What is G3: Angle facts?

Use angle facts as a chain of justified steps. Angles at a point total 360°, angles on a straight line total 180°, and vertically opposite angles are equal.

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.

Explanation

Use angle facts as a chain of justified steps. Angles at a point total $360^\circ$ , angles on a straight line total $180^\circ$ , and vertically opposite angles are equal.
When a transversal crosses parallel lines, corresponding and alternate angles are equal; interior angles on the same side total $180^\circ$ .
A triangle totals $180^\circ$ .
Drawing diagonals from one vertex divides an $n$ -gon into $n-2$ triangles, so its interior-angle sum is $(n-2)\times180^\circ$ .
In exam working, write the relevant equality or subtraction and name the angle fact, including the parallel lines that make it valid.

Alternate angles are equal when a transversal crosses parallel lines.

Worked example

A regular polygon has each exterior angle equal to $24^\circ$ . Find its number of sides and each interior angle.

1.Exterior angles of any polygon total $360^\circ$ , so $n=360\div24=15$ .
2.An interior and exterior angle on a straight line total $180^\circ$ .
3.Interior angle $=180^\circ-24^\circ=156^\circ$ .

Answer: The polygon has $15$ sides and each interior angle is $156^\circ$ .

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.

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

Other points in G Geometry and measures

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