What is a bearing in the context of Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, a bearing is a way of describing the direction of one point from another using angles. Bearings are measured in degrees, clockwise from the north direction, and are typically expressed as three-digit numbers. For example, a bearing of 045° indicates a direction 45 degrees clockwise from north.

How do you calculate the bearing between two points?

To calculate the bearing between two points, first identify the north direction at the starting point. Then, measure the angle clockwise from north to the line connecting the two points. Ensure the angle is expressed as a three-digit number. For example, if the angle is 60 degrees, the bearing is written as 060°.

Why are bearings always expressed as three-digit numbers?

Bearings are expressed as three-digit numbers to maintain consistency and avoid confusion. This format ensures clarity, especially when dealing with angles less than 100 degrees. For instance, a bearing of 5 degrees is written as 005°, making it clear that it is a bearing and not just a simple angle.

What is the difference between true bearing and magnetic bearing?

True bearing is the angle measured clockwise from true north, which is the direction towards the geographic North Pole. Magnetic bearing, on the other hand, is measured from magnetic north, which is the direction a compass needle points. The difference between true north and magnetic north is known as magnetic declination, which varies depending on your location on Earth.

How are bearings used in real-life applications?

Bearings are widely used in navigation, aviation, and maritime activities to determine directions and plot courses. They are essential for pilots and ship captains to navigate accurately. In addition, bearings are used in land surveying and map reading to describe the direction from one point to another precisely.

Why is "Measuring anticlockwise instead of clockwise" a common mistake in Bearing?

Why it happens: Defaulting to maths-style positive direction. How to avoid it: Bearings: always CLOCKWISE from north. Different from standard angle convention. Source: 0580/42 — recurring

Why is "Writing a bearing with fewer than 3 digits" a common mistake in Bearing?

Why it happens: Treating it like a normal angle. How to avoid it: Pad with zeros: 30 030, 5 005.

Why is "Using the wrong north line" a common mistake in Bearing?

Why it happens: Each point has its own north line. How to avoid it: Bearing of B from A uses the north line at A, not at B.

Why is "Subtracting $180\degree$ when you should add" a common mistake in Bearing?

Why it happens: Forgetting the 0–360 range. How to avoid it: If the bearing is less than 180, ADD 180. If it's more, SUBTRACT 180.

Cambridge IGCSE Mathematics (0580)

Bearing

0% Complete

3 tasks remaining

Short Notes - Bearing

Detailed Study Notes

Detailed notes on Trigonometry for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Bearings — Cambridge IGCSE 0580 Maths Extended (2026)

Compass directions written as three-figure angles measured CLOCKWISE from NORTH. Cambridge expects $0\degree \le$ bearing $< 360\degree$ — three digits even for small angles.

At a glance

Bearing = angle measured CLOCKWISE from NORTH.
Always quote with THREE digits: $060\degree$ not $60\degree$ .
Range: $000\degree$ to $360\degree$ (often $0 \le \theta < 360$ ).
North = $000\degree$ . East = $090\degree$ . South = $180\degree$ . West = $270\degree$ .
Bearing of $B$ from $A$ — measure at $A$ , looking toward $B$ .
Back bearing: $\theta \pm 180\degree$ (use $+180$ if $\theta < 180$ , else $-180$ ).

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E8.2 — Use the sine, cosine and tangent ratios to solve problems involving bearings.

What is a bearing?

Three-figure angle, measured clockwise from north.

A bearing is a direction expressed as a three-figure angle measured clockwise from north.

Compass cardinal points.

North → $000\degree$
North-East → $045\degree$
East → $090\degree$
South-East → $135\degree$
South → $180\degree$
South-West → $225\degree$
West → $270\degree$
North-West → $315\degree$

Why three figures? Bearings always use three digits — write $060\degree$ , not $60\degree$ . This avoids confusion with non-bearing angles.

Bearings are measured clockwise from north and always written with three digits.

Worked. A point lies due east of an observer. The bearing is $090\degree$ .

Worked. A point lies $30\degree$ west of north. The bearing is $360\degree - 30\degree = 330\degree$ .

Clockwise from north.
Three figures always.
Range $000\degree$ to $360\degree$ .
Cardinal points: N $000$ , E $090$ , S $180$ , W $270$ .

Bearing of B from A

The phrase "bearing of B FROM A" — start at A, draw a north line, measure the angle clockwise to AB.

"Bearing of B from A" means: stand at $A$ , face north, then turn clockwise until you face $B$ .

Step-by-step.

Mark point $A$ and draw a vertical line UP from $A$ (representing north).
Draw the line from $A$ to $B$ .
Measure the clockwise angle from the north line to $AB$ .
Write as a three-figure bearing.

Worked. $A$ is at the origin, $B$ is at $(4, 3)$ (i.e. east-then-north of $A$ ). Find the bearing of $B$ from $A$ .

The angle from north (the positive $y$ -axis) clockwise to $AB$ .
$\tan(\text{angle from north}) = \dfrac{4}{3}$ (east displacement over north displacement).
Angle $= \tan^{-1}(4/3) \approx 53.13\degree$ .
Bearing $= 053\degree$ .

Tip. Drawing the north arrow at the STARTING point — and ONLY at the starting point — is the key visual.

Stand at A, face north, then turn clockwise until you face B — that angle is the bearing.

Bearing of B from A: north arrow at A, clockwise to AB.
Express as a three-figure bearing.
Use right-angled trig (or sine/cosine rules) to compute the angle.

Back bearings

The back bearing of B from A is the bearing of A from B. Add $180\degree$ if the original is $< 180\degree$ , otherwise subtract $180\degree$ .

If the bearing of B from A is $\theta$ , the bearing of A from B is the back bearing: $\text{back bearing} = \begin{cases} \theta + 180\degree & \text{if } \theta < 180\degree \\ \theta - 180\degree & \text{if } \theta \ge 180\degree \end{cases}$

The two opposite directions are exactly $180\degree$ apart, but you stay in the $0$ to $360$ range.

The north lines at A and B are parallel; the back bearing is exactly 180° from the original.

Worked. Bearing of $B$ from $A$ is $075\degree$ . Find the bearing of $A$ from $B$ .

$075 + 180 = 255\degree$ . Back bearing $= 255\degree$ .

Worked. Bearing of $C$ from $D$ is $230\degree$ . Find the bearing of $D$ from $C$ .

$230 - 180 = 050\degree$ . Back bearing $= 050\degree$ .

Bearing of B from A	Back bearing (A from B)
$040\degree$	$220\degree$
$075\degree$	$255\degree$
$150\degree$	$330\degree$
$230\degree$	$050\degree$
$310\degree$	$130\degree$

Back bearing differs by $180\degree$ .
Add $180$ if original $< 180\degree$ .
Subtract $180$ if original $\ge 180\degree$ .
Stay within $000\degree$ to $360\degree$ .

Bearings combined with trigonometry

Find the angles inside the triangle, then apply right-angled trig or the sine/cosine rule.

Cambridge's hardest bearing questions combine a journey (multiple legs, each with a bearing and distance) with trigonometry to find a missing distance or angle.

Worked. A ship sails on a bearing of $060\degree$ for $40\,\text{km}$ , then on a bearing of $150\degree$ for $30\,\text{km}$ . How far is it from the start, and on what bearing?

Step 1. Draw the journey. Note that the two legs differ in bearing by $150 - 60 = 90\degree$ , so the angle between the two legs at the turning point is $90\degree$ (the journey makes a right angle).

Step 2. The displacement from start to end is the hypotenuse of a right-angled triangle:

$d = \sqrt{40^{2} + 30^{2}} = \sqrt{2500} = 50\,\text{km}$ .

Step 3. Find the bearing. The angle inside the triangle at the start, opposite the second leg (length $30$ ), is $\theta = \tan^{-1}(30/40) \approx 36.87\degree$ . The bearing from start to end is $060\degree + 36.87\degree \approx 097\degree$ .

The two bearings differ by 90°, so the legs form a right angle — the displacement is the hypotenuse.

General template.

Draw all north arrows at each waypoint.
Mark each bearing.
Use angle facts (alternate angles, co-interior, etc.) to find the interior angle of the triangle formed.
Apply right-angled trig (if a $90\degree$ pops out) or sine/cosine rule (general case — see those study notes).

Draw north arrows at each waypoint.
Identify the interior angle of any triangle that forms.
Apply right-angled trig, sine rule, or cosine rule.
Express the final answer as a three-figure bearing.

Quick recap

Bearing: clockwise from north, three figures.
N $000\degree$ , E $090\degree$ , S $180\degree$ , W $270\degree$ .
Bearing of B from A: north arrow at A, clockwise to AB.
Back bearing: $\pm 180\degree$ .
Combine with right-angled trig or sine/cosine rules.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Bearing — direction as a three-figure angle measured clockwise from north.
Back bearing — the reverse-direction bearing; differs by $180\degree$ .
Cardinal direction — N, E, S, W, at $000\degree, 090\degree, 180\degree, 270\degree$ .

How it’s examined

Bearings appear most years on Paper 4 as a 4-6 mark question, often combining bearings with trigonometry (sine/cosine rule) to find distances or angles in a journey. Examiner reports flag two errors: (i) missing leading zeros in three-figure bearings, (ii) measuring anticlockwise from north (or from the wrong axis).

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E8.2); 0580/42 Oct/Nov 2024 — Q14 (bearings + cosine rule); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Bearing

Step-by-step solutions to past-paper-style questions on bearing, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Bearing

Almost every bearing exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Graph or diagram

Recognise it by

A diagram (or a verbal description of one) showing a point relative to north, with a write down / state the bearing instruction — you read or interpret an angle rather than calculate.

How to approach it

Measure the angle clockwise from the north line at the starting point, then write it as a three-digit value between $000\degree$ and $360\degree$ .

Common trap

Measuring anticlockwise (the standard maths convention) or omitting leading zeros. Examiner reports flag both — bearings go clockwise and always use three digits.

Direct calculation

Recognise it by

A bearing is given and a back bearing is asked for, or two bearings from the same point need combining into an angle — a single arithmetic step.

How to approach it

For a back bearing add $180\degree$ if the original is below $180\degree$ , otherwise subtract $180\degree$ . For an angle between two bearings from one point, subtract them.

Common trap

Adding when you should subtract (or vice versa) and leaving the range $000\degree$ - $360\degree$ . Examiner reports flag the $180\degree$ direction error and missing leading zeros.

Word problem

Recognise it by

A navigation context — a ship or hiker travelling several legs on stated bearings — with no triangle drawn, asking for a distance and/or a final bearing.

How to approach it

Draw a north line at each point, use back-bearings to find the interior angles, then apply Pythagoras, the sine rule or the cosine rule. Add the calculated angle to the original bearing for the final answer.

Common trap

Adding the calculated angle to the back-bearing instead of the forward bearing, or misjudging the quadrant. Examiner reports stress sketching the north lines before computing.

1Read a bearing from a diagram
CoreGraph or diagram• read bearing
Question
From point $A$ , point $B$ lies $40\degree$ east of north. Write the bearing of $B$ from $A$ .
Step-by-step solution
1. Step 1
  Bearings are measured CLOCKWISE from north and given as 3 digits.
  $\text{Bearing} = 040\degree$
Answer
$040\degree$
Examiner tip
Always state bearings as THREE digits (e.g. $040\degree$ not $40\degree$ ).
2Find a back bearing
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q14• back bearing
Question
The bearing of $B$ from $A$ is $075\degree$ . Find the bearing of $A$ from $B$ .
Step-by-step solution
1. Step 1
  Add $180\degree$ if the original bearing is less than $180\degree$ .
  $075\degree + 180\degree = 255\degree$
Answer
$255\degree$
3Combine bearings with trigonometry
ExtendedWord problem• bearings + trig
Question
A ship sails $20$ km from $A$ to $B$ on a bearing of $060\degree$ . Find how far east of $A$ the ship is now.
Step-by-step solution
1. Step 1
  East = adjacent leg if we measure $60\degree$ from north (so it's the side opposite the angle from north).
2. Step 2
  Use $\sin 60\degree = \dfrac{\text{east}}{20}$ .
  $\text{east} = 20 \sin 60\degree \approx 17.3\,\text{km}$
Answer
$\approx 17.3\,\text{km}$
4Bearing between three points
ExtendedDirect calculation• three-point bearing
Question
From $A$ , the bearing of $B$ is $080\degree$ . From $A$ , the bearing of $C$ is $140\degree$ . Find $\angle BAC$ .
Step-by-step solution
1. Step 1
  Both bearings share the same north line at $A$ , so subtract.
  $\angle BAC = 140\degree - 80\degree = 60\degree$
Answer
$60\degree$
5Reverse bearing from a value over $180\degree$
CoreDirect calculation• Adapted from 0580/22 Feb/Mar 2024 Q5• back bearing
Question
The bearing of $Y$ from $X$ is $235\degree$ . Find the bearing of $X$ from $Y$ .
Step-by-step solution
1. Step 1
  Since the original bearing exceeds $180\degree$ , subtract $180\degree$ rather than add.
  $235\degree - 180\degree = 055\degree$
Answer
$055\degree$
Examiner tip
Always state the answer with three digits. Writing $55\degree$ instead of $055\degree$ commonly loses one mark in 0580 paper 2 questions.
6Navigation: ship A to B to C
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2023 Q13• navigation
Question
A ship sails $40$ km from $A$ on a bearing of $060\degree$ to $B$ . From $B$ it then sails $30$ km on a bearing of $150\degree$ to $C$ . Find the bearing of $C$ from $A$ and the distance $AC$ .
Step-by-step solution
1. Step 1
  Interior angle at $B$ : the back-bearing from $B$ to $A$ is $060\degree + 180\degree = 240\degree$ . So $\angle ABC = 240\degree - 150\degree = 90\degree$ .
2. Step 2
  Triangle $ABC$ is right-angled at $B$ . Apply Pythagoras to find $AC$ .
  $AC = \sqrt{40^{2} + 30^{2}} = \sqrt{2500} = 50\,\text{km}$
3. Step 3
  Find $\angle BAC$ using $\tan$ : opposite is $BC = 30$ , adjacent is $AB = 40$ .
  $\tan(\angle BAC) = \dfrac{30}{40} = 0.75$
4. Step 4
  Apply $\tan^{-1}$ .
  $\angle BAC = \tan^{-1}(0.75) \approx 36.87\degree$
5. Step 5
  The bearing of $C$ from $A$ is the bearing of $B$ from $A$ plus $\angle BAC$ .
  $\text{Bearing} = 060\degree + 36.87\degree \approx 096.9\degree$
Answer
$AC = 50$ km; bearing of $C$ from $A \approx 097\degree$
Examiner tip
The 2023 examiner report repeatedly warns students to add the calculated angle to the original bearing (not the back-bearing). Sketch the north lines at each point before computing.
7Bearing combined with Pythagoras
ExtendedWord problem• pythagoras
Question
From a harbour $H$ , a yacht sails on a bearing of $090\degree$ for $7$ km to $P$ , then turns and sails on a bearing of $180\degree$ for $24$ km to $Q$ . Find the distance $HQ$ and the bearing of $Q$ from $H$ .
Step-by-step solution
1. Step 1
  Bearings $090\degree$ and $180\degree$ are perpendicular. Triangle $HPQ$ is right-angled at $P$ .
  $HQ = \sqrt{7^{2} + 24^{2}} = \sqrt{625} = 25\,\text{km}$
2. Step 2
  Find the angle at $H$ between north and $HQ$ . The east-component is $7$ and the south-component is $24$ .
  $\tan\theta = \dfrac{7}{24} \Rightarrow \theta \approx 16.26\degree$
3. Step 3
  This angle is measured east of south, so the bearing from $H$ is $180\degree - 16.26\degree = 163.74\degree$ .
Answer
$HQ = 25$ km; bearing $\approx 164\degree$
Examiner tip
Watch the quadrant: the yacht is south-east of the harbour, so its bearing must be between $090\degree$ and $180\degree$ .
8Bearing combined with cosine rule
ChallengeWord problem• Adapted from 0580/42 May/Jun 2023 Q19• cosine rule, challenge
Question
Town $B$ is $58$ km from town $A$ on a bearing of $068\degree$ . Town $C$ is $42$ km from $A$ on a bearing of $156\degree$ . Find the distance $BC$ and the bearing of $C$ from $B$ .
Step-by-step solution
1. Step 1
  Angle at $A$ inside triangle $ABC$ : $\angle BAC = 156\degree - 68\degree = 88\degree$ .
2. Step 2
  Apply the cosine rule.
  $BC^{2} = 58^{2} + 42^{2} - 2(58)(42)\cos 88\degree \approx 3364 + 1764 - 4872(0.0349) \approx 5128 - 170.0 = 4958$
3. Step 3
  Square root.
  $BC \approx \sqrt{4958} \approx 70.41\,\text{km}$
4. Step 4
  Apply the sine rule for $\angle ABC$ : $\dfrac{\sin B}{42} = \dfrac{\sin 88\degree}{70.41}$ .
  $\sin B = \dfrac{42 \sin 88\degree}{70.41} \approx 0.5963 \Rightarrow \angle B \approx 36.6\degree$
5. Step 5
  Back-bearing of $A$ from $B = 068\degree + 180\degree = 248\degree$ . The bearing of $C$ from $B$ lies clockwise from this by $\angle ABC$ .
  $\text{Bearing} = 248\degree - 36.6\degree \approx 211.4\degree$
Answer
$BC \approx 70.4$ km; bearing of $C$ from $B \approx 211\degree$
Examiner tip
Decide whether to add or subtract $\angle ABC$ from the back-bearing based on the diagram orientation. Sketching it before computing avoids the most common sign error.
9Bearing in a real-world map
ExtendedWord problem• real-world, map
Question
On a map, a hiker walks from point $X$ to point $Y$ on a bearing of $325\degree$ . After $6$ km she turns and walks $4$ km due east to reach a campsite $Z$ . State the bearing of $Z$ from $Y$ , and find how far north of $X$ the campsite is.
Step-by-step solution
1. Step 1
  Due east corresponds to a bearing of $090\degree$ .
2. Step 2
  Northward component from $X$ to $Y$ : $325\degree$ is $35\degree$ west of north, so the northward leg is $6\cos 35\degree \approx 4.915$ km.
  $y_{1} = 6\cos 35\degree \approx 4.915$
3. Step 3
  From $Y$ to $Z$ the displacement is purely east, so it contributes nothing to the north component.
  $y_{\text{total}} \approx 4.915\,\text{km}$
Answer
Bearing of $Z$ from $Y$ is $090\degree$ ; campsite is $\approx 4.92$ km north of $X$ .
Examiner tip
Resolve each leg into north/east components separately and add. The east leg from $Y$ to $Z$ doesn't change the north-distance from $X$ .
10Back bearing from a south-easterly value
CoreDirect calculation• back bearing
Question
The bearing of $Q$ from $P$ is $128\degree$ . Find the bearing of $P$ from $Q$ .
Step-by-step solution
1. Step 1
  $128\degree$ is less than $180\degree$ , so add $180\degree$ .
  $\text{back bearing} = 128\degree + 180\degree = 308\degree$
Answer
$308\degree$
Examiner tip
Apply the rule consistently: if the original bearing is less than $180\degree$ , ADD $180\degree$ ; if it is at least $180\degree$ , SUBTRACT $180\degree$ .

Key Formulae — Bearing

The formulae you need to memorise for bearing on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Bearing convention
$\text{Measured clockwise from NORTH, given as 3 digits } 000\degree\text{–}360\degree$
When to use
Always.
Back bearing
$\text{back} = \theta \pm 180\degree \quad (+\,180\degree \text{ if } \theta < 180\degree)$
When to use
To reverse a bearing.

Key Definitions and Keywords — Bearing

Definitions to memorise and the exact keywords mark schemes credit for bearing answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Bearing
Examiner keyword
An angle measured clockwise from north, written as a 3-digit number between $000\degree$ and $360\degree$ .
Back bearing
Examiner keyword
The bearing in the reverse direction. Differs from the original by exactly $180\degree$ .
North line
Examiner keyword
A vertical line drawn at each point in a bearings diagram, used as the reference direction.

Common Mistakes and Misconceptions — Bearing

The traps other students keep falling into on bearing questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Measuring anticlockwise instead of clockwise
0580/42 — recurring
Why it happens
Defaulting to maths-style positive direction.
How to avoid it
Bearings: always CLOCKWISE from north. Different from standard angle convention.
✕Writing a bearing with fewer than 3 digits
Why it happens
Treating it like a normal angle.
How to avoid it
Pad with zeros: $30\degree \to 030\degree$ , $5\degree \to 005\degree$ .
✕Using the wrong north line
Why it happens
Each point has its own north line.
How to avoid it
Bearing of $B$ from $A$ uses the north line at $A$ , not at $B$ .
✕Subtracting $180\degree$ when you should add
Why it happens
Forgetting the $0$ – $360$ range.
How to avoid it
If the bearing is less than $180\degree$ , ADD $180\degree$ . If it's more, SUBTRACT $180\degree$ .

Bearing — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Bearing of B from A

Back bearing (A from B)

040\degree

220\degree

075\degree

255\degree

150\degree

330\degree

230\degree

050\degree

310\degree

130\degree

Cambridge's hardest bearing questions combine a journey (multiple legs, each with a bearing and distance) with trigonometry to find a missing distance or angle.

Worked. A ship sails on a bearing of $060\degree$ for $40\,\text{km}$ , then on a bearing of $150\degree$ for $30\,\text{km}$ . How far is it from the start, and on what bearing?

Step 2. The displacement from start to end is the hypotenuse of a right-angled triangle:

$d = \sqrt{40^{2} + 30^{2}} = \sqrt{2500} = 50\,\text{km}$ .

The two bearings differ by 90°, so the legs form a right angle — the displacement is the hypotenuse.

General template.

Draw all north arrows at each waypoint.
Mark each bearing.
Use angle facts (alternate angles, co-interior, etc.) to find the interior angle of the triangle formed.
Apply right-angled trig (if a $90\degree$ pops out) or sine/cosine rule (general case — see those study notes).

Bearing

Short Notes - Bearing

Detailed Study Notes

At a glance

What you’ll learn

What is a bearing?

Bearing of B from A

Back bearings

Bearings combined with trigonometry

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Bearing

1Read a bearing from a diagram

2Find a back bearing

3Combine bearings with trigonometry

4Bearing between three points

5Reverse bearing from a value over 180°180\degree180°

6Navigation: ship A to B to C

7Bearing combined with Pythagoras

8Bearing combined with cosine rule

9Bearing in a real-world map

10Back bearing from a south-easterly value

Bearing convention

Back bearing

Bearing

Back bearing

North line

✕Measuring anticlockwise instead of clockwise

✕Writing a bearing with fewer than 3 digits

✕Using the wrong north line

✕Subtracting 180°180\degree180° when you should add

Bearing — frequently asked questions

Bearing

Short Notes - Bearing

Detailed Study Notes

At a glance

What you’ll learn

What is a bearing?

Bearing of B from A

Back bearings

Bearings combined with trigonometry

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Bearing

1Read a bearing from a diagram

2Find a back bearing

3Combine bearings with trigonometry

4Bearing between three points

5Reverse bearing from a value over 180°180\degree180°

6Navigation: ship A to B to C

7Bearing combined with Pythagoras

8Bearing combined with cosine rule

9Bearing in a real-world map

10Back bearing from a south-easterly value

Bearing convention

Back bearing

Bearing

Back bearing

North line

✕Measuring anticlockwise instead of clockwise

✕Writing a bearing with fewer than 3 digits

✕Using the wrong north line

✕Subtracting 180°180\degree180° when you should add

Bearing — frequently asked questions

5Reverse bearing from a value over $180\degree$

✕Subtracting $180\degree$ when you should add

5Reverse bearing from a value over $180\degree$

✕Subtracting $180\degree$ when you should add