Sine, Cosine and Area Rule · Study Notes · Edexcel IGCSE 4MA1 Higher

Trigonometry for non-right-angled triangles. Sine rule, cosine rule, and the area formula. A* topic.

At a glance

Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ .
Cosine rule: $a^{2} = b^{2} + c^{2} - 2bc \cos A$ .
Area: $\dfrac{1}{2}ab \sin C$ (included angle).
Use sine rule when you have an angle-side pair.
Use cosine rule when you have 2 sides + included angle, or all 3 sides.
Watch the ambiguous case when finding an angle from sine rule.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.10 — Use sine rule to find sides and angles.
4.10 — Use cosine rule to find sides and angles.
4.10 — Use area formula $\dfrac{1}{2}ab \sin C$ .
4.10 — Choose appropriate rule for given problem.

The sine rule

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$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ .

Sine rule relates each side to the sine of its opposite angle.

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

When to use:

You have an ANGLE-SIDE PAIR (e.g. $a$ and $A$ both known).
You want to find another side or angle.

Each lower-case side sits directly opposite the capital-letter angle of the same name.

Find a side. $a = 8, A = 50°, B = 60°$ . Find $b$ .

$\dfrac{8}{\sin 50°} = \dfrac{b}{\sin 60°}$ .
$b = \dfrac{8 \sin 60°}{\sin 50°} \approx 9.04$ .

Find an angle. $a = 7, b = 10, A = 40°$ . Find $B$ .

$\sin B = \dfrac{10 \sin 40°}{7} \approx 0.918$ .
$B \approx 66.7°$ .

The AMBIGUOUS CASE.

When finding an angle from the sine rule, the calculator gives you ONE answer (acute). But the OBTUSE supplementary angle ( $180° -$ acute) might also work.

For the example above: $66.7°$ OR $180° - 66.7° = 113.3°$ .

Check which is geometrically possible:

The angles must sum to less than $180°$ .
Use the diagram if available.

Worked qualitative. Why might both angles be valid?

$\sin \theta = \sin (180° - \theta)$ .
So sine rule gives both as solutions.
The diagram or problem context disambiguates.

Edexcel tip. Always state the sine rule explicitly. Mark schemes give M1 for the rule, M1 for substitution, A1 for the answer.

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ .
Use with angle-side pairs.
Watch the ambiguous case.
Both acute and $180° -$ acute may work.

The cosine rule

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$a^{2} = b^{2} + c^{2} - 2bc \cos A$ .

Cosine rule generalises Pythagoras to non-right triangles.

For finding a SIDE:

$a^{2} = b^{2} + c^{2} - 2bc \cos A$

When to use:

Two sides and the INCLUDED ANGLE (between them).

The included angle A lies between the two known sides; the cosine rule finds the opposite side a.

Example. $b = 6, c = 8, A = 60°$ . Find $a$ .

$a^{2} = 36 + 64 - 2(6)(8)\cos 60° = 100 - 48 = 52$ .
$a \approx 7.21$ .

For finding an ANGLE (rearrange the rule):

$\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$

When to use:

All three sides known.

Example. $a = 5, b = 7, c = 9$ . Find $A$ .

$\cos A = \dfrac{49 + 81 - 25}{2 \cdot 7 \cdot 9} = \dfrac{105}{126} \approx 0.833$ .
$A \approx 33.6°$ .

Sine rule vs cosine rule:

You have	Use
Angle-side pair + another side or angle	Sine rule
Two sides + INCLUDED angle	Cosine rule
All three sides	Cosine rule (for angle)

Worked qualitative. Why does the cosine rule reduce to Pythagoras when $A = 90°$ ?

$\cos 90° = 0$ .
$a^{2} = b^{2} + c^{2} - 2bc(0) = b^{2} + c^{2}$ .
Same as Pythagoras.
So Pythagoras is a SPECIAL CASE of the cosine rule.

Edexcel tip. State the cosine rule explicitly. NOT on Edexcel formula sheet — must memorise.

$a^{2} = b^{2} + c^{2} - 2bc \cos A$ .
Generalises Pythagoras.
Used with 2 sides + included angle.
Or all 3 sides (find angle).

Area formula $\dfrac{1}{2}ab \sin C$

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Area when you have two sides and the included angle.

Area of any triangle:

$\text{Area} = \dfrac{1}{2}ab \sin C$

where $a, b$ are two sides and $C$ is the INCLUDED angle (between them).

Why this works.

$h = b \sin C$ (opposite side of right triangle formed by dropping perpendicular).
Area = $\dfrac{1}{2} \cdot a \cdot h = \dfrac{1}{2}ab \sin C$ .

Dropping the height h = b sin C turns the triangle area into ½ × base a × height.

Example. Triangle with $a = 8, b = 10, C = 40°$ .

Area $= \dfrac{1}{2}(8)(10) \sin 40° = 40 \sin 40° \approx 25.7$ .

Compared to standard $\dfrac{1}{2}bh$ :

If you have base and PERPENDICULAR HEIGHT, use $\dfrac{1}{2}bh$ . Otherwise, the sine formula avoids needing to find $h$ first.

Worked qualitative. Why use this formula?

Often you have two sides and the included angle (not the perpendicular height).
This formula is direct.
Pythagoras would be needed to find $h$ otherwise.

Edexcel tip. Mark schemes credit the formula explicitly. State 'Area = $\dfrac{1}{2}ab \sin C$ ' before substituting.

$\dfrac{1}{2}ab \sin C$ .
$C$ is the INCLUDED angle.
Avoid needing perpendicular height.
Generalises $\dfrac{1}{2}bh$ .

Quick recap

Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ .
Cosine rule: $a^{2} = b^{2} + c^{2} - 2bc \cos A$ .
Area: $\dfrac{1}{2}ab \sin C$ .
Sine: angle-side pair. Cosine: 2 sides + angle.
Watch ambiguous case for sine rule angles.
Cosine generalises Pythagoras.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ .
Cosine rule: $a^{2} = b^{2} + c^{2} - 2bc \cos A$ .
Cosine for angle: $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$ .
Area: $\dfrac{1}{2}ab \sin C$ .

How it’s examined

Sine/cosine/area rules every Higher Tier paper (5-8 marks). Often combined in a single multi-part question. Examiner reports flag (1) wrong rule choice, (2) ambiguous case missed, (3) using non-included angle for area.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on sine, cosine and area rule, written exactly the way a tutor would explain them at the board.

1Sine rule — find a side
Higher• sine rule
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Question
In triangle $ABC$ : $a = 8$ , $A = 50°$ , $B = 60°$ . Find $b$ .
Step-by-step solution
1. Step 1
  Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ .
2. Step 2
  $\dfrac{8}{\sin 50°} = \dfrac{b}{\sin 60°}$ .
3. Step 3
  $b = \dfrac{8 \sin 60°}{\sin 50°} \approx 9.04$ .
Answer
$b \approx 9.04$
2Sine rule — find an angle
Higher• Adapted from 4MA1/2H May/Jun 2024 Q20• sine rule
▼
Question
Triangle: $a = 7$ , $b = 10$ , $A = 40°$ . Find $B$ .
Step-by-step solution
1. Step 1
  $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ .
2. Step 2
  $\dfrac{7}{\sin 40°} = \dfrac{10}{\sin B}$ .
3. Step 3
  $\sin B = \dfrac{10 \sin 40°}{7} \approx 0.918$ .
4. Step 4
  $B = \sin^{-1}(0.918) \approx 66.7°$ .
Answer
$B \approx 66.7°$
Examiner tip
Watch for the AMBIGUOUS CASE: $\sin^{-1}$ gives an acute angle, but the obtuse angle ( $180° - 66.7° = 113.3°$ ) might also work. Check the triangle's geometry.
3Cosine rule — find a side
Higher• cosine rule
▼
Question
Triangle: $b = 6$ , $c = 8$ , $A = 60°$ . Find $a$ .
Step-by-step solution
1. Step 1
  Cosine rule: $a^{2} = b^{2} + c^{2} - 2bc \cos A$ .
2. Step 2
  $a^{2} = 36 + 64 - 2 \cdot 6 \cdot 8 \cos 60°$ .
3. Step 3
  $\cos 60° = 0.5$ . So $a^{2} = 100 - 48 = 52$ . $a = \sqrt{52} \approx 7.21$ .
Answer
$a \approx 7.21$
4Cosine rule — find an angle
Higher• Adapted from 4MA1/1H Jan 2024 Q17• cosine rule
▼
Question
Triangle: $a = 5$ , $b = 7$ , $c = 9$ . Find $A$ .
Step-by-step solution
1. Step 1
  Rearrange cosine rule: $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$ .
2. Step 2
  $\cos A = \dfrac{49 + 81 - 25}{2 \cdot 7 \cdot 9} = \dfrac{105}{126} = 0.833$ .
3. Step 3
  $A = \cos^{-1}(0.833) \approx 33.6°$ .
Answer
$A \approx 33.6°$
5Area of triangle — sine formula
Higher• area
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Question
Triangle with sides $a = 8$ , $b = 10$ , included angle $C = 40°$ . Find area.
Step-by-step solution
1. Step 1
  Area $= \dfrac{1}{2}ab \sin C$ .
2. Step 2
  $= \dfrac{1}{2} \times 8 \times 10 \times \sin 40° = 40 \times 0.643 \approx 25.7$ .
Answer
$\approx 25.7$ cm²
6The cosine of an obtuse angle [1 mark]
Getting started• Paper 1H/2H style — 1 mark• cosine, obtuse angles
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Question
Given that $\cos 72^{\circ} = 0.309$ correct to $3$ significant figures, write down the value of $\cos 108^{\circ}$ correct to $3$ significant figures. [1]
Step-by-step solution
1. Step 1
  $108^{\circ} = 180^{\circ} - 72^{\circ}$ . The cosine of an obtuse angle is negative, and $\cos(180^{\circ} - x) = -\cos x$ .
2. Step 2
  B1 Write down the value.
  $\cos 108^{\circ} = -\cos 72^{\circ} = -0.309$
Answer
$-0.309$
Examiner tip
Examiner tip: This is the reason the cosine rule handles obtuse angles on its own: a negative cosine can only come from an angle between $90^{\circ}$ and $180^{\circ}$ .

Common mistake: Writing $0.309$ , copying the rule for sine. Sine keeps its sign for an obtuse angle; cosine does not.
7Area of a regular octagon from its centre [2 marks]
Getting started• Paper 1H/2H style — 2 marks• area of a triangle, regular polygons
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Question
A regular octagon has centre $O$ . Each vertex of the octagon is $7$ cm from $O$ .

Work out the area of the octagon.

Give your answer correct to $3$ significant figures. [2]
Step-by-step solution
1. Step 1
  Joining $O$ to each vertex splits the octagon into $8$ congruent isosceles triangles. Each has two sides of $7$ cm, and the angle between them at $O$ is $360^{\circ} \div 8 = 45^{\circ}$ .
2. Step 2
  M1 Use area $= \dfrac{1}{2}ab\sin C$ for one triangle and multiply by $8$ .
  $\text{Area} = 8 \times \dfrac{1}{2} \times 7 \times 7 \times \sin 45^{\circ}$
3. Step 3
  A1 Evaluate and round.
  $\text{Area} = 138.59\ldots = 139 \text{ cm}^{2}$
Answer
$139$ cm $^{2}$ (awrt $139$ )
Examiner tip
Examiner tip: The angles at the centre of a regular polygon with $n$ sides are each $360^{\circ} \div n$ . The two $7$ cm lengths enclose that angle, so the area formula needs no other lengths.

Common mistake: Treating $7$ cm as the length of a side of the octagon. It is the distance from the centre to a vertex, which is why it appears twice in each triangle. Forgetting to multiply by $8$ (giving $17.3$ cm $^{2}$ ) is the other common loss.
8Cosine rule after an angle on a straight line [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• cosine rule, angles on a straight line
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Question
$ABC$ is a triangle. The side $BC$ is extended to the point $D$ , so that $BCD$ is a straight line.

$AC = 7.4$ cm, $CD = 5.6$ cm and angle $ACB = 67^{\circ}$ .

Calculate the length of $AD$ .

Give your answer correct to $3$ significant figures. [3]
Step-by-step solution
1. Step 1
  B1 $AD$ is a side of triangle $ACD$ , and the angle between $AC$ and $CD$ is angle $ACD$ . Angles on a straight line add up to $180^{\circ}$ .
  $\angle ACD = 180^{\circ} - 67^{\circ} = 113^{\circ}$
2. Step 2
  M1 Two sides and the angle between them are now known, so use the cosine rule in triangle $ACD$ .
  $AD^{2} = 7.4^{2} + 5.6^{2} - 2 \times 7.4 \times 5.6 \times \cos 113^{\circ} = 86.12 + 32.38\ldots = 118.50\ldots$
3. Step 3
  A1 Square root and round.
  $AD = \sqrt{118.50\ldots} = 10.885\ldots = 10.9 \text{ cm}$
Answer
$10.9$ cm (awrt $10.9$ )
Examiner tip
Examiner tip: Sketch the figure and mark the angle you actually need. $\cos 113^{\circ}$ is negative, so subtracting $2 \times 7.4 \times 5.6 \times \cos 113^{\circ}$ adds $32.38\ldots$ , and $AD$ comes out longer than both $AC$ and $CD$ , as it must opposite an obtuse angle.

Common mistake: Using the $67^{\circ}$ angle in the cosine rule. That angle is inside triangle $ABC$ , not between $AC$ and $CD$ , and it gives $AD = 7.33$ cm, which scores no marks.
9Sine rule for an angle, then the area [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• sine rule, area of a triangle
▼
Question
In triangle $ABC$ , $AB = 7.9$ cm, $BC = 11.6$ cm and angle $BAC = 83^{\circ}$ .

Work out the area of triangle $ABC$ .

Give your answer correct to $3$ significant figures. [4]
Step-by-step solution
1. Step 1
  M1 The known angle is not between the two known sides, so find another angle first. $AB$ is opposite angle $ACB$ and $BC$ is opposite the $83^{\circ}$ angle.
  $\sin ACB = \dfrac{7.9 \sin 83^{\circ}}{11.6} = 0.6759\ldots$
2. Step 2
  A1 Angle $ACB$ is opposite the shorter side $AB$ , so it is smaller than $83^{\circ}$ and acute.
  $\angle ACB = 42.52\ldots^{\circ}$
3. Step 3
  M1 Angle $ABC$ is between $AB$ and $BC$ . Find it from the angle sum, then use the area formula.
  $\angle ABC = 180^{\circ} - 83^{\circ} - 42.52\ldots^{\circ} = 54.47\ldots^{\circ}, \quad \text{Area} = \dfrac{1}{2} \times 7.9 \times 11.6 \times \sin 54.47\ldots^{\circ}$
4. Step 4
  A1 Evaluate and round.
  $\text{Area} = 37.28\ldots = 37.3 \text{ cm}^{2}$
Answer
$37.3$ cm $^{2}$ (awrt $37.3$ )
Examiner tip
Examiner tip: Before using $\dfrac{1}{2}ab\sin C$ , check that the angle is the one between $a$ and $b$ . If it is not, the sine rule and the angle sum will get you there.

Common mistake: Working out $\dfrac{1}{2} \times 7.9 \times 11.6 \times \sin 83^{\circ} = 45.5$ . The $83^{\circ}$ angle is at $A$ , but $BC$ does not touch $A$ , so the angle is not between these sides.
10An angle at a point on the side of a triangle [5 marks]
Stretch• Paper 1H/2H style — 5 marks• cosine rule, two triangles, angles on a straight line
▼
Question
$ABC$ is a triangle. $D$ is the point on $BC$ such that $BD = 6$ cm.

$AB = 7.5$ cm and angle $ABD = 64^{\circ}$ .

Calculate the size of angle $ADC$ .

Give your answer correct to $1$ decimal place. [5]
Step-by-step solution
1. Step 1
  M1 In triangle $ABD$ , find $AD$ with the cosine rule.
  $AD^{2} = 7.5^{2} + 6^{2} - 2 \times 7.5 \times 6 \times \cos 64^{\circ} = 92.25 - 39.45\ldots = 52.79\ldots$
2. Step 2
  A1 Square root. Keep the full value.
  $AD = 7.266\ldots \text{ cm}$
3. Step 3
  M1 Find angle $ADB$ , which is opposite $AB = 7.5$ . The cosine rule is the safe choice here, as it shows directly whether the angle is acute or obtuse.
  $\cos ADB = \dfrac{7.266\ldots^{2} + 6^{2} - 7.5^{2}}{2 \times 7.266\ldots \times 6} = 0.3732\ldots \quad \Rightarrow \quad \angle ADB = 68.08\ldots^{\circ}$
4. Step 4
  M1 $BDC$ is a straight line and angles on a straight line add up to $180^{\circ}$ . A1 Round the answer.
  $\angle ADC = 180^{\circ} - 68.08\ldots^{\circ} = 111.91\ldots = 111.9^{\circ}$
Answer
$111.9^{\circ}$ (awrt $111.9$ )
Examiner tip
Examiner tip: The length of $DC$ is not needed. Angle $ADC$ is on the other side of the straight line from angle $ADB$ , so the whole problem happens in triangle $ABD$ .

Common mistake: Giving angle $ADB = 68.1^{\circ}$ as the answer. The June 2024 Paper 1H examiner report notes that candidates who labelled their diagram incorrectly (giving an interior angle the value of the exterior angle) lost all their marks on that question. Sketch the triangle and mark which angle is wanted before you start.

Question 1
Paper 1H/2H style — 1 mark1 mark
In triangle $LMN$ , $LM = 8.5$ cm, $MN = 6.2$ cm and $LN = 10.1$ cm.

Write down the name of the largest angle of the triangle. [1]
Model answer
The largest angle is opposite the longest side, $LN$ . The vertex not on $LN$ is $M$ .

Answer: angle $LMN$
Why this scores
Mark scheme:
- B1 for angle $LMN$ (accept angle $M$ or angle $NML$ )
Why this scores full marks: The candidate names the angle opposite the longest side, not an angle at one end of it. This is the first decision in any cosine rule question: the side opposite the wanted angle is the one that is subtracted.
Question 2
Paper 1H/2H style — 2 marks2 marks
A triangle has sides of length $6$ cm, $8$ cm and $11$ cm.

Show that the triangle has an obtuse angle. [2]
Model answer
The largest angle, $\theta$ , is opposite the longest side, $11$ cm.

$\cos \theta = \dfrac{6^{2} + 8^{2} - 11^{2}}{2 \times 6 \times 8} = \dfrac{-21}{96} = -0.21875$

$\cos \theta$ is negative, so $\theta$ is between $90^{\circ}$ and $180^{\circ}$ ( $\theta = 102.6^{\circ}$ ). The triangle has an obtuse angle.
Why this scores
Mark scheme:
- M1 for $\dfrac{6^{2} + 8^{2} - 11^{2}}{2 \times 6 \times 8}$ oe, or for comparing $6^{2} + 8^{2} = 100$ with $11^{2} = 121$
- A1 for $\cos \theta = -0.21875$ oe (or $\theta = 102.6\ldots^{\circ}$ ) with a conclusion that the angle is obtuse
Why this scores full marks: The candidate chooses the only angle that could be obtuse, the one opposite the longest side, and finishes with a statement that answers the question. A negative cosine is enough: there is no need to find the angle, although giving it does no harm. A correct value with no conclusion loses the A1.
Question 3
Paper 1H/2H style — 2 marks2 marks
In triangle $ABC$ , $AB = 10$ cm, $BC = 8$ cm and angle $ACB = 30^{\circ}$ .

Show that $\sin BAC = 0.4$ [2]
Model answer
$BC$ is opposite angle $BAC$ , and $AB$ is opposite angle $ACB$ .

$\dfrac{\sin BAC}{8} = \dfrac{\sin 30^{\circ}}{10}$

$\sin BAC = \dfrac{8 \times \sin 30^{\circ}}{10} = \dfrac{8 \times 0.5}{10} = \dfrac{4}{10} = 0.4$
Why this scores
Mark scheme:
- M1 for $\dfrac{\sin BAC}{8} = \dfrac{\sin 30^{\circ}}{10}$ oe
- A1 for a complete correct argument, showing $\sin 30^{\circ} = 0.5$ (or $8 \times 0.5 = 4$ ) and reaching $0.4$
Why this scores full marks: The candidate pairs each side with the angle opposite it, then shows the exact value $\sin 30^{\circ} = 0.5$ so that every step to $0.4$ can be seen. In a show that, writing only $\dfrac{8 \sin 30^{\circ}}{10} = 0.4$ risks the A1, because the printed answer is not evidence of the working.
Question 4
Paper 1H/2H style — 3 marks3 marks
A triangle has sides of length $11$ cm, $14$ cm and $17$ cm.

Calculate the size of the smallest angle of the triangle. Give your answer correct to $1$ decimal place. [3]
Model answer
The smallest angle is opposite the shortest side, $11$ cm.

$\cos \theta = \dfrac{14^{2} + 17^{2} - 11^{2}}{2 \times 14 \times 17} = \dfrac{364}{476} = 0.7647\ldots$

$\theta = \cos^{-1}(0.7647\ldots) = 40.11\ldots^{\circ}$

Answer: $40.1^{\circ}$
Why this scores
Mark scheme:
- M1 for $\dfrac{14^{2} + 17^{2} - 11^{2}}{2 \times 14 \times 17}$ oe
- M1 for $\cos^{-1}\left(\dfrac{364}{476}\right)$
- A1 for awrt $40.1$
Why this scores full marks: The opening sentence identifies which angle is wanted before any calculation. Subtracting $17^{2}$ instead gives $84.8^{\circ}$ , the largest angle, which scores nothing.
Question 5
Paper 1H/2H style — 3 marks3 marks
In triangle $PQR$ , $PR = 13$ cm, $QR = 9$ cm and angle $QPR = 36^{\circ}$ .

There are two possible triangles that fit this information.

Find the two possible sizes of angle $PQR$ . Give your answers correct to $1$ decimal place. [3]
Model answer
$PR$ is opposite angle $PQR$ , and $QR$ is opposite angle $QPR$ .

$\dfrac{\sin PQR}{13} = \dfrac{\sin 36^{\circ}}{9}$

$\sin PQR = \dfrac{13 \sin 36^{\circ}}{9} = 0.8490\ldots$

$\angle PQR = \sin^{-1}(0.8490\ldots) = 58.10\ldots^{\circ} \quad \text{or} \quad 180^{\circ} - 58.10\ldots^{\circ} = 121.89\ldots^{\circ}$

Check: $36^{\circ} + 121.9^{\circ} = 157.9^{\circ} < 180^{\circ}$ , so both triangles exist.

Answer: $58.1^{\circ}$ and $121.9^{\circ}$
Why this scores
Mark scheme:
- M1 for $\dfrac{\sin PQR}{13} = \dfrac{\sin 36^{\circ}}{9}$ oe
- A1 for awrt $58.1$
- B1 for awrt $121.9$ (ft $180 -$ their acute angle)
Why this scores full marks: The candidate knows that $\sin^{-1}$ gives only the acute angle, and that $\sin(180^{\circ} - x) = \sin x$ gives the second. The check that $36^{\circ} + 121.9^{\circ}$ leaves room for a third angle confirms that the obtuse answer is a real triangle. Weaker scripts give $58.1^{\circ}$ only and lose the B1.
Question 6
Paper 1H/2H style — 4 marks4 marks
In triangle $PQR$ , $PQ = 10.5$ cm, $PR = 13$ cm and angle $QPR = 47^{\circ}$ .

Calculate the shortest distance from $P$ to the line $QR$ . Give your answer correct to $3$ significant figures. [4]
Model answer
$QR^{2} = 10.5^{2} + 13^{2} - 2 \times 10.5 \times 13 \times \cos 47^{\circ} = 279.25 - 186.18\ldots = 93.06\ldots$

$QR = 9.646\ldots$

The shortest distance is the perpendicular height $h$ from $P$ to $QR$ . Two expressions for the area:

$\dfrac{1}{2} \times 9.646\ldots \times h = \dfrac{1}{2} \times 10.5 \times 13 \times \sin 47^{\circ} = 49.91\ldots$

$h = \dfrac{2 \times 49.91\ldots}{9.646\ldots} = 10.34\ldots$

Answer: $10.3$ cm
Why this scores
Mark scheme:
- M1 for $10.5^{2} + 13^{2} - 2 \times 10.5 \times 13 \times \cos 47^{\circ}$
- A1 for $QR = 9.64\ldots$ to $9.65$
- M1 for $\dfrac{1}{2} \times \text{their } QR \times h = \dfrac{1}{2} \times 10.5 \times 13 \times \sin 47^{\circ}$ oe
- A1 for awrt $10.3$
Why this scores full marks: The candidate recognises that the shortest distance to a line is the perpendicular distance, which is the height of the triangle when $QR$ is the base. Working out the area in two ways links the height to lengths that can be calculated. Weaker scripts give $QR$ as the answer, or take $h = 13 \sin 47^{\circ} = 9.51$ cm, which is the distance from $R$ to the line $PQ$ , not from $P$ to $QR$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
A plot of land is in the shape of a triangle. Two sides of the plot are $23$ m and $31$ m long, and the angle between them is $76^{\circ}$ .

Work out the perimeter of the plot. Give your answer correct to $3$ significant figures. [4]
Model answer
$c^{2} = 23^{2} + 31^{2} - 2 \times 23 \times 31 \times \cos 76^{\circ} = 1490 - 344.98\ldots = 1145.01\ldots$

$c = 33.83\ldots$

$\text{Perimeter} = 23 + 31 + 33.83\ldots = 87.83\ldots$

Answer: $87.8$ m
Why this scores
Mark scheme:
- M1 for $23^{2} + 31^{2} - 2 \times 23 \times 31 \times \cos 76^{\circ}$
- A1 for third side $33.8\ldots$
- M1 for $23 + 31 + \text{their third side}$
- A1 for awrt $87.8$
Why this scores full marks: The candidate answers the question set. Stopping at $33.8$ m, the third side, is a common way to lose the last two marks.
Question 8
Paper 1H/2H style — 5 marks5 marks
In triangle $ABC$ , $AB = 6$ cm, $AC = 14.5$ cm and angle $BAC = 58^{\circ}$ .

Calculate the size of angle $ABC$ . Give your answer correct to $1$ decimal place. [5]
Model answer
$BC^{2} = 6^{2} + 14.5^{2} - 2 \times 6 \times 14.5 \times \cos 58^{\circ} = 154.04\ldots$

$BC = 12.41\ldots$

Angle $ABC$ is opposite the longest side, $AC$ , so it may be obtuse. Use the cosine rule.

$\cos ABC = \dfrac{6^{2} + 12.41\ldots^{2} - 14.5^{2}}{2 \times 6 \times 12.41\ldots} = -0.1356\ldots$

$\angle ABC = 97.79\ldots^{\circ}$

Answer: $97.8^{\circ}$
Why this scores
Mark scheme:
- M1 for $6^{2} + 14.5^{2} - 2 \times 6 \times 14.5 \times \cos 58^{\circ}$
- M1 for $\sqrt{154.04\ldots}$
- A1 for $BC = 12.4\ldots$
- M1 for $\cos ABC = \dfrac{6^{2} + 12.41\ldots^{2} - 14.5^{2}}{2 \times 6 \times 12.41\ldots}$ oe with their $BC$
- A1 for awrt $97.8$
Why this scores full marks: The candidate notices that angle $ABC$ is opposite the longest side and so might be obtuse. The sine rule gives $\sin^{-1}(0.9907\ldots) = 82.2^{\circ}$ , the acute angle with the same sine, which is wrong here and loses the final mark. The cosine rule gives a negative cosine and the obtuse angle automatically.
Question 9
Paper 1H/2H style — 5 marks5 marks
In triangle $ABC$ , $AB = 9$ cm, $BC = 14$ cm and angle $ABC$ is obtuse.

The area of triangle $ABC$ is $42$ cm $^{2}$ .

Calculate the length of $AC$ . Give your answer correct to $3$ significant figures. [5]
Model answer
$\dfrac{1}{2} \times 9 \times 14 \times \sin ABC = 42$

$\sin ABC = \dfrac{42}{63} = \dfrac{2}{3}, \quad \sin^{-1}\left(\dfrac{2}{3}\right) = 41.81\ldots^{\circ}$

Angle $ABC$ is obtuse, so $\angle ABC = 180^{\circ} - 41.81\ldots^{\circ} = 138.18\ldots^{\circ}$

$AC^{2} = 9^{2} + 14^{2} - 2 \times 9 \times 14 \times \cos 138.18\ldots^{\circ} = 277 + 187.82\ldots = 464.82\ldots$

$AC = 21.55\ldots$

Answer: $21.6$ cm
Why this scores
Mark scheme:
- M1 for $\dfrac{1}{2} \times 9 \times 14 \times \sin ABC = 42$ oe
- A1 for angle $ABC = 138.1\ldots$ to $138.2$
- M1 for $9^{2} + 14^{2} - 2 \times 9 \times 14 \times \cos(\text{their obtuse angle})$
- M1 for the square root of their $AC^{2}$
- A1 for awrt $21.6$
Why this scores full marks: The candidate acts on the word obtuse before using the angle. Using $41.8^{\circ}$ instead gives $AC = 9.44$ cm, which is the acute triangle with the same area and not the one described.
Question 10
Paper 1H/2H style — 6 marks6 marks
$ABCD$ is the plan of a patio in the shape of a quadrilateral. The diagonal $AC$ divides it into triangles $ABC$ and $ACD$ .

$AB = 8.2$ m, $BC = 10.7$ m and angle $ABC = 104^{\circ}$ .

Angle $CAD = 39^{\circ}$ and angle $ACD = 57^{\circ}$ .

Work out the perimeter of the patio. Give your answer correct to $3$ significant figures. [6]
Model answer
Triangle $ABC$ :

$AC^{2} = 8.2^{2} + 10.7^{2} - 2 \times 8.2 \times 10.7 \times \cos 104^{\circ} = 181.73 + 42.45\ldots = 224.18\ldots$

$AC = 14.97\ldots$

Triangle $ACD$ :

$\angle ADC = 180^{\circ} - 39^{\circ} - 57^{\circ} = 84^{\circ}$

$AD = \dfrac{14.97\ldots \times \sin 57^{\circ}}{\sin 84^{\circ}} = 12.62\ldots, \quad CD = \dfrac{14.97\ldots \times \sin 39^{\circ}}{\sin 84^{\circ}} = 9.474\ldots$

$\text{Perimeter} = 8.2 + 10.7 + 12.62\ldots + 9.474\ldots = 41.00\ldots$

Answer: $41.0$ m
Why this scores
Mark scheme:
- M1 for $8.2^{2} + 10.7^{2} - 2 \times 8.2 \times 10.7 \times \cos 104^{\circ}$
- A1 for $AC = 14.97\ldots$
- B1 for angle $ADC = 84^{\circ}$
- M1 for a correct sine rule for $AD$ or $CD$ using their $AC$
- M1 for correct sine rules for both $AD$ and $CD$ and adding the four outside sides
- A1 for awrt $41.0$
Why this scores full marks: The diagonal $AC$ is found but not added: the perimeter uses only the four outside edges. Carrying $AC = 14.97\ldots$ unrounded keeps the answer safe: with $AC$ rounded to $15.0$ the perimeter becomes $41.04\ldots$ , which only just survives, and cruder rounding of the intermediate lengths can move the final figure.

4MA1 Sine, Cosine and Area Rule

At a glance

What you’ll learn

The sine rule

The cosine rule

Area formula $\dfrac{1}{2}ab \sin C$

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Sine rule — find a side

2Sine rule — find an angle

3Cosine rule — find a side

4Cosine rule — find an angle

5Area of triangle — sine formula

6The cosine of an obtuse angle [1 mark]

7Area of a regular octagon from its centre [2 marks]

8Cosine rule after an angle on a straight line [3 marks]

9Sine rule for an angle, then the area [4 marks]

10An angle at a point on the side of a triangle [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Sine, Cosine and Area Rule

At a glance

What you’ll learn

The sine rule

The cosine rule

Area formula 12absin⁡C\dfrac{1}{2}ab \sin C21​absinC

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Sine rule — find a side

2Sine rule — find an angle

3Cosine rule — find a side

4Cosine rule — find an angle

5Area of triangle — sine formula

6The cosine of an obtuse angle [1 mark]

7Area of a regular octagon from its centre [2 marks]

8Cosine rule after an angle on a straight line [3 marks]

9Sine rule for an angle, then the area [4 marks]

10An angle at a point on the side of a triangle [5 marks]

Model AnswersFull mark-scheme-style answers

Area formula $\dfrac{1}{2}ab \sin C$