Lesson 18 ~25 min Unit 3 · Trigonometry +85 XP

Solving Bearing Problems

Translate bearings into right-triangle diagrams. Apply trig and Pythagoras to find distances, positions, and the resultant of multi-leg journeys.

Today's hook: A yacht sails 40 km on bearing 060°, then 25 km on bearing 150°. How far is it now from where it started?

0/5QUESTS

Printable Worksheets

Print or save as PDF — or build a custom worksheet from any module's questions.

Build Apply Master Build custom

Think First

warm-up

You travel on TWO different bearings, one after the other. The total straight-line distance from start to end is NOT just the sum of the legs (unless they're collinear). What technique do you need? (Hint: components.)

Record your answer in your workbook.

The Big Idea

+5 XP

To solve a bearing problem, break each leg into its N-S and E-W components. Add up all N-S movements; add up all E-W movements. Then use Pythagoras to find the straight-line distance, and inverse trig to find the resultant bearing.

For a leg of length $d$ on true bearing $\theta$: N-component = $d\cos\theta$, E-component = $d\sin\theta$. Negative N-components mean southward; negative E-components mean westward. Sum components separately; then resultant distance $= \sqrt{(\text{net N})^2 + (\text{net E})^2}$.

Each leg → N + E components; sum → resultant

Components

Always split each leg into N-S and E-W parts before adding.

Signs matter

Use positive N + E by convention; negative means S or W respectively.

Pythagoras finishes

Once you have net N and net E, Pythagoras gives the straight-line distance.

What You'll Master

objectives

Know

Each leg splits into N-component ($d\cos\theta$) and E-component ($d\sin\theta$)
Negative components handle S and W directions
Net N and net E with Pythagoras give the resultant distance

Understand

Why component method works for any combination of bearings
Why the resultant bearing comes from $\tan^{-1}$ on net components
When two legs cancel or augment each other

Can Do

Convert any leg to N-S and E-W components
Sum components for multi-leg journeys
Find the resultant distance and bearing

Words You Need

vocabulary

ComponentThe projection of a leg onto the N-S or E-W axis. Useful for adding multiple legs.

Net displacementTotal movement in one direction (N or E) after all legs are added with appropriate signs.

ResultantThe single straight-line displacement equivalent to all the legs together.

Multi-leg journeyA path made of two or more straight segments on different bearings.

Final positionThe end point relative to the start, often expressed as N and E displacements or as a bearing + distance.

Resultant bearingThe compass/true bearing of the final position from the start, calculated using $\tan^{-1}$ on the net components.

Spot the Trap

heads-up

Wrong: Adding distances directly: 40 + 25 = 65 km is NOT the straight-line distance unless the legs are collinear and same direction.

Right: Use components: sum N's, sum E's, then Pythagoras.

Wrong: Forgetting to track sign — southward and westward give NEGATIVE components.

Right: Convert each bearing to N-component (cos with appropriate sign) and E-component (sin with sign).

Component Calculator

+5 XP

Given any true bearing $\theta$ and distance $d$:

Bearing $\theta$	N-comp	E-comp
$0 \le \theta < 90$	$+d\cos\theta$	$+d\sin\theta$
$90 \le \theta < 180$	$-d\cos(180-\theta)$	$+d\sin(180-\theta)$
$180 \le \theta < 270$	$-d\cos(\theta - 180)$	$-d\sin(\theta - 180)$
$270 \le \theta < 360$	$+d\cos(360-\theta)$	$-d\sin(360-\theta)$

Or simply: $\text{N-comp} = d\cos\theta$, $\text{E-comp} = d\sin\theta$ if you compute in standard math mode — the signs come out automatically. (Bearing 060°: cos 60 = 0.5 positive, sin 60 positive. Bearing 150°: cos 150 negative (south), sin 150 positive (east).)

$\text{N} = d\cos\theta$ $\text{E} = d\sin\theta$ (signs from $\cos$/$\sin$)

Calculator does signs

If you use $\cos(150°)$ directly, you get $-0.866$ — correctly negative.

Net = sum

Add components from each leg with their signs.

Resultant from net

$\sqrt{N^2 + E^2}$ gives straight-line distance.

Find the Resultant Bearing

+5 XP

After summing components, find the resultant bearing using $\tan^{-1}(E/N)$ — adjusted for the quadrant.

N net	E net	Resultant bearing
+	+	$\tan^{-1}(E/N)$ (in NE quadrant)
$-$	+	$180 - \tan^{-1}(E/\|N\|)$ (SE)
$-$	$-$	$180 + \tan^{-1}(\|E\|/\|N\|)$ (SW)
+	$-$	$360 - \tan^{-1}(\|E\|/N)$ (NW)

Bearing depends on quadrant of net N/E displacement

Sketch the net

A small sketch of the resultant point relative to start clarifies the quadrant.

Pad zeros

Final bearing must be three digits.

Sanity check

Compare with average direction of the trip.

Watch Me Solve It · 3 examples

Watch Me Solve It · Two-leg journey

+15 XP per step

PROBLEM

A yacht sails 40 km on bearing 060°, then 25 km on bearing 150°. Find the straight-line distance and bearing from start to end (2 d.p.).

1

Leg 1 components

N$_1 = 40\cos 60° = 20$; E$_1 = 40\sin 60° \approx 34.64$
2

Leg 2 components

N$_2 = 25\cos 150° \approx -21.65$; E$_2 = 25\sin 150° = 12.5$
3

Sum + resultant

Net N = $20 - 21.65 = -1.65$ (S); Net E = $34.64 + 12.5 = 47.14$. Distance $= \sqrt{1.65^2 + 47.14^2} \approx 47.17$ km. Bearing: SE quadrant, $\tan^{-1}(47.14/1.65) \approx 88°$, so true bearing $\approx 180 - 88 = 092°$.

Answer$\approx 47.17$ km on bearing $\approx 092°$

Watch Me Solve It · Single leg position

+15 XP per step

PROBLEM

A walker travels 10 km on true bearing 220°. Find their final position relative to start (N/S and E/W).

1

Components

N $= 10\cos 220° \approx -7.66$ (south)
2

E

E $= 10\sin 220° \approx -6.43$ (west)
3

State

7.66 km south and 6.43 km west of start.

Answer7.66 km S, 6.43 km W

Watch Me Solve It · Three-leg journey

+15 XP per step

PROBLEM

A delivery van travels 5 km on 080°, then 8 km on 200°, then 4 km on 320°. Find the straight-line distance from start (2 d.p.).

1

Leg 1

N$_1 = 5\cos 80° \approx 0.87$; E$_1 = 5\sin 80° \approx 4.92$
2

Leg 2

N$_2 = 8\cos 200° \approx -7.52$; E$_2 = 8\sin 200° \approx -2.74$
3

Leg 3 + total

N$_3 = 4\cos 320° \approx 3.06$; E$_3 = 4\sin 320° \approx -2.57$. Net N $\approx -3.59$; Net E $\approx -0.39$. Distance $\approx \sqrt{3.59^2 + 0.39^2} \approx 3.61$ km.

Answer$\approx 3.61$ km from start

Common Pitfalls

heads-up

Adding lengths directly

40 + 25 km isn't the straight-line distance unless the legs are collinear.

Fix: Decompose into components first; then Pythagoras the net N and net E.

Sign errors

Forgetting that south or west gives negative components.

Fix: Use $\cos$/$\sin$ of the true bearing directly — the calculator handles signs.

Wrong quadrant for resultant bearing

Reporting $\tan^{-1}(E/N)$ without considering signs.

Fix: Sketch the net position; identify the quadrant; adjust the bearing accordingly.

Copy Into Your Books

Components

N $= d\cos\theta$
E $= d\sin\theta$
Use $\cos$/$\sin$ of true bearing

Multi-leg

Compute each leg
Sum N's; sum E's
Pythagoras net

Resultant

Distance $= \sqrt{N^2 + E^2}$
Bearing from $\tan^{-1}(E/N)$
Adjust by quadrant

Three-digit

Pad zeros: 045°
Clockwise from N
000-360

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Bearing Solving

4 problems

Four quick drills to lock in today's skill. Try each, then reveal the answer.

1 5 km on bearing 030°. N-component (2 d.p.)?

$5\cos 30° \approx 4.33$.$\approx 4.33$ km N
2 5 km on bearing 030°. E-component (2 d.p.)?

$5\sin 30° = 2.5$.2.5 km E
3 10 km on 270°. Direction?

$\cos 270° = 0$, $\sin 270° = -1$.10 km west
4 6 km on 045° then 4 km on 135°. Net N (2 d.p.)?

$6\cos 45° + 4\cos 135° \approx 4.24 - 2.83 \approx 1.41$.$\approx 1.41$ km N

Complete in your workbook.

Quick Check · 5 questions

To find resultant distance of a multi-leg journey:

+10 XP

6 km on bearing 090°. North-component?

+10 XP

10 km on bearing 150°. The direction is:

+10 XP

A ship sails 10 km on 050° then 6 km on 230°. The net displacement is:

+10 XP

4 km on bearing 030° then 4 km on bearing 120°. The resultant distance is (2 d.p.):

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. A car travels 12 km on true bearing 040°, then 9 km on true bearing 130°. Find (a) the net N-S and E-W displacement and (b) the straight-line distance from start (2 d.p.).

Answer in your workbook.

ApplyEasy2 MARKS

Q7. A boat sails 25 km on bearing 080°. Find its (a) east and (b) north displacement from start (2 d.p.).

Answer in your workbook.

ReasonHard4 MARKS

Q8. A plane flies on bearing 060° for 200 km, then turns to bearing 130° and flies 150 km. (a) Find the plane's position relative to start (N and E km, 2 d.p.). (b) Find the plane's straight-line distance from start. (c) Find the true bearing on which the plane would fly directly home.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — Component method.

2. A — $\cos 90° = 0$.

3. C — SE quadrant.

4. D — 4 km on 050°.

5. B — Perpendicular → $\sqrt{32} \approx 5.66$.

Show Your Working Model Answers

Q6 (3 marks): Leg 1: N $\approx 9.19$, E $\approx 7.71$. Leg 2: N $\approx -5.79$, E $\approx 6.89$ [1]. Net N $\approx 3.40$, net E $\approx 14.60$ [1]. Distance $= \sqrt{3.40^2 + 14.60^2} \approx 14.99$ km [1].

Q7 (2 marks): East = $25\sin 80° \approx 24.62$ km [1]. North = $25\cos 80° \approx 4.34$ km [1].

Q8 (4 marks): (a) Leg 1: N = $200\cos 60° = 100$, E = $200\sin 60° \approx 173.21$ [1]. Leg 2: N = $150\cos 130° \approx -96.42$, E = $150\sin 130° \approx 114.91$ [1]. Net: N $\approx 3.58$, E $\approx 288.12$ km [1]. (b) Distance $= \sqrt{3.58^2 + 288.12^2} \approx 288.14$ km. (c) Bearing from start to plane: $\tan^{-1}(288.12/3.58) \approx 89.3°$ — nearly due east, so $\approx 089°$. Reverse for the trip home: $089 + 180 = 269°$ [1].

Stretch Challenge · +25 XP, +10 coins

Triangular tour

A walker travels 4 km on bearing 030°, then 5 km on bearing 120°, then 4 km on bearing 240°. How far from the start are they at the end (2 d.p.)?

Reveal solution

Leg 1: N = $4\cos 30 \approx 3.46$, E = $4\sin 30 = 2$. Leg 2: N = $5\cos 120 = -2.5$, E = $5\sin 120 \approx 4.33$. Leg 3: N = $4\cos 240 = -2$, E = $4\sin 240 \approx -3.46$. Net N $\approx 3.46 - 2.5 - 2 = -1.04$; Net E $\approx 2 + 4.33 - 3.46 = 2.87$. Distance $\approx \sqrt{1.08 + 8.24} \approx 3.05$ km.

Quick Review

Components

N = $d\cos\theta$, E = $d\sin\theta$

Sum N, sum E

Per leg, with signs

Pythagoras

$d = \sqrt{N^2 + E^2}$

Resultant bearing

$\tan^{-1}(E/N)$ + quadrant

Multi-leg

Components handle any combo

Sketch

Visualise the final position

Your Badges

0 of 6

First Steps

3-Day Streak

3 in a Row

Lesson Ace

Stretch Seeker

Daily Warrior

Mark lesson as complete

Tick when you've finished Learn, Practice and the Stretch. Earns +85 XP and +25 coins.

Unit overview · Maths subject page