertical line pointing upward drawn at every reference point in a bearing diagram — the baseline from which all bearings are measured clockwise
Back bearingThe 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: A bearing of 045° and a compass direction of NE are always interchangeable
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Right: Compass directions (NE, SW, etc.) are approximate 45° sectors. Bearings give precise clockwise angles from north — 045° is NE only approximately. Always use bearings for precise navigation problems.
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 1Converting 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
2(b) N65°W is in the NW quadrant. True bearing $= 360° - 65° = \mathbf{295°T}$NW quadrant: subtract the compass angle from 360°
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.
Mark the starting point and draw a vertical North line through it
Measure the bearing angle clockwise from North; draw the direction of travel
Mark the distance along that line to the next point
At the new point, draw a new North line (parallel to the first)
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 2Single 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
1130°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)
2East component $= 85\sin50° = 85 \times 0.76604... \approx 65.11$ kmEast = $d \times \sin(\text{angle from S toward E})$; or equivalently $d\sin(130°)$ since $\sin130° = \sin50°$
3South component $= 85\cos50° = 85 \times 0.64278... \approx 54.64$ kmSouth = $d \times \cos(\text{angle from S toward E})$
Two-Leg Journey Problems
Worked Example 3N 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$AC^2 = AB^2 + BC^2 = 12^2 + 9^2 = 144 + 81 = 225$; $AC = 15$ kmRight angle at B (North leg and East leg are perpendicular); Pythagoras
2$\tan\theta = \dfrac{BC}{AB} = \dfrac{9}{12} = 0.75$; $\theta = \tan^{-1}(0.75) \approx 37°$$\theta$ = angle at A between North (AB) and the direction AC; opposite = East leg 9 km, adjacent = North leg 12 km
3C is NE of A; bearing $= 037°T$NE quadrant; $\theta$ from North = 37°; true bearing $= 037°$T (three digits)
Worked Example 4Two 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
1Bearings 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
2$PC^2 = 20^2 + 15^2 = 625$; $PC = 25$ kmPythagoras on the 20-15-25 right triangle
3Find bearing from P to C using components: N component: $20\cos40° - 15\cos50° = 15.321 - 9.642 = 5.679$ km (N) E component: $20\sin40° + 15\sin50° = 12.856 + 11.491 = 24.347$ km (E)Add N-S and E-W displacements of both legs; southward components are negative
4$\theta = \tan^{-1}\!\left(\dfrac{24.347}{5.679}\right) \approx \tan^{-1}(4.286) \approx 77°$; bearing P to C $= 077°$T Back bearing C to P $= 077° + 180° = \mathbf{257°T}$NE quadrant (N and E positive) → bearing = angle from North; add 180° for back bearing
Practice Questions
Draw the diagram first with North lines at every point. Label all distances and angles before calculating.
Find the back bearing: (a) 045°T (b) 200°T (c) 310°T (d) 090°T
Section B — Single Leg Components
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.).
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
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.
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.
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 $= 150\sin40° \approx \mathbf{96.42 \text{ km}}$
(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
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
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
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 42 marks
A plane flies from airport A on a bearing of 112°T for 380 km to airport B.
SA 53 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)
Work in your book
Saved
(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 64 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.
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.