What is G22: Sine and cosine rule?

Higher tier only. The sine rule a/sin A=b/sin B=c/sin C pairs each side with its opposite angle.

How many marks are G22 questions worth?

In verified past-paper appearances, G22 questions on Edexcel GCSE Maths 1MA1 have been worth 3–5 marks.

Which past exam papers has G22 appeared in?

G22 has appeared in 14 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

G22

Know and apply the sine rule a/sin A = b/sin B = c/sin C, and cosine rule a² = b² + c² - 2bc cos A, to find unknown lengths and angles [Higher only]

Higher only

Sine and cosine rule

Worked answers, methods and verified real exam appearances for G22 on Edexcel GCSE Maths 1MA1.

Explanation

Higher tier only. The sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ pairs each side with its opposite angle.
Use it when a complete opposite side-angle pair is known. The cosine rule $a^2=b^2+c^2-2bc\cos A$ finds a side from two sides and their included angle, or an angle from three sides.
Label the triangle before substituting so the lowercase side is opposite its matching capital angle.
When an inverse sine gives an angle, check the supplementary value against the triangle angle sum and other information.
Examiners expect an unrounded substitution followed by the requested final accuracy.

In the sine and cosine rules, each lowercase side is opposite its matching capital angle.

Worked example

Two sides of a triangle are $8$ cm and $11$ cm, with included angle $47^\circ$ . Find the third side to $1$ decimal place.

1.Use the cosine rule with the required side opposite $47^\circ$ : $c^2=8^2+11^2-2(8)(11)\cos47^\circ$ .
2. $c^2=64.968\ldots$ , so $c=\sqrt{64.968\ldots}=8.060\ldots$ .
3.Round the final length to $1$ decimal place.

Answer: $8.1$ cm.

Common mistakes

Don't pair side $a$ with an angle other than its opposite angle $A$ in the sine rule.
Don't use the cosine rule with an angle that is not included between the two substituted sides.

Exam tip

Sketch and label $a$ opposite $A$ before choosing a rule; this prevents most substitution errors.

Worked practice

Tier 1 · Easy

In triangle

ABC

, side

a=7\text{ cm}

, angle

A=30^\circ

and angle

B=90^\circ

. Work out side

b

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$14\text{ cm}$	2	By the sine rule, $\frac{b}{\sin90^\circ}=\frac{7}{\sin30^\circ}$ . Hence $b=7\times\frac{1}{1/2}=14\text{ cm}$ .

Tier 2 · Standard

Two sides of a triangle are

7\text{ cm}

and

10\text{ cm}

, and the included angle is

60^\circ

. Work out the exact length of the third side.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$\sqrt{79}\text{ cm}$	3	Using the cosine rule, $c^2=7^2+10^2-2(7)(10)\cos60^\circ=49+100-70=79$ . Therefore $c=\sqrt{79}\text{ cm}$ .

Tier 3 · Hard

In triangle

ABC

A=35^\circ

a=8\text{ cm}

and

b=11\text{ cm}

. Find all possible values of angles

B

and

C

. Give each angle to

1

decimal place.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$B=52.1^\circ$ , $C=92.9^\circ$ $B=127.9^\circ$ , $C=17.1^\circ$	5	The sine rule gives $\sin B=\frac{11\sin35^\circ}{8}=0.7887\ldots$ . Therefore $B=52.1^\circ$ or $B=180^\circ-52.1^\circ=127.9^\circ$ . The corresponding third angles are $C=180^\circ-35^\circ-52.1^\circ=92.9^\circ$ and $C=180^\circ-35^\circ-127.9^\circ=17.1^\circ$ .

Tier 1 · Easy

A triangle has side lengths

5\text{ cm}

13\text{ cm}

and

15\text{ cm}

. Work out the angle opposite the

15\text{ cm}

side. Give the angle correct to

1

decimal place.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$103.8^\circ$	3	By the cosine rule, $15^2=5^2+13^2-2(5)(13)\cos C$ . Therefore $\cos C=\frac{5^2+13^2-15^2}{2(5)(13)}=-\frac{31}{130}$ , so $C=103.795\ldots^\circ=103.8^\circ$ correct to $1$ decimal place.

Tier 2 · Standard

In triangle

ABC

A=45^\circ

a=7\sqrt2\text{ cm}

and

b=7\text{ cm}

. Work out angle

B

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$B=30^\circ$	3	By the sine rule, $\frac{\sin B}{7}=\frac{\sin45^\circ}{7\sqrt2}$ . Hence $\sin B=\frac{7(\sqrt2/2)}{7\sqrt2}=\frac12$ , so $B=30^\circ$ or $150^\circ$ . Since $45^\circ+150^\circ>180^\circ$ , only $B=30^\circ$ is possible.

Tier 3 · Hard

Two sides of a triangle are

8\text{ cm}

and

10\text{ cm}

, and their included angle is

120^\circ

. Work out the third side and the angle opposite the

10\text{ cm}

side. Give both answers correct to

1

decimal place.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	Third side $=15.6\text{ cm}$ Angle $=33.7^\circ$	5	By the cosine rule, the third side $c$ satisfies $c^2=8^2+10^2-2(8)(10)\cos120^\circ=244$ , so $c=\sqrt{244}=15.620\ldots\text{ cm}$ . Let $B$ be the angle opposite the $10\text{ cm}$ side. By the sine rule, $\frac{\sin B}{10}=\frac{\sin120^\circ}{\sqrt{244}}$ , so $B=\sin^{-1}\left(\frac{10\sin120^\circ}{\sqrt{244}}\right)=33.670\ldots^\circ$ . The supplementary value is impossible with the $120^\circ$ angle. Therefore the answers are $15.6\text{ cm}$ and $33.7^\circ$ .

Tier 2 · Standard

In triangle

ABC

A=38^\circ

B=79^\circ

and the side opposite angle

A

8.4\text{ cm}

. Work out the length of the side opposite angle

B

, 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	$13.4\text{ cm}$	3	By the sine rule, $\frac{b}{\sin79^\circ}=\frac{8.4}{\sin38^\circ}$ . Hence $b=\frac{8.4\sin79^\circ}{\sin38^\circ}=13.3931\ldots\text{ cm}$ , which is $13.4\text{ cm}$ correct to $1$ decimal place.

Tier 3 · Hard

Convex quadrilateral

ABCD

has diagonal

AC

, with

B

and

D

on opposite sides of

AC

. Given that

AB=7\text{ cm}

BC=9\text{ cm}

\angle ABC=58^\circ

AD=6\text{ cm}

and

CD=10\text{ cm}

, work out

\angle ADC

, correct to

1

decimal place.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$52.7^\circ$	5	In triangle $ABC$ , the cosine rule gives $AC^2=7^2+9^2-2(7)(9)\cos58^\circ=63.2301\ldots$ . In triangle $ACD$ , $\cos\angle ADC=\frac{6^2+10^2-AC^2}{2(6)(10)}=0.6064\ldots$ . Therefore $\angle ADC=52.6692\ldots^\circ=52.7^\circ$ correct to $1$ decimal place.

Tier 3 · Hard

A triangle has sides of lengths

(x+1)\text{ cm}

(x+3)\text{ cm}

and

2\sqrt{13}\text{ cm}

. The angle between the sides of lengths

(x+1)\text{ cm}

and

(x+3)\text{ cm}

60^\circ

. Work out

x

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$x=5$	4	By the cosine rule, $52=(x+1)^2+(x+3)^2-2(x+1)(x+3)\cos60^\circ$ . Since $2\cos60^\circ=1$ , this simplifies to $52=x^2+4x+7$ . Hence $x^2+4x-45=0$ , so $(x+9)(x-5)=0$ . The value $x=-9$ gives negative side lengths, so $x=5$ .

Q10

Tier 3 · Hard

A triangle has side lengths

7\text{ cm}

9\text{ cm}

and

11\text{ cm}

. Work out the difference between its largest angle and its smallest angle, 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	$46.5^\circ$	4	The largest angle $C$ is opposite the $11\text{ cm}$ side. The cosine rule gives $\cos C=\frac{7^2+9^2-11^2}{2(7)(9)}=\frac1{14}$ , so $C=85.9039\ldots^\circ$ . The smallest angle $A$ is opposite the $7\text{ cm}$ side, and $\cos A=\frac{9^2+11^2-7^2}{2(9)(11)}=\frac{17}{22}$ , so $A=39.4005\ldots^\circ$ . The difference is $46.5033\ldots^\circ$ , which is $46.5^\circ$ correct to $1$ decimal place.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	3H	Q18	5	Allowed	Higher	QP MS
2019-06	3H	Q23	5	Allowed	Higher	QP MS
2023-06	2H	Q13	4	Allowed	Higher	QP MS
2023-11	2H	Q16	4	Allowed	Higher	QP MS
2023-11	1H	Q20	4	Non-calculator	Higher	QP MS
2024-11	3H	Q17	4	Allowed	Higher	QP MS
2021-11	3H	Q23	4	Allowed	Higher	QP MS
2024-11	1H	Q14	3	Non-calculator	Higher	QP MS
2019-11	3H	Q18	5	Allowed	Higher	QP MS
2022-11	3H	Q26	5	Allowed	Higher	QP MS
2022-06	2H	Q18	4	Allowed	Higher	QP MS
2022-11	1H	Q22	4	Non-calculator	Higher	QP MS
2021-11	2H	Q15	5	Allowed	Higher	QP MS
2023-06	3H	Q14	4	Allowed	Higher	QP MS

Other points in G Geometry and measures

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