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hscscience Maths Std · Y11
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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?
0/5QUESTS
1

Orient to bearings

Meet true and compass bearings, set the goal and settle the key terms.

Worksheets

Practise this lesson

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

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall, your gut answer first
+5 XP warm-up

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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02
The bearing system, two ways to express direction
+5 XP to read

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°.

True bearings are measured clockwise from north; the back bearing differs by 180°.
$\text{Back bearing} = \theta \pm 180°$ (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.
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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
04
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°.
2

Convert between the two systems

Move between a true bearing and a compass bearing without losing the direction.

05
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.

True bearings are measured clockwise from north (000° to 360°). Compass bearings use N/S as base directions with an angle (e.g., N40°E). To convert: a true bearing of 040° is the same direction as N40°E.

Pause, copy the true bearing convention (measured clockwise from north, written as a three-digit number e.g., 040°) and the compass bearing format (N/S as base, then angle, then E/W e.g., N40°E), plus one worked conversion showing the same direction in both systems into your book.

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

3

Work three bearing examples

Follow a conversion, a single-leg journey and a two-leg journey.

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.

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(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.
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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.

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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).
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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.

1
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.
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True or false: When drawing a bearing diagram with multiple legs, you only need to draw a North line at the starting point.

4

Avoid the three traps

Spot the full-bearing slip, the missing diagram and the back-bearing error.

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°.

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

5

Drill bearings

Run the quick-fire conversions and journeys until the setup is automatic.

Quick-fire practice · 5 calculations

1

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

2

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

3

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.).

4

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.

5

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°T
  • 120°T
  • 245°T
  • 310°T
  • N50°W
  • S65°W
  • S60°E
  • N35°E
6

Revisit your thinking

Return to your opening answer and name what has changed.

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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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7

Practise bearings

Answer the question bank, then write full short-answer responses.

01
Multiple choice
+5 XP per correct · +25 XP all-correct

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

02
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].