What is G3: Angle facts?

Angles at a point total 360°, angles on a straight line total 180°, and vertically opposite angles are equal. Where a line crosses two parallel lines, corresponding angles are equal, alternate angles are equal, and co-interior (allied) angles add up to 180°.

How many marks are G3 questions worth?

In verified past-paper appearances, G3 questions on Edexcel GCSE Maths 1MA1 have been worth 1–5 marks.

Which past exam papers has G3 appeared in?

G3 has appeared in 39 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Apply angles at a point, on a straight line, vertically opposite angles; use alternate and corresponding angles on parallel lines; derive and use the angle sum of a triangle and of any polygon

Angle facts

Worked answers, methods and verified real exam appearances for G3 on Edexcel GCSE Maths 1MA1.

Definition

Specification G3: apply the properties of angles at a point, angles at a point on a straight line and vertically opposite angles; understand and use alternate and corresponding angles on parallel lines; derive and use the sum of angles in a triangle, and use it to deduce the angle sum in any polygon and the properties of regular polygons.

Explanation

Angles at a point total $360^\circ$ , angles on a straight line total $180^\circ$ , and vertically opposite angles are equal. Where a line crosses two parallel lines, corresponding angles are equal, alternate angles are equal, and co-interior (allied) angles add up to $180^\circ$ .
Angles in a triangle add up to $180^\circ$ , and an exterior angle of a triangle equals the sum of the two interior opposite angles.
Splitting an $n$ -sided polygon into $n-2$ triangles gives an interior-angle sum of $(n-2)\times180^\circ$ ; the exterior angles of any polygon total $360^\circ$ , so a regular polygon has exterior angle $360^\circ\div n$ .
In a reasoning question each step needs a calculation and the reason for it, with the key words in the reason and the reason tied to the parallel lines or triangle actually used.
Angles found should be marked on the diagram or named.

Alternate angles are equal when a transversal crosses parallel lines.

Worked example

Each exterior angle of a regular polygon is

24^\circ

. Work out the number of sides of the polygon and the size of each interior angle.

(3)

(Total for Question 1 is 3 marks)

Working

1.The exterior angles of any polygon add up to $360^\circ$ , so the number of sides is $360\div24=15$ .
2.An interior angle and its exterior angle lie on a straight line, so they add up to $180^\circ$ .
3.Each interior angle is $180-24=156^\circ$ .

Answer: $15$ sides; each interior angle is $156^\circ$ .

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for $360\div24$	M1	oe
	for $15$	A1	cao
	for $156$	B1	ft $180-24$

Common mistakes

Don't claim alternate or corresponding angles are equal without establishing that the two lines are parallel.
Don't use $n\times180^\circ$ instead of $(n-2)\times180^\circ$ for an interior-angle sum.

Mark-scheme wording to use

“Corresponding angles are equal … Allied angles / Co-interior angles add up to 180 … Angles on a straight line add up to 180 … Angles in a triangle add up to 180” — Pearson 1MA1, June 2020 Paper 1H, Q6
“When reasons are given the key words underlined must be present. Reasons need to be linked to their method; any reasons not linked, do not credit. There should be no incorrect reasons given” — Pearson 1MA1, June 2020 Paper 1H, Q6
“M1 for finding an exterior angle eg 360 ÷ 6 (= 60) or 360 ÷ 5 (= 72) or an interior angle eg 180 × 4 ÷ 6 (= 120) or 180 × 3 ÷ 5 (= 108) … M1 for a complete method eg 360 – “120” – “108” or “60” + ”72” A1 cao Answer only award no marks” — Pearson 1MA1, June 2022 Paper 1H, Q5

What the examiner wants

The correct name for the fact used. The November 2024 Paper 1H report says the reasoning mark was for an appropriate parallel-lines reason, usually “alternate angles”, and that “corresponding angles” was a common wrong reason for the same angle; the June 2024 Paper 2H report warns that showing the angles sum to $360^\circ$ does not show a shape is a trapezium.

Exam tip

A 'give a reason' angle question needs the named fact, such as 'alternate angles, parallel lines', not just the arithmetic.

Worked practice

Tier 1 · Easy

Two adjacent angles on a straight line are

63^\circ

and

x^\circ

. Work out

x

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$117^\circ$	1	Angles on a straight line total $180^\circ$ , so $x=180^\circ-63^\circ=117^\circ$ .

Tier 2 · Standard

Two parallel lines are crossed by a transversal. A pair of alternate angles are

(4x+7)^\circ

and

(7x-38)^\circ

. Find

x

and the size of these angles.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$x=15$ $67^\circ$	3	Alternate angles between parallel lines are equal, so $4x+7=7x-38$ . Hence $45=3x$ and $x=15$ . Substitution gives $4(15)+7=67^\circ$ .

Tier 3 · Hard

A polygon has

13

sides. Twelve of its interior angles are each

155^\circ

. Work out the remaining interior angle.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$120^\circ$	4	The interior-angle sum is $(13-2)\times180^\circ=1980^\circ$ . The twelve known angles total $12\times155^\circ=1860^\circ$ . The remaining angle is $1980^\circ-1860^\circ=120^\circ$ .

Tier 1 · Easy

Two straight lines cross. One angle at the intersection is

38^\circ

. Write down the vertically opposite angle.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$38^\circ$	1	Vertically opposite angles are equal, so the required angle is $38^\circ$ .

Tier 2 · Standard

An exterior angle of a triangle is

126^\circ

. One of the two opposite interior angles is

49^\circ

. Work out the other opposite interior angle.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$77^\circ$	2	An exterior angle equals the sum of the two opposite interior angles. Therefore the missing angle is $126^\circ-49^\circ=77^\circ$ .

Tier 3 · Hard

A straight line

DAE

passes through vertex

A

of triangle

ABC

, with

DE\parallel BC

. Given

\angle DAB=68^\circ

and

\angle CAE=47^\circ

, work out all three angles of triangle

ABC

. For each stage of your working, give a reason.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\angle ABC=68^\circ$ (alternate angles, $DE\parallel BC$ ) $\angle ACB=47^\circ$ (alternate angles, $DE\parallel BC$ ) $\angle BAC=65^\circ$ (angles in a triangle total $180^\circ$ )	4	Since $DE\parallel BC$ , $\angle ABC=68^\circ$ and $\angle ACB=47^\circ$ by alternate angles. Angles in a triangle total $180^\circ$ , so $\angle BAC=180^\circ-68^\circ-47^\circ=65^\circ$ .

Tier 2 · Standard

Three angles at a point are

(2x+10)^\circ

(x+35)^\circ

and

135^\circ

. Work out

x

and the size of the smallest angle.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$x=60$ ; the smallest angle is $95^\circ$	3	Angles at a point total $360^\circ$ , so $2x+10+x+35+135=360$ . This gives $3x=180$ and $x=60$ . The three angles are $130^\circ$ , $95^\circ$ and $135^\circ$ , so the smallest is $95^\circ$ .

Tier 3 · Hard

The five exterior angles of a convex pentagon are

54^\circ

68^\circ

70^\circ

x^\circ

and

2x^\circ

. Work out

x

and the largest interior angle of the pentagon.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$x=56$ ; the largest interior angle is $126^\circ$	4	Exterior angles total $360^\circ$ , so $54+68+70+x+2x=360$ . Hence $3x=168$ and $x=56$ . The largest interior angle is paired with the smallest exterior angle, $54^\circ$ , so it is $180^\circ-54^\circ=126^\circ$ .

Tier 3 · Hard

Rays

OA

OB

OC

and

OD

occur in that clockwise order.

AOC

is a straight line. Angles

AOB

and

BOC

are

(3x+7)^\circ

and

(5x-3)^\circ

. The reflex angle

BOD

270^\circ

. Work out

x

and the smaller angle

COD

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$x=22$ ; $\angle COD=163^\circ$	4	Angles $AOB$ and $BOC$ make the straight angle $AOC$ , so $(3x+7)+(5x-3)=180$ . This gives $8x+4=180$ and $x=22$ . Then $\angle BOC=107^\circ$ . The clockwise reflex angle from $OB$ to $OD$ passes through $OC$ , so $\angle COD=270-107=163^\circ$ . The stated ray order fixes this as the unique smaller angle.

Q10

Tier 3 · Hard

The vertices of convex trapezium

ABCD

are in order, with

AB\parallel CD

. Angle

DAB=(4x+9)^\circ

and angle

CDA=(7x-5)^\circ

. Angle

ABC

49^\circ

larger than angle

DAB

. Work out

x

and all four interior angles of the trapezium.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$x=16$ ; $\angle DAB=73^\circ$ , $\angle ABC=122^\circ$ , $\angle BCD=58^\circ$ , $\angle CDA=107^\circ$	4	Co-interior angles on transversal $AD$ total $180^\circ$ , so $(4x+9)+(7x-5)=180$ . Hence $11x+4=180$ and $x=16$ . Therefore $\angle DAB=73^\circ$ and $\angle CDA=107^\circ$ . The given difference makes $\angle ABC=122^\circ$ . Co-interior angles on $BC$ then give $\angle BCD=180-122=58^\circ$ . For example, $A=(0,0)$ , $B=(10,0)$ , $D=(4\cot73^\circ,4)$ and $C=(10+4\cot58^\circ,4)$ give a convex trapezium with exactly these angles; the linear equation and co-interior pairs fix the four angle values uniquely.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	3F	Q20	2	Allowed	Foundation	QP MS
2024-06	3F	Q20	5	Allowed	Foundation	QP MS
2023-06	1H	Q10	4	Non-calculator	Higher	QP MS
2022-11	1F	Q9	2	Non-calculator	Foundation	QP MS
2019-11	1F	Q24	4	Non-calculator	Foundation	QP MS
2024-11	1F	Q23	4	Non-calculator	Foundation	QP MS
2021-11	2F	Q13	3	Allowed	Foundation	QP MS
2024-11	1H	Q6	4	Non-calculator	Higher	QP MS
2023-11	1F	Q23	4	Non-calculator	Foundation	QP MS
2024-06	1F	Q8	3	Non-calculator	Foundation	QP MS
2022-06	2F	Q11	3	Allowed	Foundation	QP MS
2024-11	2F	Q13	4	Allowed	Foundation	QP MS
2022-06	3F	Q20	5	Allowed	Foundation	QP MS
2023-11	1H	Q6	4	Non-calculator	Higher	QP MS
2023-06	2H	Q13	4	Allowed	Higher	QP MS
2024-11	1H	Q4	3	Non-calculator	Higher	QP MS
2019-06	1F	Q12	5	Non-calculator	Foundation	QP MS
2019-11	3H	Q8	4	Allowed	Higher	QP MS
2019-06	3H	Q5	4	Allowed	Higher	QP MS
2019-11	2F	Q17	4	Allowed	Foundation	QP MS
2019-11	3F	Q29	4	Allowed	Foundation	QP MS
2019-11	1F	Q9	2	Non-calculator	Foundation	QP MS
2019-11	1F	Q28	3	Non-calculator	Foundation	QP MS
2022-06	1H	Q5	3	Non-calculator	Higher	QP MS
2019-11	1H	Q5	4	Non-calculator	Higher	QP MS
2021-11	3F	Q15	1	Allowed	Foundation	QP MS
2022-06	1F	Q27	3	Non-calculator	Foundation	QP MS
2022-11	3H	Q26	5	Allowed	Higher	QP MS
2022-11	3F	Q13	3	Allowed	Foundation	QP MS
2024-06	2H	Q7	4	Allowed	Higher	QP MS
2024-06	2F	Q26	4	Allowed	Foundation	QP MS
2023-06	1F	Q8	3	Non-calculator	Foundation	QP MS
2024-11	1F	Q21	3	Non-calculator	Foundation	QP MS
2023-11	2F	Q19	4	Allowed	Foundation	QP MS
2023-06	3F	Q22	5	Allowed	Foundation	QP MS
2024-11	2H	Q9	3	Allowed	Higher	QP MS
2022-11	3H	Q16	4	Allowed	Higher	QP MS
2023-06	3H	Q3	5	Allowed	Higher	QP MS
2019-06	3F	Q28	4	Allowed	Foundation	QP MS

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