Maths · 5 Geometry

5.2 Angle measurement in degrees

Before you start

A full turn is $360 °$ , a half turn is $180 °$ and a quarter turn (a right angle) is $90 °$ .
Reading a ruler in centimetres and millimetres, and that $1$ cm $= 10$ mm.
Using a scale such as 1 cm to 2 km: multiply to go from the drawing to real life, divide to go from real life to the drawing.
Adding and subtracting whole numbers of degrees without a calculator.

Syllabus objectives covered (2)

Measure and draw lines and angles.
Use and interpret three-figure bearings.

Part 1 of 7

Measuring and drawing a straight line

A ruler is marked in centimetres, and each centimetre is split into $10$ millimetres. So a length of $7.3$ cm is $73$ mm. Measure every length to the nearest millimetre.
Put the $0$ mark on the start of the line, not the end of the ruler. On many rulers the $0$ is a little way in from the edge, so lining up the edge adds that gap to every length you measure.
Look straight down at the mark you are reading. If you look from the side, the end of the line seems to sit against a different mark.
To draw a line of a given length, say $6.4$ cm: mark a point, put $0$ on it, make a small mark at $6.4$ , and join the two marks with the ruler. Label each end with its letter.
Use a ruler for every straight edge. The syllabus says so, and a freehand line has no exact length or direction for anyone to check.
Use a sharp pencil. A blunt pencil line can be about a millimetre wide, which is the whole size of the unit you are measuring to, so you cannot tell which edge to measure from.

Part 2 of 7

An angle is an amount of turn

An angle measures the turn between two straight lines, called arms, that meet at a point called the vertex. Longer arms do not make a bigger angle: the size depends only on the directions of the arms.
Angles are measured in degrees. A full turn is $360 °$ , a half turn is $180 °$ and a quarter turn is $90 °$ , a right angle.
Angles are named by size. Acute: less than $90 °$ . Right: exactly $90 °$ . Obtuse: between $90 °$ and $180 °$ . Straight: exactly $180 °$ . Reflex: between $180 °$ and $360 °$ .
Angle $A B C$ means the angle at $B$ , between the arms $B A$ and $B C$ . The middle letter is always the vertex.
Where two arms meet there are two angles: the turn one way round and the turn the other way round. Together they make a full turn, so if one is $130 °$ the other is $360 ° - 130 ° = 230 °$ . 'Angle $A B C$ ' means the one less than $180 °$ ; 'reflex angle $A B C$ ' means the bigger one.
graph diagram — not drawn yet
Two arms BA and BC meeting at vertex B, with arm BA pointing right and arm BC pointing up and to the left; a small arc between the arms on the inside labelled 130° and a larger arc going round the outside from BA to BC labelled 230°.
Every vertex has two angles that add up to $360 °$ ; the reflex one goes the long way round.
Before you measure any angle, decide by eye whether it is acute, obtuse or reflex. That estimate decides which protractor scale you read, and it catches a wrong reading straight away.

Part 3 of 7

Measuring an angle with a protractor

A protractor is a half circle marked from $0 °$ to $180 °$ . Its straight edge is the baseline, and the small cross or mark at the middle of the baseline is its centre.
Put the centre exactly on the vertex. Then turn the protractor so that the baseline lies along one arm.
The protractor has two scales. One starts at $0$ at the right-hand end of the baseline, the other starts at $0$ at the left-hand end. Any line through the centre crosses the two scales at two numbers that add up to $180$ .
Read the scale that shows $0$ on the arm lying along the baseline, and count up from that $0$ to where the other arm crosses.
Example: the other arm crosses the scales at $65$ and $115$ . The angle looks acute, so it is $65 °$ . The $115$ is the angle between that arm and the other half of the baseline, which is not the angle you want.
graph diagram — not drawn yet
A semicircular protractor with its centre on vertex O and its baseline along arm OA, which points to the right; arm OB rises up and to the right and crosses the scale whose 0 is at A's end at 65 and the other scale at 115; the 65 is circled.
Count from the $0$ that sits on arm $O A$ : the angle is $65 °$ , not $115 °$ .
If an arm is too short to reach the scale, extend it with a ruler first. Extending an arm does not change its direction, so it does not change the angle.
A $180 °$ protractor cannot measure a reflex angle in one go. Measure the other angle at the same vertex, which is less than $180 °$ , and subtract it from $360 °$ .
$reflex angle = 360 ° - the other angle$
Example: the other angle measures $128 °$ , so the reflex angle is $360 ° - 128 ° = 232 °$ .

Part 4 of 7

Drawing an angle and drawing a triangle

To draw an angle of $52 °$ at $A$ on a line $A B$ : draw $A B$ with the ruler, put the centre on $A$ and the baseline along $A B$ . Find the $0$ that lies on $A B$ , count up to $52$ on that scale and make a small dot. Take the protractor away and join $A$ to the dot with the ruler.
Check by eye. $52 °$ is acute, a little more than half a right angle. If your angle looks obtuse, you counted from the wrong $0$ and drew $128 °$ .
To draw a reflex angle such as $250 °$ , draw the other angle, $360 ° - 250 ° = 110 °$ , then mark the reflex side with an arc and label it $250 °$ , so it is clear which of the two angles you mean.
A triangle from two sides and the angle between them, e.g. $A B = 8$ cm, $A C = 5$ cm and angle $B A C = 50 °$ : draw $A B$ , draw the $50 °$ angle at $A$ , measure $5$ cm along the new arm from $A$ and mark $C$ , then join $B$ to $C$ with the ruler.
A triangle from one side and the angles at both ends: draw the side, draw each angle at its own end, and extend both new arms until they cross. The crossing point is the third vertex.
Leave your arms and marks on the page and label the vertices. When a question then asks you to find a side or an angle by measuring, measure your own drawing.
The marking rules give no marks for a scale drawing where a calculation is required. A measured value earns marks only when the question asks you to draw and measure.

Part 5 of 7

What a three-figure bearing is

A bearing gives a direction as an angle. It is measured from north, turning clockwise, at the point you start from.
A bearing is written with three figures before any decimal point. A bearing of $45 °$ is written $045 °$ and a bearing of $7 °$ is written $007 °$ . A bearing of $170 °$ already has three figures.
The four compass points as bearings: north is $000 °$ , east is $090 °$ , south is $180 °$ and west is $270 °$ . A full turn of $360 °$ brings you back to north, so $360 °$ points the same way as $000 °$ .
'Due' means exactly. 'Point $D$ is due east of point $C$ ' means the bearing of $D$ from $C$ is $090 °$ .
Read the words carefully. 'The bearing of $B$ from $A$ ': the point after 'from' is where you stand. Put the north line at $A$ , face north, and turn clockwise until you face $B$ . The angle you turn through is the bearing.
graph diagram — not drawn yet
Point A with a vertical north line drawn up from it and labelled N; point B up and to the right of A, joined to A by a straight line; a clockwise arc at A from the north line to AB labelled 062°.
The north line and the arc are at $A$ , the point after 'from'; the turn is clockwise.
The size of a bearing tells you its rough direction, which is a quick check. Between $000 °$ and $090 °$ points between north and east; between $090 °$ and $180 °$ , between east and south; between $180 °$ and $270 °$ , between south and west; between $270 °$ and $360 °$ , between west and north.

Part 6 of 7

Measuring and drawing bearings, to scale

First draw a north line at the 'from' point, with a ruler and parallel to the north arrow on the diagram. On a map or a scale drawing, the north lines at different points are all parallel.
For a bearing up to $180 °$ : centre on the point, baseline along the north line, and read the scale that has $0$ on the north line. Measure clockwise from north.
For a bearing over $180 °$ : measure the anticlockwise angle from north round to the line, then subtract it from $360 °$ . If that angle is $70 °$ , the bearing is $290 °$ .
$360 ° - 70 ° = 290 °$
graph diagram — not drawn yet
Point P with a vertical north line labelled N; line PQ going up and to the left of P; a small anticlockwise arc from the north line to PQ labelled 70°, and a large clockwise arc from the north line all the way round to PQ labelled 290°.
The small angle is easy to measure; the bearing is the clockwise turn, $360 ° - 70 °$ .
Another way for a bearing over $180 °$ : measure clockwise from south and add $180 °$ , because south is already $180 °$ round from north. A line $35 °$ clockwise past south is on a bearing of $180 ° + 35 ° = 215 °$ .
In a scale drawing the scale changes lengths only, never angles. With 1 cm to 2 km, a $12$ km journey is drawn $12 \div 2 = 6$ cm long, and its bearing is drawn exactly as given.
For a journey with two legs, measure the second bearing from a new north line drawn at the turning point. Do not measure it from the first north line or from the line you arrived along.
graph diagram — not drawn yet
A scale drawing at 1 cm to 2 km: start point S with a north line, a 4 cm line on a bearing of 070° to point P with the 70° arc marked at S, a second north line drawn at P parallel to the first, and a 3 cm line on a bearing of 160° from P to point F with the 160° clockwise arc marked at P.
Each leg gets its own north line at its own starting point.

Part 7 of 7

Turning a bearing round, and angles from bearings

The bearing of $A$ from $B$ is the direction you face at $B$ looking back at $A$ . That is exactly the opposite way to facing $B$ from $A$ , and opposite directions are a half turn, $180 °$ , apart.
So: if the bearing is less than $180 °$ , add $180 °$ . If it is $180 °$ or more, subtract $180 °$ . Either way the answer stays between $000 °$ and $360 °$ .
Example: the bearing of $B$ from $A$ is $025 °$ , so the bearing of $A$ from $B$ is $025 ° + 180 ° = 205 °$ . Check by going back: $205 ° - 180 ° = 025 °$ .
graph diagram — not drawn yet
Points A (lower left) and B (upper right) joined by a straight line, with a parallel vertical north line at each; at A a clockwise arc from north to AB labelled 025°; at B a clockwise arc from north round through south to BA labelled 205°, with the part from north to south labelled 180° and the extra part past south labelled 25°.
The north lines are parallel, so the turn at $B$ is a half turn plus the same $25 °$ .
Two bearings measured at the same point give the angle between the two directions: subtract one from the other. If the bearing of $Q$ from $P$ is $040 °$ and the bearing of $R$ from $P$ is $110 °$ , then angle $Q P R = 110 ° - 40 ° = 70 °$ .
If the bearings are measured at different points, turn one round first so that both start at the same point. The bearing of $B$ from $A$ is $070 °$ and the bearing of $C$ from $B$ is $160 °$ . At $B$ , the bearing of $A$ is $070 ° + 180 ° = 250 °$ , so angle $A B C = 250 ° - 160 ° = 90 °$ .
If a bearing comes out of a calculation as a decimal, the accuracy rule for angles applies: give it to $1$ decimal place, still with three figures before the point, e.g. $061.9 °$ .

Worked examples

Example 1. You put the centre of a protractor on the vertex $Q$ of angle $P Q R$ , with the baseline along the arm $Q P$ . The arm $Q P$ points to $0$ on the outer scale. The arm $Q R$ crosses the protractor at $48$ on the inner scale and $132$ on the outer scale. By eye, angle $P Q R$ is bigger than a right angle. Find (a) angle $P Q R$ (b) the reflex angle $P Q R$ .

Show the working

Choose the scale from the arm on the baseline. $Q P$ points to $0$ on the outer scale, so count on the outer scale. On that scale, $Q R$ crosses at $132$ .
Check with the estimate. $132 °$ is between $90 °$ and $180 °$ , so it is obtuse, which matches 'bigger than a right angle'. The $48$ on the inner scale is the angle between $Q R$ and the other half of the baseline.
(b) The two angles at $Q$ make a full turn, so subtract from $360 °$ .
$360 ° - 132 ° = 228 °$

Answer: (a) $132 °$ (b) $228 °$

Example 2. Draw triangle $A B C$ with $A B = 7$ cm, angle $C A B = 40 °$ and angle $C B A = 65 °$ . Then find the length of $A C$ by measuring.

Show the working

Start with the side you are given, because both angles sit on its ends. Draw $A B = 7$ cm with the ruler and label $A$ and $B$ .
At $A$ : centre on $A$ , baseline along $A B$ . Count from the $0$ that lies on $A B$ up to $40$ , make a dot, and rule a long line from $A$ through it. $40 °$ should look a little less than half a right angle.
At $B$ : centre on $B$ , baseline along $B A$ . The arm $B A$ points the opposite way to $A B$ , so the $0$ you need is at the other end of the baseline, on the other scale. Count up to $65$ , make a dot and rule a line from $B$ through it.
The two new lines cross at $C$ . Label it and leave all the lines on the page.
Check the drawing before you use it. The angles of a triangle add up to $180 °$ , so angle $A C B$ should be $180 ° - 40 ° - 65 ° = 75 °$ . If you measure about $75 °$ at $C$ , the drawing is accurate.
Put $0$ of the ruler on $A$ and read the length of $A C$ to the nearest millimetre.

Answer: $A C = 6.6$ cm (a careful drawing gives a value within about a millimetre of this).

Example 3. The bearing of $Q$ from $P$ is $238 °$ . Point $R$ is due north of $P$ , and point $S$ is due east of $R$ . Find (a) the bearing of $P$ from $Q$ (b) the bearing of $P$ from $R$ (c) the bearing of $R$ from $S$ .

Show the working

(a) Turn the bearing round. $238 °$ is $180 °$ or more, so subtract $180 °$ , then write the answer with three figures.
$238 ° - 180 ° = 58 °, written 058 °$
Sense-check: $238 °$ is between $180 °$ and $270 °$ , so $Q$ is between south and west of $P$ . Then $P$ must be between north and east of $Q$ , and $058 °$ is.
(b) $R$ is due north of $P$ , so the bearing of $R$ from $P$ is $000 °$ . Turn it round: $000 ° + 180 ° = 180 °$ . $P$ is due south of $R$ .
(c) $S$ is due east of $R$ , so the bearing of $S$ from $R$ is $090 °$ . Turn it round: $090 ° + 180 ° = 270 °$ . $R$ is due west of $S$ .

Answer: (a) $058 °$ (b) $180 °$ (c) $270 °$

Example 4. A plane flies $300$ km from airport $A$ on a bearing of $110 °$ to $B$ . It then flies $400$ km on a bearing of $200 °$ to $C$ . Make a scale drawing with a scale of 1 cm to 50 km, and use it to find the distance of $C$ from $A$ and the bearing of $C$ from $A$ .

Show the working

Convert the lengths, not the angles. Divide each distance by $50$ to get centimetres.
$300 \div 50 = 6 cm, 400 \div 50 = 8 cm$
Mark $A$ and rule a north line up from it. $110 °$ is between $090 °$ and $180 °$ , so $B$ is between east and south of $A$ . Measure $110 °$ clockwise from north at $A$ and draw $A B = 6$ cm.
Rule a new north line at $B$ , parallel to the first. $200 °$ is more than $180 °$ , so measure $200 ° - 180 ° = 20 °$ clockwise past south (or $360 ° - 200 ° = 160 °$ anticlockwise from north). Draw $B C = 8$ cm.
Check the shape. At $B$ , the bearing of $A$ is $110 ° + 180 ° = 290 °$ and the bearing of $C$ is $200 °$ , so angle $A B C = 290 ° - 200 ° = 90 °$ . Your drawing should show a right angle at $B$ .
Join $A$ to $C$ and measure it: $10.0$ cm. Multiply by the scale to get the real distance.
$10.0 \times 50 = 500 km$
For the bearing of $C$ from $A$ , use the north line at $A$ and measure clockwise from north to $A C$ . You should read $163 °$ , which is between east and south, as $C$ is on the drawing.

Answer: $A C$ measures $10.0$ cm, so $C$ is $500$ km from $A$ , on a bearing of $163 °$ .

Where people go wrong

Mistake: Reading the wrong protractor scale, e.g. giving $115 °$ for an angle that is clearly acute.

Instead: Read the scale that shows $0$ on the arm along the baseline, then check the answer against your acute, obtuse or reflex estimate.

Both scales cross the second arm. The other scale gives $180 °$ minus the angle you want, which is the angle with the other half of the baseline.

Mistake: Measuring a bearing anticlockwise, or putting the north line at the wrong point.

Instead: For 'the bearing of $B$ from $A$ ', put the north line at $A$ , the point after 'from', and turn clockwise to $B$ .

A bearing is defined as clockwise from north at the starting point. Measured at $B$ instead, you get the bearing of $A$ from $B$ , which differs by $180 °$ .

Mistake: Writing the bearing $045 °$ as $45 °$ .

Instead: Write it with three figures: $045 °$ .

The objective names three-figure bearings: every bearing is written with three figures before any decimal point, so $45 °$ is not in that form.

Mistake: Turning a bearing round the wrong way, e.g. $025 ° - 180 ° = - 155 °$ , or $238 ° + 180 ° = 418 °$ .

Instead: Add $180 °$ to a bearing under $180 °$ , and subtract $180 °$ from one of $180 °$ or more: $025 ° + 180 ° = 205 °$ and $238 ° - 180 ° = 058 °$ .

Every bearing is between $000 °$ and $360 °$ . A negative answer or one over $360 °$ is a sign you chose the wrong one of add and subtract.

Mistake: Measuring the second bearing of a journey from the first north line, or from the line you arrived along.

Instead: Rule a new north line at the turning point, parallel to the first, and measure clockwise from it.

A bearing is measured at the point you start from, and each leg of a journey starts from a different point.

Check yourself

Write $52 °$ , $8 °$ and $170 °$ as three-figure bearings.
Answer
$052 °$ , $008 °$ and $170 °$ .
The baseline of a protractor lies along one arm of an angle, and that arm points to $0$ on the inner scale. The other arm crosses $35$ on the inner scale and $145$ on the outer scale. The angle looks acute. How big is it?
Answer
$35 °$ . Read the inner scale, because its $0$ is on the baseline arm, and $35 °$ is acute as expected.
You need to draw a reflex angle of $205 °$ with a $180 °$ protractor. What angle do you actually draw with the protractor?
Answer
$360 ° - 205 ° = 155 °$ . Then mark the reflex side with an arc and label it $205 °$ .
On a drawing, the line from $P$ to $Q$ is $40 °$ anticlockwise from the north line at $P$ . What is the bearing of $Q$ from $P$ ?
Answer
$360 ° - 40 ° = 320 °$ .
The bearing of $Y$ from $X$ is $312 °$ . Work out the bearing of $X$ from $Y$ .
Answer
$312 °$ is $180 °$ or more, so subtract: $312 ° - 180 ° = 132 °$ .
Point $D$ is due east of point $C$ . Write down the bearing of $D$ from $C$ and the bearing of $C$ from $D$ .
Answer
$090 °$ and $270 °$ .
The bearing of $B$ from $A$ is $065 °$ and the bearing of $C$ from $A$ is $150 °$ . Work out angle $B A C$ .
Answer
Both bearings are measured at $A$ , so subtract: $150 ° - 65 ° = 85 °$ .
A scale drawing uses 1 cm to 5 km. How long is the line for a distance of $18$ km, and what real distance does a line of $6.4$ cm represent?
Answer
$18 \div 5 = 3.6$ cm, and $6.4 \times 5 = 32$ km.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. You need a ruler and a protractor for this question.

(a) Draw triangle $A B C$ with $A B = 7.5$ cm, angle $B A C = 64 °$ and $A C = 6$ cm. Label each vertex. 3 marks No calculator

Mark scheme

B1 AB ruled, 7.5 cm, accept 7.4 to 7.6 cm
B1 angle BAC = 64°, accept 62° to 66°
B1 AC = 6 cm, accept 5.9 to 6.1 cm, with B joined to C by a ruled line and the vertices labelled

(b) By measuring your drawing, find the length of $B C$ . 1 mark No calculator

Mark scheme

B1 7.3 [cm], accept 7.1 to 7.5

(c) By measuring your drawing, find the size of angle $A B C$ . 1 mark No calculator

Mark scheme

B1 48°, accept 46° to 50°

Question 2. Three buoys $A$ , $B$ and $C$ float in a bay. $B$ is due east of $A$ . The bearing of $C$ from $A$ is $140 °$ . The bearing of $C$ from $B$ is $215 °$ .

(a) Write down the bearing of $B$ from $A$ . 1 mark No calculator

Mark scheme

B1 090° cao

(b) Write down the bearing of $A$ from $B$ . 1 mark No calculator

Mark scheme

B1 270° cao

(c) Work out the bearing of $A$ from $C$ . 2 marks No calculator

Mark scheme

M1 140 + 180 oe
A1 320°

(d) Work out the size of angle $A B C$ . 2 marks No calculator

Mark scheme

M1 270 − 215 oe, FT their (b)
A1 55°

Question 3. A boat sails from a harbour $H$ for $12$ km on a bearing of $040 °$ to a point $P$ . It then sails $10$ km on a bearing of $125 °$ to a point $Q$ . North is straight up the page.

(a) Using a scale of 1 cm to represent 2 km, draw an accurate scale drawing of the boat's journey. Show a north line at each point where you measure a bearing. 4 marks No calculator

Mark scheme

B1 HP = 6 cm, accept 5.9 to 6.1 cm
B1 bearing of P from H = 040°, accept 038° to 042°
B1 PQ = 5 cm, accept 4.9 to 5.1 cm
B1 bearing of Q from P = 125°, accept 123° to 127°, measured from a north line drawn at P

(b) By measuring your drawing, find the distance $H Q$ in kilometres. 2 marks No calculator

Mark scheme

M1 their measured HQ in cm × 2
A1 16.3 [km], accept 15.8 to 16.8

(c) By measuring your drawing, find the bearing of $Q$ from $H$ . 1 mark No calculator

Mark scheme

B1 078°, accept 076° to 080°

(d) Write down the bearing of $H$ from $Q$ . 1 mark No calculator

Mark scheme

B1 258°, FT their (c) + 180

How this is examined

Drawing a two-leg journey to scale from distances and bearings: A Draw part worth about 4 marks: two legs, each given as a distance and a bearing, with a scale such as 1 cm to 2 km Each leg the right length to scale ( $14$ km at 1 cm to 2 km is $7$ cm), and each bearing measured clockwise from a north line drawn at the start of that leg. The second bearing needs its own north line at the turning point.
Drawing a triangle with ruler and protractor, then measuring a side or angle you were not given: A Draw part for the triangle, followed by 1-mark Find parts that ask you to measure your own drawing Ruled lines, each given length to the nearest millimetre and each given angle to the nearest degree, the triangle closed and the vertices labelled. The measured value is then read from your own drawing, with its unit.
Reading a distance and a bearing back off a scale drawing: Find parts after a scale drawing: the real distance between two points and the bearing of one from the other The measured length converted with the scale and given a unit ( $8.1$ cm $\times 2 = 16.2$ km). The bearing measured at the point after 'from', clockwise from north, written with three figures.
Turning a bearing round: the bearing of A from B when you know the bearing of B from A: A 1- or 2-mark Write down or Work out, with the first bearing given in the stem $180 °$ added to a bearing under $180 °$ , or subtracted from one of $180 °$ or more, with the answer in three figures: $025 °$ gives $205 °$ .
An angle inside a triangle found from bearings: A 2-mark Work out of an angle such as angle $P Q R$ , when bearings at the corners are given Both directions written as bearings at the same point (from $Q$ , $P$ is on $300 °$ and $R$ is on $235 °$ ), then their difference: $300 ° - 235 ° = 65 °$ . A bearing given at another point is turned round first.
Due north, east, south and west as bearings: A 1-mark Write down, e.g. the bearing of $D$ from $C$ when $D$ is due east of $C$ The bearing in three figures: due north $000 °$ , due east $090 °$ , due south $180 °$ , due west $270 °$ .

Words to know

Angle: The amount of turn between two arms that meet at a point, measured in degrees. A full turn is $360 °$ .
Vertex: The point where the two arms of an angle meet. In angle $A B C$ it is the middle letter, $B$ .
Acute, obtuse and reflex angles: Acute: less than $90 °$ . Obtuse: between $90 °$ and $180 °$ . Reflex: between $180 °$ and $360 °$ .
Right angle: A quarter turn, exactly $90 °$ .
Protractor: A half-circle instrument marked from $0 °$ to $180 °$ on two scales, used to measure and draw angles. Its centre goes on the vertex and its baseline along one arm.
Bearing: A direction given as an angle measured clockwise from north at the starting point, from $000 °$ to $360 °$ .
Three-figure bearing: A bearing written with three figures before any decimal point, e.g. $045 °$ rather than $45 °$ .
Due: Exactly in that direction. Due east is a bearing of $090 °$ .
North line: A ruled line pointing north, drawn at the point a bearing is measured from. North lines at different points are parallel.
Scale drawing: An accurate drawing in which every length is the real length multiplied by the same scale, e.g. 1 cm to 2 km. Angles are drawn at their real size.

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