Geometry and measures

Sine rule, cosine rule and area of a triangle

Q: Is the sine rule on the Foundation paper?

No. The sine rule, cosine rule and \frac{1}{2}ab\sin C are Higher tier only.

19 past-paper questions, easiest first. Try each one, then reveal the answer.

Higher only

Revision guide by The Full Marks

Sine rule, cosine rule and area of a triangle: what you need to know

The sine rule and cosine rule let you find sides and angles in triangles that have no right angle, and $\frac{1}{2}ab\sin C$ finds their area. This is Higher only and often appears as a multi-step problem.

Key facts

Label sides with lower case letters opposite the matching angle: side $a$ is opposite angle $A$ .
Sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ (flip it to $\frac{\sin A}{a} = \frac{\sin B}{b}$ to find an angle).
Cosine rule: $a^2 = b^2 + c^2 - 2bc\cos A$ , or for an angle: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ .
Area of a triangle $= \frac{1}{2}ab\sin C$ , where $C$ is the angle between sides $a$ and $b$ .
Non-calculator questions use exact values: $\sin 30^\circ = \frac{1}{2}$ , $\sin 45^\circ = \frac{\sqrt{2}}{2}$ , $\sin 60^\circ = \frac{\sqrt{3}}{2}$ .

Choosing the right rule

Know a side and its opposite angle, plus one more side or angle? Use the sine rule. (Missing an angle? Angles in a triangle add up to $180^\circ$ .)
Know two sides and the angle between them, or all three sides? Use the cosine rule.
Need the area and know two sides with the angle between them? Use $\frac{1}{2}ab\sin C$ . Given the area? Substitute it in and rearrange to find a side or angle.
In two-part problems, find a shared side with one rule, then use it in the next triangle.

Worked examples

Example 1 Higher · grade 6

Triangle $PQR$ has $PQ = 7$ cm, $PR = 9$ cm and angle $QPR = 40^\circ$ . Find its area, to 3 significant figures.

Show the solution

The $40^\circ$ angle is between the two given sides.
Area $= \frac{1}{2} \times 7 \times 9 \times \sin 40^\circ$
$= 20.247\ldots$
Area $= 20.2$ cm $^2$

Example 2 Higher · grade 7

In triangle $ABC$ , $AB = 8$ cm, $AC = 11$ cm and angle $BAC = 65^\circ$ . Find $BC$ , then find angle $ABC$ . Give answers to 1 decimal place.

Show the solution

Two sides and the angle between them, so use the cosine rule.
$BC^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos 65^\circ$
$= 185 - 74.38\ldots = 110.61\ldots$
$BC = 10.517\ldots = 10.5$ cm
Now use the sine rule. $AC = 11$ is opposite angle $B$ .
$\sin B = \frac{11 \times \sin 65^\circ}{10.517\ldots} = 0.9479\ldots$
$B = 71.4^\circ$

Common mistakes

Pairing a side with the wrong angle in the sine rule. Each side must go with the angle opposite it.
Working out $b^2 + c^2 - 2bc$ first and then multiplying by $\cos A$ . Do $2bc\cos A$ first, then subtract it.
Using $\frac{1}{2} \times$ base $\times$ height with two sloping sides when there is no right angle.

Exam tips

Learn the sine rule, the cosine rule and $\frac{1}{2}ab\sin C$ so you can recall and use them quickly.
Keep the full calculator value between steps; rounding $BC$ early can push the final angle out of range.
Inverse sine always gives an acute angle. If the angle you want could be obtuse, find it with the cosine rule instead. Check your calculator is in degrees.

Questions people ask

When do you use the sine rule and when the cosine rule?

Use the sine rule when you know a side and the angle opposite it. Use the cosine rule when you know two sides and the angle between them, or all three sides.

Is the sine rule on the Foundation paper?

No. The sine rule, cosine rule and $\frac{1}{2}ab\sin C$ are Higher tier only.

Exam questions: Sine rule, cosine rule and area of a triangle (19, easiest first)

Sine rule, cosine rule and area of a triangle2 marksGrade 6CalculatorHigher

November 2022 · Paper 2 Higher · Q18

Find the area of a triangle from two sides and the angle between them.

Show answer

Answer	$21.3\text{ cm}^2$	2 marks

Worked solution Checked against the mark scheme

Area $= \frac{1}{2}ab\sin C$

$= \frac{1}{2} \times 11.2 \times 4.3 \times \sin 118° = 21.26...$

Area $= 21.3\text{ cm}^2$ (3 s.f.)

Watch out: Using ½ × base × height with the two given sides.

Sine rule, cosine rule and area of a triangle3 marksGrade 7No calculatorHigher

November 2024 · Paper 1 Higher · Q14

Use the sine rule with exact values of sin 30° and sin 45° to find a side of a triangle without a calculator.

Show answer

Answer	$6$ cm	3 marks

Worked solution Checked against the mark scheme

Sine rule: $\frac{AB}{\sin 45°} = \frac{3\sqrt{2}}{\sin 30°}$

$\sin 45° = \frac{\sqrt{2}}{2}$ , $\sin 30° = \frac{1}{2}$

$AB = 3\sqrt{2} \times \frac{\sqrt{2}}{2} \div \frac{1}{2} = 3\sqrt{2} \times \sqrt{2} = 6$ cm

Watch out: Matching each side with the wrong opposite angle.

Sine rule, cosine rule and area of a triangle4 marksGrade 7CalculatorHigher

November 2024 · Paper 3 Higher · Q17

Find the area of a triangle given all three sides, using the cosine rule then ½ab sin C.

Show answer

Answer	$56.0$ cm²	4 marks

Worked solution Checked against the mark scheme

Angle A is between 14.6 and 7.9 cm:

$\cos A = \frac{14.6^2 + 7.9^2 - 18.2^2}{2 \times 14.6 \times 7.9} = -0.2413...$ , so $A = 103.97^\circ$

Area $= \frac{1}{2} \times 14.6 \times 7.9 \times \sin 103.97^\circ = 55.97 \approx 56.0$ cm²

Watch out: Pairing an angle with the wrong two sides in the area formula.

Sine rule, cosine rule and area of a triangle4 marksGrade 7CalculatorHigher

June 2023 · Paper 2 Higher · Q13

Use the sine rule to find one angle of a non-right-angled triangle, then angles in a triangle to find another.

Show answer

Answer	$58.5°$	4 marks

Worked solution Checked against the mark scheme

Angle $C$ is opposite $AB = 15$ ; angle $B = 70°$ is opposite $AC = 18$ .

$\frac{\sin C}{15} = \frac{\sin 70°}{18}$ , so $\sin C = \frac{15 \sin 70°}{18} = 0.7830...$

$C = 51.54...°$

$BAC = 180 - 70 - 51.54... = 58.5°$

Watch out: Stopping at angle C (51.5°) instead of going on to find angle A.

Sine rule, cosine rule and area of a triangle4 marksGrade 7CalculatorHigher

November 2023 · Paper 2 Higher · Q16

In a trapezium, use the sine rule then the sine area formula to find the area of a triangle.

Show answer

Answer	$56.0$ cm²	4 marks

Worked solution Checked against the mark scheme

In triangle $ABD$ : $\frac{BD}{\sin 112°} = \frac{7.9}{\sin 42°}$ , so $BD = 10.946...$ cm.

AD is parallel to BC, so angle $DBC = 42°$ (alternate angles).

Area of $BCD = \frac{1}{2} \times 10.946... \times 15.3 \times \sin 42° = 56.03... \approx 56.0$ cm²

Watch out: Not spotting that angle DBC equals 42° by alternate angles.

Sine rule, cosine rule and area of a triangle4 marksGrade 7CalculatorHigher

June 2023 · Paper 3 Higher · Q14

Use the cosine rule to find a radius, then work out the area of a sector.

Show answer

Answer	19.9 cm²	4 marks

Worked solution Checked against the mark scheme

Cosine rule: $OB^2 = 9^2 + 6^2 - 2 \times 9 \times 6 \cos 35° = 28.53...$

$OB = 5.341...$ cm

Sector area $= \dfrac{80}{360} \times \pi \times 28.53... = 19.918...$

$= 19.9$ cm² (3 s.f.)

Watch out: Working out $81 + 36 - 108$ before multiplying by $\cos 35°$ .

Sine rule, cosine rule and area of a triangle4 marksGrade 7No calculatorHigher

November 2022 · Paper 1 Higher · Q22

Use the sine rule without a calculator to find the exact sine of an angle in a triangle.

Show answer

Answer	$\frac{65}{214}$	4 marks

Worked solution Checked against the mark scheme

Sine rule: $\frac{\sin ABC}{6.5} = \frac{\sin 30°}{10.7}$

$\sin ABC = \frac{6.5 \times 0.5}{10.7} = \frac{3.25}{10.7}$

Multiply top and bottom by 20: $\frac{65}{214}$

Watch out: Pairing 6.5 with the 30° angle instead of 10.7.

Sine rule, cosine rule and area of a triangle4 marksGrade 7CalculatorHigher

June 2022 · Paper 2 Higher · Q18

Use the cosine rule then the sine rule across two triangles to find a length.

Show answer

Answer	$7.63\text{ cm}$	4 marks

Worked solution Checked against the mark scheme

Triangle PQR: $QR^2 = 11^2 + 9.4^2 - 2 \times 11 \times 9.4 \cos 27^\circ = 25.09...$ , so $QR = 5.0099...$

Triangle QRS: $\frac{QS}{\sin 88^\circ} = \frac{QR}{\sin 41^\circ}$

$QS = \frac{5.0099... \times \sin 88^\circ}{\sin 41^\circ} = 7.6318...$ , so $QS = 7.63$ cm

Watch out: Pairing QS with the wrong angle in the sine rule (it is opposite the 88° angle).

Sine rule, cosine rule and area of a triangle5 marksGrade 7CalculatorHigher

November 2021 · Paper 2 Higher · Q15

Use the cosine rule to find a side of a triangle, then find its area using ½ab sin C.

Show answer

(a)	$11.4$ cm	3 marks
(b)	$41.8$ cm²	2 marks

Worked solution Checked against the mark scheme

(a) $BC^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \cos 72° = 185 - 54.38 = 130.61...$

$BC = \sqrt{130.61...} = 11.4$ cm.

(b) Area $= \frac{1}{2} \times 8 \times 11 \times \sin 72° = 41.8$ cm².

Watch out: Working out $185 - 176$ first and then multiplying by cos 72.

Sine rule, cosine rule and area of a triangle3 marksGrade 7CalculatorHigher

November 2020 · Paper 2 Higher · Q13

Find a side of a non-right-angled triangle using the sine rule after working out the missing angle.

Show answer

Answer	15.4	3 marks

Worked solution Checked against the mark scheme

Angle $A = 180° - 34° - 26° = 120°$ , which is opposite the 23.8 cm side.

$\frac{AB}{\sin 34°} = \frac{23.8}{\sin 120°}$

$AB = \frac{23.8 \times \sin 34°}{\sin 120°} = 15.367...$

$AB = 15.4$ cm (1 d.p.)

Watch out: Pairing 23.8 cm with 26° or 34° instead of the opposite angle, 120°.

Sine rule, cosine rule and area of a triangle4 marksGrade 7CalculatorHigher

June 2019 · Paper 3 Higher · Q14

Use the sine area formula for two congruent triangles and a side ratio to find a rectangle's length.

Show answer

Answer	36 cm	4 marks

Worked solution Checked against the mark scheme

Area of triangle $AFE = \frac{1}{2} \times 24 \times AE \times \sin 30° = 6\,AE$

Two triangles: $12\,AE$

Rectangle: $AB \times AE = 12\,AE$ , so $AB = 12$

$AE = 3 \times AB = 36$ cm

Watch out: Using ½ × base × height with 24 as the height.

Sine rule, cosine rule and area of a triangle5 marksGrade 7CalculatorHigher

June 2018 · Paper 3 Higher · Q17

Use the sine rule in one triangle and then the cosine rule in the adjoining triangle to find a length.

Show answer

Answer	$13.1$ cm	5 marks

Worked solution Checked against the mark scheme

In triangle $BCD$ : $\frac{BD}{\sin 34°} = \frac{12.5}{\sin 109°}$

$BD = \frac{12.5 \sin 34°}{\sin 109°} = 7.393$ cm

In triangle $ABD$ : $AD^2 = 11.4^2 + 7.393^2 - 2(11.4)(7.393)\cos 86°$

$AD^2 = 129.96 + 54.65 - 11.76 = 172.85$

$AD = 13.1$ cm

Watch out: Working out $(129.96 + 54.65 - 2 \times 11.4 \times 7.393) \times \cos 86°$ in the wrong order.

Sine rule, cosine rule and area of a triangle5 marksGrade 8CalculatorHigher

June 2024 · Paper 3 Higher · Q18

Use a triangle's area to find a shared side, then the sine rule and area formula to find another triangle's area.

Show answer

Answer	$52.5$ cm² (accept 52.4 to 52.52)	5 marks

Worked solution Checked against the mark scheme

$\frac12 \times 17 \times AC \times \sin 35° = 54 \Rightarrow AC = 11.076$ cm

Angle $CAD = 180° - 48° - 57° = 75°$

$CD = \dfrac{11.076 \sin 75°}{\sin 57°} = 12.757$ cm

Area $= \frac12 \times 11.076 \times 12.757 \times \sin 48° = 52.5$ cm²

Watch out: Using 17 cm as a side of triangle ACD, or rounding AC too early.

Sine rule, cosine rule and area of a triangle4 marksGrade 8No calculatorHigher

November 2023 · Paper 1 Higher · Q20

Use the cosine rule with a surd side length to find an angle in a triangle without a calculator.

Show answer

Answer	$120°$	4 marks

Worked solution Checked against the mark scheme

$(5\sqrt{7})^2 = 25 \times 7 = 175$

$\cos B = \frac{10^2 + 5^2 - 175}{2 \times 10 \times 5} = \frac{125 - 175}{100} = -\frac{1}{2}$

$\cos 60° = \frac{1}{2}$ , so $B = 180° - 60° = 120°$

Watch out: Squaring 5√7 as 35 or 5 × 49.

Sine rule, cosine rule and area of a triangle5 marksGrade 8CalculatorHigher

November 2022 · Paper 3 Higher · Q26

Find a diagonal of a regular heptagon from the area of a triangle formed by two sides.

Show answer

Answer	15.8 cm	5 marks

Worked solution Checked against the mark scheme

Interior angle: $180 - 360 \div 7 \approx 128.57^\circ$

Area: $\frac{1}{2}s^2 \sin 128.57^\circ = 30$ , so $s = \sqrt{60 \div \sin 128.57^\circ} \approx 8.760$ cm

Cosine rule: $GB^2 = 2s^2 - 2s^2 \cos 128.57^\circ \approx 249.2$

$GB \approx 15.8$ cm

Watch out: Using the exterior angle (51.4°) in the area formula.

Sine rule, cosine rule and area of a triangle5 marksGrade 8CalculatorHigher

June 2019 · Paper 3 Higher · Q23

Use bearings, the cosine rule and the sine rule to find the bearing of one town from another.

Show answer

Answer	$098.6°$	5 marks

Worked solution Checked against the mark scheme

The back-bearing of $A$ from $B$ is $217°$ , so $ABC = 217 - 150 = 67°$ .

$AC^2 = 8^2 + 9^2 - 2 \times 8 \times 9 \cos 67° = 88.73$ , $AC = 9.420$ km

$\sin BAC = \frac{9 \sin 67°}{9.420} \Rightarrow BAC = 61.58°$

Bearing $= 37 + 61.58 = 098.6°$

Watch out: Using 150° − 37° = 113° as angle ABC.

Sine rule, cosine rule and area of a triangle5 marksGrade 8CalculatorHigher

November 2019 · Paper 3 Higher · Q18

Use the cosine rule then the sine rule to find a length in a triangle split by a line from one vertex.

Show answer

Answer	$1.95$ cm	5 marks

Worked solution Checked against the mark scheme

Cosine rule: $\cos C = \dfrac{6.1^2 + 6.2^2 - 3.4^2}{2 \times 6.1 \times 6.2}$ , so $C = 32.08°$

Angle DAC $= \frac{2}{5} \times 32.08 = 12.83°$

Angle ADC $= 180 - 32.08 - 12.83 = 135.09°$

Sine rule: $DC = \dfrac{6.2 \sin 12.83°}{\sin 135.09°} = 1.95$ cm

Watch out: Using 6.1 cm (BC) as the side opposite angle ADC instead of AC.

Sine rule, cosine rule and area of a triangle5 marksGrade 8CalculatorHigher

November 2018 · Paper 3 Higher · Q16

Find the perimeter of a shape made of a triangle and a major sector, using the sine rule for a missing side.

Show answer

Answer	$39.9$ cm	5 marks

Worked solution Checked against the mark scheme

The sector's outer arc uses $360 - 140 = 220°$ :

Arc $= \frac{220}{360} \times 2\pi \times 6 = 23.04$ cm

Sine rule in triangle AOD: $\frac{OD}{\sin 24°} = \frac{14}{\sin 140°}$ , so $OD = 8.859$ cm

$CD = OD - 6 = 2.859$ cm

Perimeter $= 23.04 + 14 + 2.859 = 39.9$ cm

Watch out: Using the 140° minor arc instead of the 220° major arc.

Sine rule, cosine rule and area of a triangle4 marksGrade 9CalculatorHigher

November 2021 · Paper 3 Higher · Q23

Prove that an angle bisector in a triangle splits the opposite side in the same ratio as the two adjacent sides.

Show answer

Answer	Proof using the sine rule in both small triangles and $\sin(180^\circ - y) = \sin y$	4 marks

Worked solution Checked against the mark scheme

Let angle $ADB = y$ , so angle $ADC = 180^\circ - y$ .

Sine rule in $ABD$ : $\frac{AB}{\sin y} = \frac{BD}{\sin x}$ , so $\frac{AB}{BD} = \frac{\sin y}{\sin x}$

Sine rule in $ADC$ : $\frac{AC}{DC} = \frac{\sin(180^\circ - y)}{\sin x}$

Since $\sin(180^\circ - y) = \sin y$ , both ratios equal $\frac{\sin y}{\sin x}$ , so $\frac{AB}{BD} = \frac{AC}{DC}$ .

Watch out: Not justifying why the sines of the two angles at D are equal.

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

June 2022 · Paper 1 Higher · Q21: Find the exact area of the two shaded regions formed inside the middle of three overlapping equal circles.
November 2020 · Paper 3 Higher · Q19: Show that the area between a regular hexagon and its enlargement by scale factor p is a given expression.
November 2019 · Paper 2 Higher · Q23: Form a quadratic inequality from the area of a parallelogram using ½ab sin C, then solve it.
November 2018 · Paper 2 Higher · Q13: Use a circle's circumference to find its diameter, then find the area of an equilateral triangle built on that diameter.