Bearings · Study Notes · Edexcel IGCSE 4MA1 Higher

Three-digit bearings measured CLOCKWISE from NORTH. Standard for navigation; combined with trigonometry.

At a glance

Always 3 digits ( $047°$ , not $47°$ ).
Clockwise from NORTH.
North = $000°/360°$ . East = $090°$ . South = $180°$ . West = $270°$ .
Back bearing: $\pm 180°$ .
Bearings combine with right-angle trig for navigation.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.9 — Read and use bearings.
4.9 — Find back bearings.
4.9 — Combine bearings with trigonometry for navigation problems.

Reading bearings

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$3$ digits. Clockwise from North.

Bearing = angle measured CLOCKWISE from NORTH.

Always 3 digits. $047°$ , $123°$ , $310°$ . (Not $47°$ , $123°$ , $310°$ .)

Compass directions:

Direction	Bearing
North	$000°$ or $360°$
North-East	$045°$
East	$090°$
South-East	$135°$
South	$180°$
South-West	$225°$
West	$270°$
North-West	$315°$

Bearings are measured clockwise from North, always written with three digits.

Back bearing. The bearing in the OPPOSITE direction.

If original < $180°$ : ADD $180°$ . If original > $180°$ : SUBTRACT $180°$ .

(So back bearing always in $[0°, 360°]$ .)

Examples:

$060°$ → back $240°$ .
$250°$ → back $070°$ .
$090°$ → back $270°$ .

Worked qualitative. Why is North = $000°$ ?

Bearings start at North and rotate clockwise.
North is the starting point: $0°$ rotation.
Convention from compass orientation.

Edexcel tip. ALWAYS three digits. Mark schemes deduct for missing leading zeros.

Three digits.
Clockwise from North.
N = $000°$ , E = $090°$ , S = $180°$ , W = $270°$ .
Back bearing: $\pm 180°$ .

Bearings with trigonometry

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Combine with right-angle trig for navigation problems.

Common scenario: A ship/aeroplane travels on a bearing for a distance. Find positions or new bearings.

Method:

Sketch the diagram.
Identify right triangle with the bearing as one angle.
Hypotenuse = distance travelled.
North/South = adjacent or opposite (depending on bearing).
East/West = opposite or adjacent.

Worked example. Ship sails $20$ km on bearing $050°$ .

Right triangle setup:

Hypotenuse: $20$ km.
Angle from North: $50°$ .
North component (adjacent to bearing): $20 \cos 50° \approx 12.86$ km.
East component (opposite to bearing): $20 \sin 50° \approx 15.32$ km.

So the ship has moved $12.86$ km north and $15.32$ km east.

The bearing angle sits between North and the path; cos gives the North leg, sin gives the East leg.

Multi-leg journeys. Add the components leg by leg.

Example: Leg 1 north $5$ + east $3$ . Leg 2 north $-2$ + east $7$ . Net: north $3$ , east $10$ .

Final position can be converted back to bearing via $\tan^{-1}$ .

Worked qualitative. A ship sails $5$ km north then $12$ km east. What's its straight-line distance and bearing from start?

Distance: Pythagoras. $\sqrt{25 + 144} = 13$ km.
Bearing: angle from North. $\tan \theta = 12/5 = 2.4$ , so $\theta \approx 67.4°$ .
Bearing: $067°$ .

Edexcel tip. Sketch the diagram. Identify the right triangle. Use SOH CAH TOA. Mark schemes credit each step.

Sketch first.
Right triangle with bearing as one angle.
N/S and E/W components.
Use Pythagoras for distances.

Quick recap

Bearings: 3 digits, clockwise from N.
N = $000°$ , E = $090°$ , S = $180°$ , W = $270°$ .
Back bearing: $\pm 180°$ .
Combine with right-angle trig for navigation.
Sketch + identify right triangle.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Always 3 digits.
Clockwise from North.
Back bearing: $\pm 180°$ .
Compass cardinal: N $000°$ , E $090°$ , S $180°$ , W $270°$ .

How it’s examined

Bearings on Higher Tier (3-6 marks). Often combined with trigonometry. Examiner reports flag (1) missing leading zeros, (2) measuring anticlockwise, (3) wrong back-bearing direction.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024. Last reviewed 2026-05-07.

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

1Read a bearing
Foundation• bearing
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Question
$B$ is due East of $A$ . What is the bearing of $B$ from $A$ ?
Step-by-step solution
1. Step 1
  Bearings are measured CLOCKWISE from NORTH.
2. Step 2
  East is $90°$ clockwise from North.
3. Step 3
  Bearing: $090°$ (always 3 digits).
Answer
$090°$
Examiner tip
Bearings are ALWAYS three digits: $090°$ , not $90°$ . Mark schemes deduct.
2Back bearing
Higher• Adapted from 4MA1/1H May/Jun 2024 Q14• back bearing
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Question
Bearing of $B$ from $A$ is $060°$ . Find the bearing of $A$ from $B$ .
Step-by-step solution
1. Step 1
  Back bearing: add $180°$ if original $< 180°$ , subtract $180°$ if original $> 180°$ .
2. Step 2
  $060° + 180° = 240°$ .
Answer
$240°$
Examiner tip
Back bearing: $\pm 180°$ . Choose the one that gives a 3-digit angle in $[0°, 360°]$ .
3Bearing combined with trigonometry
Higher• bearing, trigonometry
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Question
A ship sails on a bearing of $050°$ for $20$ km. How far north and east has it travelled?
Step-by-step solution
1. Step 1
  Bearing $050°$ = $50°$ clockwise from North.
2. Step 2
  Right triangle: hypotenuse $20$ , angle $50°$ from North-South line.
3. Step 3
  North component (adjacent): $20 \cos 50° \approx 12.86$ km.
4. Step 4
  East component (opposite): $20 \sin 50° \approx 15.32$ km.
Answer
North $\approx 12.86$ km. East $\approx 15.32$ km.
Examiner tip
Choose right triangle with the bearing as one of its angles. The hypotenuse is the distance travelled.
4A new bearing after a right-angle turn [1 mark]
Getting started• Paper 1H/2H style — 1 mark• bearings, turning
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Question
A ship is sailing on a bearing of 320°. It turns $90^{\circ}$ clockwise.

Write down the new bearing of the ship. [1]
Step-by-step solution
1. Step 1
  Turning clockwise adds to the bearing.
  $320^{\circ} + 90^{\circ} = 410^{\circ}$
2. Step 2
  B1 A bearing must be less than $360^{\circ}$ , so subtract one full turn.
  $410^{\circ} - 360^{\circ} = 050^{\circ}$
Answer
050°
Examiner tip
Examiner tip: Passing north resets the count. From 320° it is $40^{\circ}$ to north, and the remaining $50^{\circ}$ of the turn gives 050°.

Common mistake: Leaving the answer as 410°, or subtracting and giving 230°, which is a $90^{\circ}$ turn anticlockwise.
5Distance from speed and time, then a scale-drawing length [2 marks]
Getting started• Paper 1H/2H style — 2 marks• bearings, speed, scale drawing
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Question
A ferry sails from a port on a bearing of 118° at an average speed of $16$ km/h for 2 hours 15 minutes.

(a) Work out the distance the ferry sails. [1]

(b) A scale drawing of the journey uses a scale of $1$ cm to $5$ km. Work out the length of the line that represents the journey on the scale drawing. [1]
Step-by-step solution
1. Step 1
  Convert the time to hours: $15$ minutes is $\dfrac{15}{60} = 0.25$ hours.
  $2\text{ h } 15\text{ min} = 2.25\text{ h}$
2. Step 2
  B1 (a) Distance = speed × time.
  $16 \times 2.25 = 36\text{ km}$
3. Step 3
  B1 (b) Every $5$ km is $1$ cm on the drawing.
  $36 \div 5 = 7.2\text{ cm}$
Answer
(a) $36$ km (b) $7.2$ cm
Examiner tip
Examiner tip: On the drawing itself, the line would start at a north line through the port and be measured 118° clockwise with a protractor. The bearing sets the direction; the scale sets the length.

Common mistake: Writing 2 hours 15 minutes as $2.15$ hours, which gives $34.4$ km. Minutes must be divided by $60$ .
6The angle between two bearings from one point [2 marks]
Getting started• Paper 1H/2H style — 2 marks• bearings, angles at a point
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Question
From a tower $T$ , the bearing of a church $C$ is 074° and the bearing of a mast $M$ is 312°.

Work out the size of the smaller angle $CTM$ . [2]
Step-by-step solution
1. Step 1
  M1 Going clockwise from $C$ to $M$ is $312^{\circ} - 74^{\circ} = 238^{\circ}$ , which is the reflex angle. Angles around a point add up to $360^{\circ}$ .
  $360^{\circ} - (312^{\circ} - 74^{\circ})$
2. Step 2
  A1 Evaluate.
  $= 122^{\circ}$
Answer
$122^{\circ}$
Examiner tip
Examiner tip: The smaller angle passes through north here: from $M$ it is $360^{\circ} - 312^{\circ} = 48^{\circ}$ to north, then $74^{\circ}$ more to $C$ , and $48^{\circ} + 74^{\circ} = 122^{\circ}$ .

Common mistake: Giving $238^{\circ}$ , the reflex angle, when the question asks for the smaller angle.
7Deciding which of two bearings is correct [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• bearings, trigonometry, reasoning
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Question
A helicopter flies in a straight line from $P$ to $Q$ . $Q$ is $42$ km north and $17$ km west of $P$ .

Tom says the bearing of $Q$ from $P$ is 022°. Ann says it is 338°.

Who is correct? Show working to support your answer. [3]
Step-by-step solution
1. Step 1
  M1 In the right-angled triangle, let $\theta$ be the angle at $P$ between due north and $PQ$ . The $17$ km side is opposite $\theta$ and the $42$ km side is adjacent.
  $\tan \theta = \dfrac{17}{42}$
2. Step 2
  A1 Evaluate.
  $\theta = 22.03\ldots^{\circ}$
3. Step 3
  B1 $Q$ is west of north, so $PQ$ is $22.0^{\circ}$ anticlockwise from north. A bearing is measured clockwise: $360^{\circ} - 22.03\ldots^{\circ} = 337.96\ldots^{\circ}$ , so Ann is correct.
Answer
Ann is correct: the bearing is $360^{\circ} - 22.0^{\circ} \approx 338^{\circ}$ . Tom has measured anticlockwise.
Examiner tip
Examiner tip: Before any trigonometry, place the answer in a quarter. North and west means the bearing is between 270° and 360°, which settles the argument before you calculate.

Common mistake: Finding $22.0^{\circ}$ and agreeing with Tom. The final B1 needs both the correct choice and the reason (the angle is measured anticlockwise from north, so it must be subtracted from $360^{\circ}$ ).
8A right angle from due north and due east [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• bearings, trigonometry, angle facts
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Question
A boat sails $14$ km from $A$ on a bearing of 128° to $B$ . It then sails due north to $C$ . $C$ is due east of $A$ .

Calculate the distance $AC$ . Give your answer correct to 3 significant figures. [3]
Step-by-step solution
1. Step 1
  $AC$ runs due east and $CB$ runs due north-south, so angle $ACB = 90^{\circ}$ . Triangle $ABC$ is right-angled at $C$ , with hypotenuse $AB = 14$ km.
2. Step 2
  B1 $AC$ is on a bearing of 090°, so the angle at $A$ between $AC$ and $AB$ is
  $\angle BAC = 128^{\circ} - 90^{\circ} = 38^{\circ}$
3. Step 3
  M1 $AC$ is adjacent to the $38^{\circ}$ angle.
  $AC = 14 \cos 38^{\circ}$
4. Step 4
  A1 Evaluate.
  $AC = 11.03\ldots = 11.0\text{ km (3 s.f.)}$
Answer
$11.0$ km
Examiner tip
Examiner tip: "Due east" and "due north" are always at right angles, so a journey that uses them gives you a right-angled triangle for free. Mark the right angle at $C$ on your sketch.

Common mistake: Using $14 \cos 128^{\circ}$ straight from the bearing. That gives a negative number, and the bearing is not an angle in the triangle.
9How far apart after walking on two bearings [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• bearings, cosine rule, speed
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Question
Two hikers leave a camp $C$ at the same time. Hiker $A$ walks at $5$ km/h on a bearing of 020°. Hiker $B$ walks at $4$ km/h on a bearing of 135°. They both walk in straight lines for 2 hours.

Calculate the distance between the two hikers after 2 hours. Give your answer correct to 3 significant figures. [4]
Step-by-step solution
1. Step 1
  B1 Distances walked in 2 hours.
  $CA = 5 \times 2 = 10\text{ km}, \qquad CB = 4 \times 2 = 8\text{ km}$
2. Step 2
  B1 Both bearings are measured from the same north line at $C$ , so the angle between the paths is their difference.
  $\angle ACB = 135^{\circ} - 20^{\circ} = 115^{\circ}$
3. Step 3
  M1 Use the cosine rule with the included angle.
  $AB^{2} = 10^{2} + 8^{2} - 2 \times 10 \times 8 \times \cos 115^{\circ} = 231.61\ldots$
4. Step 4
  A1 Square root.
  $AB = 15.21\ldots = 15.2\text{ km (3 s.f.)}$
Answer
$15.2$ km
Examiner tip
Examiner tip: $\cos 115^{\circ}$ is negative, so the last term is added: $164 + 67.6\ldots$ . The distance apart must be more than either path when the angle is obtuse, which is a quick check.

Common mistake: Using the speeds, $5$ and $4$ , as the sides of the triangle instead of the distances walked. That gives $7.61$ km, which is the distance apart after 1 hour.
10The bearing of one moving ship from another [5 marks]
Stretch• Paper 1H/2H style — 5 marks• bearings, speed, trigonometry
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Question
Two ships leave a harbour $H$ at 10:00. Ship $A$ sails at $18$ km/h on a bearing of 050°. Ship $B$ sails at $12$ km/h on a bearing of 140°. Both ships sail in straight lines at constant speed.

Calculate the bearing of ship $B$ from ship $A$ at 11:30. Give your answer correct to the nearest degree. [5]
Step-by-step solution
1. Step 1
  B1 From 10:00 to 11:30 is $1.5$ hours, so the distances from $H$ are
  $HA = 18 \times 1.5 = 27\text{ km}, \qquad HB = 12 \times 1.5 = 18\text{ km}$
2. Step 2
  B1 Both bearings are measured from the same north line at $H$ , so the angle between the two paths is their difference. Triangle $AHB$ is right-angled at $H$ .
  $\angle AHB = 140^{\circ} - 50^{\circ} = 90^{\circ}$
3. Step 3
  M1 In the right-angled triangle, $HB$ is opposite angle $HAB$ and $HA$ is adjacent.
  $\tan HAB = \dfrac{18}{27} \quad \Rightarrow \quad \angle HAB = 33.69\ldots^{\circ}$
4. Step 4
  M1 The north lines at $H$ and $A$ are parallel, so the bearing of $H$ from $A$ is the back bearing $050^{\circ} + 180^{\circ} = 230^{\circ}$ . Seen from $A$ , $B$ is anticlockwise from $H$ , so subtract angle $HAB$ .
  $230^{\circ} - 33.69\ldots^{\circ} = 196.30\ldots^{\circ}$
5. Step 5
  A1 Round to the nearest degree.
  $196^{\circ}$
Answer
196°
Examiner tip
Examiner tip: Draw the sketch before any trigonometry. $B$ is south of $A$ and a little to the west, so the bearing must be just over 180°. Checking this catches most direction errors.

Common mistake: Adding the angle to the back bearing, $230^{\circ} + 33.7^{\circ} = 264^{\circ}$ , or giving 016°, which is the bearing of $A$ from $B$ . Using the speeds $18$ and $12$ as the sides still gives $33.7^{\circ}$ here, because both are multiplied by the same time, but only the distances earn the first B1.

Question 1
Paper 1H/2H style — 1 mark1 mark
$C$ is due south of $D$ .

Write down the bearing of $D$ from $C$ . [1]
Model answer
$D$ is due north of $C$ .

Answer: 000°
Why this scores
Mark scheme:
- B1 for 000° (accept 360°)
Why this scores full marks: The script reverses the direction first (south of $D$ means $D$ is north of $C$ ) and then writes due north as a three-figure bearing. The usual wrong answer is 180°, the bearing of $C$ from $D$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
A ship changes course from a bearing of 285° to a bearing of 040° by turning clockwise.

Work out the angle through which the ship turns. [2]
Model answer
From 285° clockwise to north: $360^{\circ} - 285^{\circ} = 75^{\circ}$

From north clockwise to 040°: $40^{\circ}$

$75^{\circ} + 40^{\circ} = 115^{\circ}$

Answer: $115^{\circ}$
Why this scores
Mark scheme:
- M1 for $360 - 285 + 40$ oe
- A1 for $115$ cao
Why this scores full marks: The script splits the turn at north, where the bearings restart from zero. Subtracting the bearings, $285 - 40 = 245$ , gives the anticlockwise turn, which is not what the question describes.
Question 3
Paper 1H/2H style — 2 marks2 marks
A buoy $B$ is $7$ km east and $7$ km south of a harbour $H$ .

Find the bearing of $B$ from $H$ . [2]
Model answer
The right-angled triangle has two equal sides of $7$ km, so it is isosceles and its other two angles are each $45^{\circ}$ .

$HB$ is $45^{\circ}$ clockwise from due east.

$090^{\circ} + 45^{\circ} = 135^{\circ}$

Answer: 135°
Why this scores
Mark scheme:
- M1 for the angle $45^{\circ}$ in the triangle, from $\tan^{-1}\left(\dfrac{7}{7}\right)$ or the isosceles right-angled triangle
- A1 for 135° cao
Why this scores full marks: The script finds the angle and then turns it into a bearing measured from north. Stopping at $45^{\circ}$ , or giving 045° (which points north-east), loses the A mark.
Question 4
Paper 1H/2H style — 3 marks3 marks
A lighthouse $L$ is $3.2$ km due north of a harbour $H$ . A boat $B$ is due east of $H$ . The bearing of $B$ from $L$ is 152°.

Calculate the distance $HB$ . Give your answer correct to 3 significant figures. [3]
Model answer
From $L$ , due south points to $H$ . $LB$ is $180^{\circ} - 152^{\circ} = 28^{\circ}$ east of south, so angle $HLB = 28^{\circ}$ .

Angle $LHB = 90^{\circ}$ (north and east are perpendicular). $HB$ is opposite the $28^{\circ}$ angle and $LH$ is adjacent.

$HB = 3.2 \tan 28^{\circ} = 1.701\ldots$

Answer: $1.70$ km
Why this scores
Mark scheme:
- B1 for angle $HLB = 28^{\circ}$
- M1 for $3.2 \tan 28^{\circ}$ oe
- A1 for $1.70$ (awrt $1.70$ )
Why this scores full marks: The script converts the bearing into the angle inside the triangle at $L$ before any trigonometry, and states the right angle at $H$ . Using $\tan 152^{\circ}$ directly gives a negative length.
Question 5
Paper 1H/2H style — 3 marks3 marks
From a lookout $P$ , a buoy $Q$ is on a bearing of 035° and a buoy $R$ is on a bearing of 105°. $PQ = PR$ .

Work out the bearing of $R$ from $Q$ . [3]
Model answer
Angle $QPR = 105^{\circ} - 35^{\circ} = 70^{\circ}$ .

Triangle $PQR$ is isosceles ( $PQ = PR$ ), so its base angles are equal:

$\angle PQR = \dfrac{180^{\circ} - 70^{\circ}}{2} = 55^{\circ}$

The bearing of $P$ from $Q$ is $035^{\circ} + 180^{\circ} = 215^{\circ}$ . From $Q$ , $R$ is $55^{\circ}$ anticlockwise from $QP$ .

$215^{\circ} - 55^{\circ} = 160^{\circ}$

Answer: 160°
Why this scores
Mark scheme:
- B1 for angle $PQR = 55^{\circ}$ (with $QPR = 70^{\circ}$ )
- M1 for the back bearing $215^{\circ}$ used with their $55^{\circ}$
- A1 for 160° cao
Why this scores full marks: The script works in three clear stages (angle at $P$ , base angle, bearing from $Q$ ), each with its reason. A sketch shows that $R$ is south-east of $Q$ , so the answer must lie between 090° and 180°; adding instead ( $215 + 55 = 270$ ) points due west and fails that check.
Question 6
Paper 1H/2H style — 4 marks4 marks
A boat leaves a harbour at 09:15 and sails on a bearing of 150° at a constant speed of $12$ km/h. It stops at 10:45.

Calculate how far south of the harbour the boat stops. Give your answer correct to 3 significant figures. [4]
Model answer
Time: 09:15 to 10:45 is 1 hour 30 minutes $= 1.5$ hours.

Distance $= 12 \times 1.5 = 18$ km.

The path is $180^{\circ} - 150^{\circ} = 30^{\circ}$ east of south, so the distance south is adjacent to the $30^{\circ}$ angle.

$\text{south} = 18 \cos 30^{\circ} = 15.58\ldots$

Answer: $15.6$ km
Why this scores
Mark scheme:
- B1 for $1.5$ hours
- B1 for $18$ km (ft their time)
- M1 for $18 \cos 30^{\circ}$ oe (e.g. $18 \sin 60^{\circ}$ )
- A1 for $15.6$ (awrt $15.6$ )
Why this scores full marks: The time is converted to decimal hours before multiplying, and the bearing is turned into an angle from due south. Writing 1 hour 30 minutes as $1.3$ hours gives $15.6 \times \cos 30^{\circ} = 13.5$ km and loses two marks.
Question 7
Paper 1H/2H style — 4 marks4 marks
$B$ is $10$ km due east of $A$ . A lighthouse $C$ is on a bearing of 040° from $A$ and on a bearing of 300° from $B$ .

Calculate the distance $AC$ . Give your answer correct to 3 significant figures. [4]
Model answer
Angle $CAB = 90^{\circ} - 40^{\circ} = 50^{\circ}$ ( $AB$ is due east, a bearing of 090°).

Angle $CBA = 300^{\circ} - 270^{\circ} = 30^{\circ}$ ( $BA$ is due west, a bearing of 270°).

Angle $ACB = 180^{\circ} - 50^{\circ} - 30^{\circ} = 100^{\circ}$ .

$\dfrac{AC}{\sin 30^{\circ}} = \dfrac{10}{\sin 100^{\circ}}$

$AC = \dfrac{10 \sin 30^{\circ}}{\sin 100^{\circ}} = 5.077\ldots$

Answer: $5.08$ km
Why this scores
Mark scheme:
- B1 for angles $50^{\circ}$ and $30^{\circ}$
- B1 for angle $ACB = 100^{\circ}$
- M1 for a correct sine rule statement for $AC$
- A1 for $5.08$ (awrt $5.08$ )
Why this scores full marks: Each bearing is converted to an angle inside the triangle using the east-west line $AB$ . The side $AC$ is opposite the $30^{\circ}$ angle at $B$ , and the known side $AB = 10$ is opposite the $100^{\circ}$ angle, so the sine rule pairs are matched correctly.
Question 8
Paper 1H/2H style — 4 marks4 marks
From a tower $T$ , buoy $A$ is $4$ km away on a bearing of 050° and buoy $B$ is on a bearing of 130°. The area of triangle $TAB$ is $15$ km².

Calculate the distance $TB$ . Give your answer correct to 3 significant figures. [4]
Model answer
Angle $ATB = 130^{\circ} - 50^{\circ} = 80^{\circ}$ .

$\dfrac{1}{2} \times 4 \times TB \times \sin 80^{\circ} = 15$

$TB = \dfrac{15}{2 \sin 80^{\circ}} = 7.615\ldots$

Answer: $7.62$ km
Why this scores
Mark scheme:
- B1 for angle $ATB = 80^{\circ}$
- M1 for $\dfrac{1}{2} \times 4 \times TB \times \sin 80^{\circ} = 15$
- M1 for rearranging to $TB = \dfrac{15}{2 \sin 80^{\circ}}$ oe
- A1 for $7.62$ (awrt $7.62$ )
Why this scores full marks: The angle between the two known sides is the included angle, and it comes straight from the two bearings measured at $T$ . The formula $\dfrac{1}{2}ab \sin C$ is on the Higher formula sheet; this question runs it backwards, so the script writes the equation first and only then rearranges.
Question 9
Paper 1H/2H style — 5 marks5 marks
A plane flies $20$ km from $O$ on a bearing of 070°. It then flies $12$ km on a bearing of 350°.

Calculate the bearing of the plane from $O$ now. Give your answer correct to the nearest degree. [5]
Model answer
East: $20 \sin 70^{\circ} - 12 \sin 10^{\circ} = 18.793\ldots - 2.083\ldots = 16.710\ldots$ km

North: $20 \cos 70^{\circ} + 12 \cos 10^{\circ} = 6.840\ldots + 11.817\ldots = 18.658\ldots$ km

The plane is north and east of $O$ , so the bearing is the angle from north:

$\tan \theta = \dfrac{16.710\ldots}{18.658\ldots} \quad \Rightarrow \quad \theta = 41.84\ldots^{\circ}$

Answer: 042°
Why this scores
Mark scheme:
- M1 for the total distance east, $20 \sin 70^{\circ} - 12 \sin 10^{\circ}$ oe
- M1 for the total distance north, $20 \cos 70^{\circ} + 12 \cos 10^{\circ}$ oe
- A1 for $16.7$ and $18.7$ (both awrt)
- M1 for $\tan^{-1}\left(\dfrac{\text{east}}{\text{north}}\right)$ with their values
- A1 for 042° (awrt $41.8$ rounded to 042)
Why this scores full marks: Each leg is split into east and north parts, with the second leg ( $10^{\circ}$ west of north) taking the plane back west. The final tangent uses east over north because a bearing is measured from north. The cosine rule then the sine rule is an equally valid route (the angle at the turn is $100^{\circ}$ ).
Question 10
Paper 1H/2H style — 5 marks5 marks
A ship is at $A$ . The bearing of a lighthouse $L$ from $A$ is 052°. The ship sails due east at $12$ km/h for 30 minutes to $B$ . The bearing of $L$ from $B$ is 018°.

Calculate the distance $BL$ . Give your answer correct to 3 significant figures. [5]
Model answer
$AB = 12 \times 0.5 = 6$ km.

Angle $LAB = 90^{\circ} - 52^{\circ} = 38^{\circ}$ ( $AB$ is due east).

Angle $ABL = 90^{\circ} + 18^{\circ} = 108^{\circ}$ (from due west at $B$ , round through north to $BL$ ).

Angle $ALB = 180^{\circ} - 38^{\circ} - 108^{\circ} = 34^{\circ}$ .

$\dfrac{BL}{\sin 38^{\circ}} = \dfrac{6}{\sin 34^{\circ}}$

$BL = \dfrac{6 \sin 38^{\circ}}{\sin 34^{\circ}} = 6.605\ldots$

Answer: $6.61$ km
Why this scores
Mark scheme:
- B1 for $AB = 6$ km
- B1 for angle $LAB = 38^{\circ}$
- B1 for angle $ABL = 108^{\circ}$ or angle $ALB = 34^{\circ}$
- M1 for a correct sine rule statement for $BL$
- A1 for $6.61$ (awrt $6.61$ )
Why this scores full marks: The script finds all three angles of the triangle before the sine rule, so each side can be paired with the angle opposite it: $BL$ with $38^{\circ}$ at $A$ , and $AB$ with $34^{\circ}$ at $L$ . The usual error is to take angle $ABL$ as $90^{\circ} - 18^{\circ} = 72^{\circ}$ , measuring from due east instead of from the line $BA$ , which points west.

4MA1 Bearings

At a glance

What you’ll learn

Reading bearings

Bearings with trigonometry

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Read a bearing

2Back bearing

3Bearing combined with trigonometry

4A new bearing after a right-angle turn [1 mark]

5Distance from speed and time, then a scale-drawing length [2 marks]

6The angle between two bearings from one point [2 marks]

7Deciding which of two bearings is correct [3 marks]

8A right angle from due north and due east [3 marks]

9How far apart after walking on two bearings [4 marks]

10The bearing of one moving ship from another [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10