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G22–G23Geometry & MeasuresHigherGrades 7–9

Sine & Cosine Rules

SOHCAHTOA only works in right-angled triangles. The sine rule, the cosine rule and the area formula 12absin⁡C\tfrac12 ab\sin C work in every triangle. They let you find missing sides, angles and areas in surveying, navigation and 3D problems.

50 min AQA GCSE Maths

The big picture

Label the triangle with capital letters for angles and matching lower-case letters for the opposite sides: side aa is opposite angle AA. With that labelling, every formula is easy to state and check.

The choice of rule depends on what you know. A side and its opposite angle as a pair means the sine rule. Two sides and the angle between them, or all three sides, means the cosine rule.

What you'll learn

  • Use the sine rule to find a missing side or angle
  • Use the cosine rule to find a missing side or angle
  • Find the area of a triangle using 12absin⁡C\tfrac12 ab\sin C
  • Choose the correct rule and solve multi-step problems
1

The sine rule

asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}. For angles, flip it: sin⁡Aa=sin⁡Bb\dfrac{\sin A}{a} = \dfrac{\sin B}{b}.

Use it when you know a side and its opposite angle.

Example 1In triangle ABCABC, A=40∘A = 40^{\circ}, B=75∘B = 75^{\circ} and a=8a = 8 cm. Find bb.
1bsin⁡75∘=8sin⁡40∘\dfrac{b}{\sin 75^{\circ}} = \dfrac{8}{\sin 40^{\circ}}.
2b=8sin⁡75∘sin⁡40∘=12.02…b = \dfrac{8\sin 75^{\circ}}{\sin 40^{\circ}} = 12.02\dots
Answerb=12.0b = 12.0 cm (3 s.f.)
Your turnSame method, new numbers — try it before moving on.

In triangle ABCABC, A=35∘A = 35^\circ, B=70∘B = 70^\circ and a=6a = 6 cm. Find bb to 1 decimal place.

Show worked solution
Solution
  1. 1bsin⁡70∘=6sin⁡35∘\dfrac{b}{\sin 70^\circ} = \dfrac{6}{\sin 35^\circ}.
  2. 2b=6sin⁡70∘sin⁡35∘=9.829…b = \dfrac{6 \sin 70^\circ}{\sin 35^\circ} = 9.829\ldots

Answer: 9.89.8 cm

Example 2In triangle ABCABC, a=12a = 12 cm, b=9b = 9 cm and A=50∘A = 50^{\circ}. Find angle BB.
1sin⁡B9=sin⁡50∘12\dfrac{\sin B}{9} = \dfrac{\sin 50^{\circ}}{12}.
2sin⁡B=0.5745…\sin B = 0.5745\dots, so B=sin⁡−1(0.5745)B = \sin^{-1}(0.5745).
AnswerB=35.1∘B = 35.1^{\circ}
2

The cosine rule

For a side: a2=b2+c2−2bccos⁡Aa^{2} = b^{2} + c^{2} - 2bc\cos A.

For an angle: cos⁡A=b2+c2−a22bc\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}.

Use it with two sides and the included angle, or with three sides.

Example 1Two sides of a triangle are 9 cm and 14 cm with an angle of 65∘65^{\circ} between them. Find the third side.
1x2=92+142−2(9)(14)cos⁡65∘x^{2} = 9^{2} + 14^{2} - 2(9)(14)\cos 65^{\circ}.
2x2=277−106.5⋯=170.5…x^{2} = 277 - 106.5\dots = 170.5\dots
Answerx=13.1x = 13.1 cm (3 s.f.)
Your turnSame method, new numbers — try it before moving on.

Two sides of a triangle are 7 cm and 11 cm, with an angle of 48∘48^\circ between them. Find the third side to 1 decimal place.

Show worked solution
Solution
  1. 1c2=72+112−2×7×11cos⁡48∘=66.95…c^2 = 7^2 + 11^2 - 2 \times 7 \times 11 \cos 48^\circ = 66.95\ldots
  2. 2c=8.18…c = 8.18\ldots, so 8.28.2 cm.

Answer: 8.28.2 cm

Example 2A triangle has sides 5 cm, 8 cm and 10 cm. Find the largest angle.
1The largest angle is opposite the 10 cm side.
2cos⁡θ=52+82−1022(5)(8)=−1180\cos \theta = \dfrac{5^{2} + 8^{2} - 10^{2}}{2(5)(8)} = -\dfrac{11}{80}.
Answerθ=97.9∘\theta = 97.9^{\circ}
Why this works

A negative cosine means the angle is obtuse. The cosine rule handles this automatically, which is why it is the safer choice for the largest angle of a triangle.

3

Area of a triangle

Area =12absin⁡C= \dfrac12 ab\sin C, where CC is the angle between sides aa and bb.

Example 1Find the area of a triangle with sides 7 m and 10 m and an included angle of 115∘115^{\circ}.
1Area =12×7×10×sin⁡115∘= \dfrac12 \times 7 \times 10 \times \sin 115^{\circ}.
Answer31.731.7 m² (3 s.f.)
Your turnSame method, new numbers — try it before moving on.

Find the area of a triangle with sides 8 cm and 12 cm and an included angle of 30∘30^\circ, in cm².

Show worked solution
Solution
  1. 1Area =12×8×12×sin⁡30∘= \frac{1}{2} \times 8 \times 12 \times \sin 30^\circ.
  2. 2=48×0.5=24= 48 \times 0.5 = 24 cm².

Answer: 2424 cm²

Tip — Keep calculator values unrounded between steps and round only the final answer.

Think like an examiner

  • Write the formula with values substituted before calculating.
  • Check your calculator is in degree mode.
  • In multi-step problems, state which triangle and rule you are using at each step.

Remember these

asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}
Sine rule.
a2=b2+c2−2bccos⁡Aa^{2} = b^{2} + c^{2} - 2bc\cos A
Cosine rule.
Area=12absin⁡C\text{Area} = \tfrac12 ab\sin C
Area of any triangle.

Watch out for these

Calculating b2+c2−2bcb^{2} + c^{2} - 2bc first and then multiplying by cos⁡A\cos A.
Work out 2bccos⁡A2bc\cos A as one term, then subtract.
Using the sine rule without a side–angle pair.
With two sides and the included angle, use the cosine rule.
Using a side that is not next to the angle in 12absin⁡C\tfrac12 ab\sin C.
The angle must be between the two sides.
Rounding an intermediate side, then using it again.
Use the stored calculator value.

Stretch yourself

A parallelogram has sides 6 cm and 9 cm with an angle of 70∘70^{\circ} between them. Find its area and the lengths of both diagonals.

Hint — The parallelogram is two congruent triangles. Adjacent angles add to 180∘180^{\circ}.

Reveal solution
SolutionArea =2×12×6×9×sin⁡70∘=50.7= 2 \times \tfrac12 \times 6 \times 9 \times \sin 70^{\circ} = 50.7 cm². Short diagonal (opposite the 70∘70^{\circ} angle): d2=36+81−108cos⁡70∘=80.06d^{2} = 36 + 81 - 108\cos 70^{\circ} = 80.06, so d=8.95d = 8.95 cm. Long diagonal (opposite the 110∘110^{\circ} angle): D2=36+81−108cos⁡110∘=153.94D^{2} = 36 + 81 - 108\cos 110^{\circ} = 153.94, so D=12.4D = 12.4 cm. ■\blacksquare

Questions students ask

Do the rules work for right-angled triangles?
Yes. With a 90∘90^{\circ} angle the cosine rule becomes Pythagoras, since cos⁡90∘=0\cos 90^{\circ} = 0.
Which form of the sine rule should I use?
Put the unknown on top: sides on top to find a side, sines on top to find an angle.
Can the sine rule give two answers?
When finding an angle it can, because sin⁡x=sin⁡(180∘−x)\sin x = \sin(180^{\circ} - x). Check whether an obtuse answer is possible.

Key takeaways

  • Side aa is opposite angle AA.
  • Sine rule: when you have a side and its opposite angle.
  • Cosine rule: two sides and the included angle, or three sides.
  • Area =12absin⁡C= \tfrac12 ab\sin C with the included angle.

Exam-style questions

Practise it the way it is examined

3 original questions · 11 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 75 marks

In triangle PQRPQR, PQ=9PQ = 9 cm, QR=12QR = 12 cm and angle PQR=105∘PQR = 105^\circ.

  1. (a)Work out the length of PRPR. Give your answer to 3 significant figures.[3]
    Show mark scheme (a)

    Answer: 16.816.8 cm

  2. (b)Work out the area of the triangle.[2]
    Show mark scheme (b)

    Answer: 52.252.2 cm²

Answer on paper first, then mark it.
Question 2Grade 83 marks
  1. A triangle has sides of 6 cm, 9 cm and 13 cm. Work out the size of its largest angle.[3]
    Show mark scheme

    Answer: 118.8∘118.8^\circ

Answer on paper first, then mark it.
Question 3Grade 93 marks
  1. A triangle has sides of 8 cm and 10 cm and an area of 30 cm². The angle between these sides is acute. Find the angle.[3]
    Show mark scheme

    Answer: 48.6∘48.6^\circ

Answer on paper first, then mark it.

Unlimited practice

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