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Module 2 · L17 of 22 ~55 min ⚡ +95 XP available

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.

Today's hook — A ship sails northeast for 20 km, then southeast for 20 km. Where does it end up relative to the port? How far is it from port, and what bearing does it need to return?

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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

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Recall — your gut answer first

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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?

Without calculating — make a prediction and explain your reasoning.

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The bearing system — two ways to express direction

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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.

True bearing rules: always three digits, measured clockwise from North. 000°T = North; 090°T = East; 180°T = South; 270°T = West. Write with the degree symbol and T suffix.

Back bearing rule: add 180° if bearing < 180°; subtract 180° if bearing ≥ 180°. Result must be in 000°–360°.

$\text{Back bearing} = \theta \pm 180°$ (add or subtract so result is in 0°–360°)

True bearing

Three-digit angle, clockwise from North. 000°–360°T. Write with a T: 045°T, 270°T.

Compass bearing

Deviation from N or S toward E or W. Format: N[angle]E, S[angle]W, etc.

N-S / E-W components

N-S component $= d\cos\alpha$; E-W component $= d\sin\alpha$ where $\alpha$ is the angle from the N-S line.

What you'll master

Know

Key facts

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

Concepts

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

Skills

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

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°.

Converting between bearing systems

core concept

To convert a true bearing to a compass bearing, identify which quadrant it falls in and apply the appropriate rule:

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

To convert a compass bearing back to a true bearing, reverse the process. For N-S components of a journey leg: N-S $= d\cos\alpha$ and E-W $= d\sin\alpha$ where $\alpha$ is the angle from the N or S direction.

Real-world anchor — Marine navigation. Ships and aircraft use true bearings for all navigation. When a search-and-rescue team reports a bearing of 235°T from a rescue base, they mean: face North, then rotate 235° clockwise. The back bearing (055°T) tells the ship's crew the direction back to base.

What to write in your book

True bearing: three-digit clockwise from North. 045°T (NE), 135°T (SE), 225°T (SW), 315°T (NW).
Back bearing: add 180° if < 180°; subtract 180° if ≥ 180°. Always stays in 000°–360°.
To convert 155°T: SE quadrant → S(180°−155°)E = S25°E.
To convert N65°W: NW quadrant → 360°−65° = 295°T.

Quick check: Which true bearing is equivalent to the compass bearing S40°E?

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · CONVERTING BEARINGS

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

(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.

PROBLEM 2 · SINGLE LEG COMPONENTS

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.

130°T is SE quadrant. Angle from South line $= 180° - 130° = 50°$. Angle from East line $= 130° - 90° = 40°$.

Draw the direction: 40° past East (or 50° from South toward East).

PROBLEM 3 · TWO-LEG JOURNEY

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.

Bearings differ by $130° - 040° = 90°$, so legs PB and BC are perpendicular. $PC^2 = 20^2 + 15^2 = 625$; $PC = \mathbf{25 \text{ km}}$

When two consecutive bearings differ by 90°, the legs form a right angle — Pythagoras applies directly.

What to write in your book

Draw North lines at every point — parallel, all pointing the same direction.
The trig angle in the triangle is not the true bearing — read the angle within the triangle from your diagram.
For two-leg problems: find N and E components of each leg, add (or subtract) them, then use Pythagoras and $\tan^{-1}$ for the resultant.
Always check your final bearing is in 000°–360°.

True or false: When drawing a bearing diagram with multiple legs, you only need to draw a North line at the starting point.

Common errors · the 3 traps that cost marks

Trap 01

Using the full bearing angle directly in sin or cos

The bearing (e.g. 130°T) is not the angle in the triangle. The trig angle is measured from the N or S direction — read it from your diagram. For 130°T: angle from South = 50°, so use sin 50° and cos 50°.

Trap 02

Forgetting North lines at intermediate points

Every turning point needs its own North line parallel to all others. Omitting it makes it impossible to measure the next bearing correctly, and angles become confused with the previous leg's direction.

Trap 03

Back bearing direction error

The back bearing is always exactly 180° different. A bearing of 045°T has a back bearing of 225°T (add 180°). A bearing of 300°T has a back bearing of 120°T (subtract 180°). Result must stay in 000°–360°.

What to write in your book

Check: is the answer direction consistent with the quadrant in your diagram?
For a back bearing: if forward bearing < 180°, add 180°. If ≥ 180°, subtract 180°.
The bearing 045°T and its back bearing 225°T always sum to 360° — use this as a sanity check.
Pythagoras only applies directly when the two legs are perpendicular (bearings differ by 90°). Otherwise use components.

Fill the gap: The back bearing of 112°T is found by adding °, giving °T.

Quick-fire practice · 5 calculations

Convert to compass bearings: (a) 070°T (b) 160°T (c) 240°T (d) 320°T

Find the back bearing for: (a) 045°T (b) 200°T (c) 310°T (d) 090°T

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.).

From A, walk 8 km due North to B, then 6 km due East to C. Find (a) distance AC and (b) bearing from A to C.

A helicopter flies 50 km on bearing 025°T, then 70 km on bearing 115°T. Show the legs are perpendicular, then find the distance from start to finish.

Match each true bearing to the correct compass bearing:

035°TN35°E

120°TS60°E

245°TS65°W

310°TN50°W

Revisit your thinking

Earlier you predicted where the ship ended up after sailing 20 km NE then 20 km SE. Let's check:

NE = bearing 045°T: N component $= 20\cos45° = 14.14$ km, E component $= 20\sin45° = 14.14$ km.

SE = bearing 135°T: S component $= 20\cos45° = 14.14$ km (cancels N), E component $= 20\sin45° = 14.14$ km.

Total displacement: 0 km North, 28.28 km East — the ship is due East of port! Distance $= 28.28$ km. Return bearing $= 270°T$ (due West).

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. 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. (b) State the bearing from B back to A. (2 marks)

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ApplyBand 43 marks

Q2. 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. (b) How far south does the boat travel from the buoy to its final position (to 2 d.p.)? (c) How far east of the marina is the boat's final position (to 2 d.p.)? (3 marks)

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AnalyseBand 54 marks

Q3. Two bushwalkers start from base camp B. Walker 1 travels 24 km on bearing 050°T to reach P. Walker 2 travels 24 km on bearing 140°T to reach Q. (a) Show that angle PBQ = 90°. (b) Find PQ. (c) Find the bearing from P to Q, correct to the nearest degree. (4 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: (a) N70°E (b) S20°E (c) S60°W (d) N40°W · Drill 2: (a) 225°T (b) 020°T (c) 130°T (d) 270°T

Drill 3: Angle from N = 60°; N $= 200\cos60° = 100$ km; E $= 200\sin60° \approx 173.21$ km

Drill 4: (a) $\sqrt{8^2+6^2} = 10$ km (b) $\tan\theta = 6/8 \Rightarrow \theta \approx 37°$; bearing = 037°T

Drill 5: $115°-025°=90°$ ✓; $HQ = \sqrt{50^2+70^2} = \sqrt{7400} \approx 86.02$ km

Q1 (2 marks): (a) SE quadrant; $S(180°-112°)E = S68°E$ [1]. (b) $112° < 180°$ so add 180°: $112°+180° = 292°T$ [1].

Q2 (3 marks): (a) Marina M at bottom-left; North line at M; line 18 km at 55° from N to buoy B; vertical south line from B to C; right angle at C; dotted horizontal M to C [1]. (b) North component $= 18\cos55° \approx 10.32$ km southward [1]. (c) East $= 18\sin55° \approx 14.74$ km [1].

Q3 (4 marks): (a) Angle PBQ $= 140°-050° = 90°$ [1]. (b) $PQ = \sqrt{24^2+24^2} = 24\sqrt{2} \approx 33.94$ km [1]. (c) Coordinates (B origin): P $= (24\sin50°,24\cos50°) = (18.39,15.43)$; Q $= (24\sin140°,24\cos140°) = (15.43,-18.39)$; vector P→Q: $\Delta x=-2.96$ (W), $\Delta y=-33.82$ (S); $\phi=\tan^{-1}(2.96/33.82)\approx5°$ W of S; bearing $= 180°+5° = 185°T$ [2].

Boss battle · Bearings Final

earn bronze · silver · gold

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

Science Jump · platform challenge

Climb platforms by answering bearing questions. Pool: lesson 17.

Mark lesson as complete

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Module overview · Year 11 · Maths Standard