Bearings and directions | Year 10

01 / LEARN

Understand the idea.

Start at north

A bearing turns clockwise from north at the starting point. East is $090^{\circ}$ , south $180^{\circ}$ and west $270^{\circ}$ . Draw a north arrow before marking the journey's direction.

Reversing a journey turns its direction through $180^{\circ}$ . Add or subtract $180^{\circ}$ so the result lies from $000^{\circ}$ up to, but excluding, $360^{\circ}$ . Keep the leading zero when needed.

Connect direction with components

North and east movements form perpendicular components of a journey. If a bearing $θ$ lies in the north-east quadrant, the north component is $d cos θ$ and the east component is $d sin θ$ .

For other quadrants, first identify the compass directions and a useful acute reference angle. Component magnitudes are positive lengths; signed north/east coordinates must reflect whether movement is south or west.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Combine movements

A hiker moves 6 km east then 8 km north. Find the direct distance and bearing from the start.

Worked solution

1
Find the direct distance.
$d = \sqrt{6^{2} + 8^{2}} = 10 km$
East and north movements are perpendicular.
2
Measure the angle from north.
$tan θ = \frac{6}{8}$
The east component is opposite the bearing angle and north is adjacent.
3
Calculate and format the bearing.
$θ = {tan}^{- 1} (\frac{6}{8}) \approx {36.9}^{\circ}$
The destination lies north-east, so this angle is the bearing ${036.9}^{\circ}$ .

Answer

10 km on bearing 036.9°

EXAMPLE 02

A south-east journey

A boat travels 120 m on bearing 150°. Find eastward and southward movements to 0.1 m and its return bearing.

Worked solution

1
Find the acute angle east of south.
$180^{\circ} - 150^{\circ} = 30^{\circ}$
The bearing lies between east and south.
2
Calculate the eastward component.
$E = 120 sin 30^{\circ} = 60.0 m$
Eastward displacement is opposite the reference angle.
3
Calculate the southward component.
$S = 120 cos 30^{\circ} \approx 103.9 m$
Southward displacement is adjacent to the reference angle.
4
Reverse the direction.
$150^{\circ} + 180^{\circ} = 330^{\circ}$
The return journey follows the same line in the opposite direction.

Answer

60.0 m east; 103.9 m south; 330°

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the reverse bearing of 065°.

Show answer & reasoning

245°

The reverse direction is half a turn away: $065^{\circ} + 180^{\circ} = 245^{\circ}$ , already within the bearing interval.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the reverse bearing of $045^{\circ}$ .
2 marks
A walker goes 9 km north and 12 km east. Find direct distance and bearing to ${0.1}^{\circ}$ .
3 marks
Find the northward/eastward movements of a 100 m journey on bearing $030^{\circ}$ , to 0.1 m.
3 marks
The bearing of $B$ from $A$ is $230^{\circ}$ . Find the bearing of $A$ from $B$ .
2 marks
Paths leave one point on bearings $035^{\circ}$ and $110^{\circ}$ . Find the smaller included angle.
2 marks
A student says bearing $300^{\circ}$ travels east. Correct the direction and explain.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the reverse bearing of $045^{\circ}$ .
2 marks
Add $180^{\circ}$ .
$225^{\circ}$
A walker goes 9 km north and 12 km east. Find direct distance and bearing to ${0.1}^{\circ}$ .
3 marks
$d = \sqrt{9^{2} + 12^{2}} = 15$ km.
$θ = {tan}^{- 1} (\frac{12}{9}) = 53.130^{\dots \circ}$ .
15 km on bearing ${053.1}^{\circ}$
Find the northward/eastward movements of a 100 m journey on bearing $030^{\circ}$ , to 0.1 m.
3 marks
North $= 100 cos 30^{\circ} = 86.602 \dots$ m.
East $= 100 sin 30^{\circ} = 50$ m.
86.6 m north; 50.0 m east
The bearing of $B$ from $A$ is $230^{\circ}$ . Find the bearing of $A$ from $B$ .
2 marks
Subtract $180^{\circ}$ and use three digits.
$050^{\circ}$
Paths leave one point on bearings $035^{\circ}$ and $110^{\circ}$ . Find the smaller included angle.
2 marks
$110^{\circ} - 35^{\circ} = 75^{\circ}$ .
$75^{\circ}$
A student says bearing $300^{\circ}$ travels east. Correct the direction and explain.
2 marks
$300^{\circ}$ lies between west ( $270^{\circ}$ ) and north ( $360^{\circ}$ ).
Its eastward component is negative.
North-west, with a westward component.

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian school course scope

Right-triangle applications include AC9M10M03. The approved chapter also develops non-right-triangle methods, exact values and trigonometric functions within one normal Year 10 course.

Official curriculum source ↗

In this lesson

Understand the idea.

Start at north

Connect direction with components

See the method in action.

Combine movements

Find the direct distance.

Measure the angle from north.

Calculate and format the bearing.

A south-east journey

Find the acute angle east of south.

Calculate the eastward component.

Calculate the southward component.

Reverse the direction.

Try it yourself.

Check your understanding

Keep practising