What is G20: Pythagoras and trigonometry?

In a right-angled triangle, Pythagoras gives a^2+b^2=c^2, with c opposite the right angle. Relative to angle θ, sinθ=opposite/hypotenuse, cosθ=adjacent/hypotenuse and tanθ=opposite/adjacent.

How many marks are G20 questions worth?

In verified past-paper appearances, G20 questions on Edexcel GCSE Maths 1MA1 have been worth 2–6 marks.

Which past exam papers has G20 appeared in?

G20 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.

G20

Know Pythagoras' theorem a² + b² = c² and the trigonometric ratios sin, cos and tan; apply them to find angles and lengths in right-angled and, where possible, general triangles in 2D and 3D figures

Pythagoras and trigonometry

Worked answers, methods and verified real exam appearances for G20 on Edexcel GCSE Maths 1MA1.

Explanation

In a right-angled triangle, Pythagoras gives $a^2+b^2=c^2$ , with $c$ opposite the right angle. Relative to angle $\theta$ , $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ and $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ .
Label the sides first, choose the relationship containing the known and required values, then rearrange. Use an inverse trigonometric function for an angle.
Foundation questions use right-angled triangles in two dimensions.
Higher tier: identify an appropriate right-angled cross-section in three dimensions or use the later general-triangle rules.
Examiners expect a diagram, substitution, correct units, sensible rounding and a clear statement of what the result represents.

Opposite, adjacent and hypotenuse are labelled relative to the chosen angle.

Worked example

A ladder is $8$ m long and its foot is $3$ m from a vertical wall. Find the height reached and the angle with the ground, both to $1$ decimal place.

1.The ladder is the hypotenuse, so $h=\sqrt{8^2-3^2}=\sqrt{55}=7.416\ldots$ m.
2.For ground angle $\theta$ , $\cos\theta=3/8$ .
3. $\theta=\cos^{-1}(3/8)=67.975\ldots^\circ$ ; round only at the end.

Answer: Height $=7.4$ m; angle $=68.0^\circ$ .

Common mistakes

Don't label the side opposite the right angle as adjacent instead of hypotenuse.
Don't use ordinary sine or cosine when an inverse function is needed to find an angle.

Exam tip

For a multi-step triangle question, keep unrounded calculator values for later steps and round only the final answers.

Worked practice

Tier 1 · Easy

A right-angled triangle has perpendicular sides

6\text{ cm}

and

8\text{ cm}

. Work out the hypotenuse.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$10\text{ cm}$	2	By Pythagoras, $c^2=6^2+8^2=36+64=100$ . Since a length is positive, $c=\sqrt{100}=10\text{ cm}$ .

Tier 2 · Standard

A straight ladder of length

7.2\text{ m}

rests against a vertical wall. Its foot is

2.1\text{ m}

from the wall. Work out the height reached by the ladder and the angle it makes with the ground. Give each answer to

1

decimal place.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Height $=6.9\text{ m}$ Angle $=73.0^\circ$	4	The ladder is the hypotenuse, so the height is $\sqrt{7.2^2-2.1^2}=6.886\ldots\text{ m}=6.9\text{ m}$ . If the ground angle is $\theta$ , then $\cos\theta=2.1/7.2$ . Hence $\theta=\cos^{-1}(2.1/7.2)=73.0^\circ$ to $1$ decimal place.

Tier 3 · Hard

Isosceles triangle

ABC

has

AB=AC=13\text{ cm}

and

BC=10\text{ cm}

. Work out the perpendicular height from

A

BC

and angle

ABC

. Give the angle to

1

decimal place.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Height $=12\text{ cm}$ $\angle ABC=67.4^\circ$	4	The perpendicular from $A$ bisects the $10\text{ cm}$ base, giving a right-angled triangle with hypotenuse $13\text{ cm}$ and base $5\text{ cm}$ . Its height is $\sqrt{13^2-5^2}=12\text{ cm}$ . For $\angle ABC=\theta$ , $\tan\theta=12/5$ , so $\theta=67.4^\circ$ to $1$ decimal place.

Tier 1 · Easy

A right-angled triangle has hypotenuse

17\text{ cm}

and one shorter side

15\text{ cm}

. Work out the length of the other shorter side.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$8\text{ cm}$	2	By Pythagoras, the missing length is $\sqrt{17^2-15^2}=\sqrt{289-225}=\sqrt{64}=8\text{ cm}$ .

Tier 2 · Standard

In a right-angled triangle, the side opposite angle

x

7\text{ cm}

and the side adjacent to

x

24\text{ cm}

. Work out

x

. Give the value of

x

correct to

1

decimal place.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$16.3^\circ$	2	$\tan x=\frac{7}{24}$ , so $x=\tan^{-1}(7/24)=16.260\ldots^\circ$ . To $1$ decimal place, $x=16.3^\circ$ .

Tier 3 · Hard

A cuboid measures

6\text{ cm}

9\text{ cm}

14\text{ cm}

. Work out the length of its space diagonal. Give the diagonal length correct to

1

decimal place.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$17.7\text{ cm}$	3	First find a base diagonal: $\sqrt{6^2+9^2}=\sqrt{117}\text{ cm}$ . The space diagonal is then $\sqrt{117+14^2}=\sqrt{313}=17.691\ldots\text{ cm}$ , which is $17.7\text{ cm}$ correct to $1$ decimal place.

Tier 2 · Standard

A kite is held by a straight taut string

25\text{ m}

long. The string makes an angle of

38^\circ

with the horizontal. The person's hand is

1.4\text{ m}

above level ground. Work out the height of the kite above the ground, correct to

1

decimal place.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$16.8\text{ m}$	3	The vertical height from the hand to the kite is $25\sin38^\circ=15.3915\ldots\text{ m}$ . Adding the hand height gives $15.3915\ldots+1.4=16.7915\ldots\text{ m}$ , which is $16.8\text{ m}$ correct to $1$ decimal place.

Tier 3 · Hard

A rhombus has diagonals that meet at right angles and bisect each other. One diagonal is

10\text{ cm}

long and each side of the rhombus is

13\text{ cm}

long. Work out the length of the other diagonal and the area of the rhombus.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	Other diagonal $=24\text{ cm}$ Area $=120\text{ cm}^2$	4	Half of the known diagonal is $5\text{ cm}$ . If half of the other diagonal is $x$ , then $x^2+5^2=13^2$ , so $x=12\text{ cm}$ . The full diagonal is $24\text{ cm}$ . The four right-angled triangles have total area $4\times\frac12\times5\times12=120\text{ cm}^2$ .

Tier 3 · Hard

A rectangle has area

240\text{ cm}^2

and diagonal length

26\text{ cm}

. Work out the perimeter of the rectangle.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$68\text{ cm}$	4	Let the side lengths be $a\text{ cm}$ and $b\text{ cm}$ . Pythagoras gives $a^2+b^2=26^2=676$ , and the area gives $ab=240$ . Therefore $(a+b)^2=a^2+b^2+2ab=676+480=1156$ , so $a+b=34$ . The perimeter is $2(a+b)=68\text{ cm}$ .

Q10

Tier 3 · Hard

Rectangle

ABCD

has

AB=16\text{ cm}

and

BC=12\text{ cm}

. Point

M

is the midpoint of

CD

. Work out

AM

in exact form and work out

\angle MAB

correct to

1

decimal place.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$AM=4\sqrt{13}\text{ cm}$ $\angle MAB=56.3^\circ$	4	The horizontal displacement from $A$ to $M$ is $8\text{ cm}$ and the vertical displacement is $12\text{ cm}$ . Hence $AM=\sqrt{8^2+12^2}=\sqrt{208}=4\sqrt{13}\text{ cm}$ . Also $\tan\angle MAB=12/8$ , so $\angle MAB=56.3099\ldots^\circ=56.3^\circ$ correct to $1$ decimal place.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	2H	Q20	3	Allowed	Higher	QP MS
2021-11	2H	Q6	4	Allowed	Higher	QP MS
2023-06	3H	Q10	3	Allowed	Higher	QP MS
2021-11	2F	Q25	4	Allowed	Foundation	QP MS
2021-11	2H	Q5	4	Allowed	Higher	QP MS
2023-11	3H	Q24	4	Allowed	Higher	QP MS
2019-06	2H	Q5	2	Allowed	Higher	QP MS
2021-11	3F	Q19	4	Allowed	Foundation	QP MS
2024-06	2H	Q1	2	Allowed	Higher	QP MS
2022-11	2H	Q13	2	Allowed	Higher	QP MS
2023-11	2H	Q12	3	Allowed	Higher	QP MS
2024-11	3H	Q19	4	Allowed	Higher	QP MS
2024-11	2H	Q7	5	Allowed	Higher	QP MS
2019-06	2F	Q24	2	Allowed	Foundation	QP MS
2023-06	1H	Q22	2	Non-calculator	Higher	QP MS
2022-11	2H	Q5	2	Allowed	Higher	QP MS
2019-06	3H	Q8	2	Allowed	Higher	QP MS
2022-11	1H	Q24	6	Non-calculator	Higher	QP MS
2022-06	1H	Q13	3	Non-calculator	Higher	QP MS
2021-11	2F	Q26	4	Allowed	Foundation	QP MS
2021-11	3H	Q10	4	Allowed	Higher	QP MS
2023-11	2F	Q29	2	Allowed	Foundation	QP MS
2024-11	3F	Q24	2	Allowed	Foundation	QP MS
2022-11	2H	Q20	4	Allowed	Higher	QP MS
2022-06	3H	Q12	3	Allowed	Higher	QP MS
2019-11	3H	Q23	5	Allowed	Higher	QP MS
2024-06	2H	Q10	3	Allowed	Higher	QP MS
2023-11	3F	Q29	4	Allowed	Foundation	QP MS
2023-06	2H	Q19	3	Allowed	Higher	QP MS
2019-06	2H	Q19	4	Allowed	Higher	QP MS
2022-06	3H	Q1	2	Allowed	Higher	QP MS
2023-06	2H	Q12	4	Allowed	Higher	QP MS
2024-11	3H	Q8	2	Allowed	Higher	QP MS
2024-06	2F	Q20	2	Allowed	Foundation	QP MS
2022-06	3H	Q18	5	Allowed	Higher	QP MS
2024-11	2F	Q26	5	Allowed	Foundation	QP MS
2022-11	2F	Q22	2	Allowed	Foundation	QP MS
2022-06	3F	Q22	2	Allowed	Foundation	QP MS
2019-11	2H	Q12	4	Allowed	Higher	QP MS

Other points in G Geometry and measures

Want help turning this into marks?

Bring G20 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call