Sine, Cosine Rule & Area of Triangles (AQA GCSE Further Maths)

Revision Note

Paul

Written by: Paul

Reviewed by: Dan Finlay

  • Pythagoras' theorem and SOHCAHTOA only work in right-angled triangles

  • The Sine Rule and Cosine Rule allow us to answer triangle questions for ANY triangle

  • The Area of a Triangle Formula is an alternative to 1 half b h

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Sine & Cosine Rules

What are the sine rule and cosine rule formulae?

  • Before applying these formulas it is useful to label the triangle using a system of 'opposites'

General Triangle labelled, IGCSE & GCSE Maths revision notes
  • Note how lowercase letters are used for side lengths and capital letters for angles

    • make sure an angle and the side opposite it have the same letter

  • Sine Rule

fraction numerator bold italic a over denominator bold sin bold space bold italic A end fraction bold equals fraction numerator bold italic b over denominator bold sin bold space bold italic B end fraction bold equals fraction numerator bold italic c over denominator bold sin bold space bold italic C end fraction

  • sin 90° = 1 so if one of the angles is 90°, this becomes ‘SOH’ from SOHCAHTOA

  • The sine rule can also be 'flipped over'

fraction numerator bold sin bold space bold italic A over denominator bold italic a end fraction bold equals fraction numerator bold sin bold space bold italic B over denominator bold italic b end fraction bold equals fraction numerator bold sin bold space bold italic C over denominator bold italic c end fraction

  • This is more useful when we are using the rule to find angles

  • Cosine Rule

bold italic a to the power of bold 2 bold equals bold italic b to the power of bold 2 bold plus bold italic c to the power of bold 2 bold minus bold 2 bold italic b bold italic c bold cos bold space bold italic A

  • cos 90° = 0 so if A = 90°, this becomes Pythagoras’ Theorem

  • The cosine rule is 'cyclic' so there are two other versions of it

bold italic b to the power of bold 2 bold equals bold italic a to the power of bold 2 bold plus bold italic c to the power of bold 2 bold minus bold 2 bold italic a bold italic c bold cos bold space bold italic B
bold italic c to the power of bold 2 bold equals bold italic a to the power of bold 2 bold plus bold italic b to the power of bold 2 bold minus bold 2 bold italic a bold italic b bold cos bold space bold italic C

  • but once you are used to using cosine rule, there's only really one version

    • you just need to 'cycle' a, b and c around accordingly for a specific question

  • Note that it's always the angle between the two sides in the final term

  • Questions can become algebraic when side lengths are replaced by letters and expressions

    • Use the same rules but put brackets around expressions like (x + 3)

What is the ambiguous case of the sine rule?

  • In some circumstances, it is possible to get two possible answers using sine rule

    • this arises when the angle you are trying to find could be either acute or obtuse

  • Consider the triangle below

Sine-Rule-Ambiguous-case, IGCSE & GCSE Maths revision notes
  • You know the pair a and A and half of the pair c and C (c = 34)

    • However there are two possible locations for the third vertex of the triangle (labelled C1 and C2)

    • So there are two possible angles for C (x and y on the diagram)

  • (Most) calculators will only give one answer when using inverse sine (sin to the power of negative 1 end exponent)

    • this will be the acute angle

    • if the obtuse angle is required, subtract the acute angle from 180°

  • Questions will state, or imply, whether the acute or obtuse angle is required

    • or could ask you to find both possibilities!

  • There is no 'ambiguous' case for the cosine rule

How do I know whether to use sine rule or cosine rule?

  • Sine rule is used when you know a 'pair' and half of the pair involving the unknown we are asked to find

    • e.g.  a question may give us side a and angles A and B, then ask us to find side b
      (you know the pair a and A, and half of the pair b and B)

  • Cosine rule is used in two circumstances

    • when questions gives us an angle and the two sides that make it

      • this allows us to find the side opposite the given angle

      • e.g.  a question may give us sides b, c and angle A
        (this would allow us to find the side a)

    • when questions give us all three sides of a triangle and we are required to find an angle

      • in this case the rearranged version of cosine rule may be helpful

cos space A equals fraction numerator b squared plus c squared minus 2 b c space cos space A over denominator 2 b c end fraction

  • then use inverse cos (cos to the power of negative 1 end exponent) to find A

  • you may prefer to substitute known values into the 'regular' version of cosine rule and then rearrange

Examiner Tips and Tricks

  • In trickier exam questions, you may have to use both the cosine rule and the sine rule

  • For the sine rule, if you are finding a side, use the version of the formula that has the sides on the numerators; if you are finding an angle, put the sines on the numerators

  • If your calculator gives you a ‘Maths ERROR’ message when trying to find an angle using the cosine rule, you probably subtracted things the wrong way around when you rearranged the formula

Worked Example

The diagram below shows a sketch of a triangle, where angle ACB is obtuse.

General-Triangle-with-values-2, IGCSE & GCSE Maths revision notes

a) table row space row space end tableShow that the cosine of angle ABC is exactly 487 over 592.

You know all three lengths and want to show a result involving cosine - so use cosine rule!
Substitute values into cosine rule

4.4 squared equals 4.8 squared plus 7.4 squared minus 2 cross times 4.8 cross times 7.4 cross times cos space A B C

Evaluate where possible, ensuring not to round as an exact answer is required later

19.36 equals 77.8 minus 71.04 space cos space A B C

Rearrange to find an expression for cos space A B C

table row cell 71.04 space cos space A B C end cell equals cell 77.8 minus 19.36 end cell row cell cos space A B C end cell equals cell fraction numerator 58.44 over denominator 71.04 end fraction end cell end table

Use your calculator to write/check this fraction is the same as the required answer

bold therefore table row blank blank bold cos end table table row blank blank cell bold space bold italic A end cell end table table row blank blank cell bold italic B bold italic C end cell end table table row blank bold equals blank end table table row blank blank cell bold 487 over bold 592 end cell end table

b) A student uses the sine rule and the answer to part (a) to calculate angle ACB.
Their final answer, rounded to one decimal place is 73.0°.
Explain why the student's final answer cannot be correct and find the correct size for angle ACB.

Angle ACB is obtuse so 90 degree less than A C B less than 180 degree

The answer 73.0° comes from using sin to the power of negative 1 end exponent space open parentheses... close parentheses and a calculator will always give an acute angle
The question says angle ACB is obtuse
To find the correct angle we need to subtract 73.0° from 180°

180° - 73.0° = 107.0°

Angle ACB = 107.0° (1 d.p.)

This is commonly known as the ambiguous case of the sine rule

Area of a Triangle

What is the area of a triangle formula?

  • The Area of a Triangle formula applies to any triangle set up using the system of 'opposites'

General Triangle labelled, IGCSE & GCSE Maths revision notes

bold Area bold equals bold 1 over bold 2 bold italic a bold italic b bold sin bold space bold italic C

  • As with cosine rule, there are 'cyclic' alternatives to this

    • but rather than rememeber three formulae, remember one and change a, b and c accordingly

    • e.g.  1 half b c sin space A

  • sin 90° =1 so if = 90°, this becomes Area equals 1 half b h

  • In genreal 1 half a b sin space C allows us to find the area without having to know or find the height

    • it is particularly useful in non-right angled triangles

  • The questions encountered can be more algebra-based when side lengths are given as letters or expressions

    • use brackets around expressions like (x + 3)

    • check for any exact trigonometric values, like sin space 30 equals 1 half

Examiner Tips and Tricks

  • Try not to get bogged down in remembering different versions of the area of a triangle formula

    • Spot that to use 1 half a b sin space C you need an angle and the two sides that form it

Worked Example

The diagram below shows triangle PQR where side PQ is 2x and angle P is 30°.
Given that the ratio of the sides PQ to PR is 2 : 3, find an expression for the area of the triangle in terms of x.

picture-1

To use the area of a triangle formula we need two sides and the angle between them
So you need to find, in terms of x, the side PR using the given ratio

table row blank blank cell space space P space colon space R end cell row blank blank cell space space 2 space colon space 3 end cell row blank blank cell 2 x space colon space 3 x end cell end table

therefore space P R equals 3 x

Now apply the area of a triangle formula

Area equals 1 half open parentheses 2 x close parentheses open parentheses 3 x close parentheses sin space 30 degree

Simplifying the right-hand side including evaluating sin space 30 degree

Area equals 1 half open parentheses 6 x squared close parentheses open parentheses 1 half close parentheses

bold therefore bold space bold Area bold equals bold 3 over bold 2 bold italic x to the power of bold 2  (square units)

Applications of Trigonometry

How do I choose which rule or formula to use

  • It is important to be able to decide which Rule or Formula to use to answer a question

  • This table summarises the possibilities:

Sine & Cosine Rules, Area of Triangle – Harder table, IGCSE & GCSE Maths revision notes
Non-Right-Angled Triangles Diagram 2

What problems will I be asked to solve using trigonometry?

  • Problems in trigonometry may involve any, or more than one

    • Pythagoras' theorem

    • SOHCAHTOA

    • Sine rule

    • Cosine rule

    • Area of a triangle

  • Beyond 2D triangles you may be asked to solve problems involving

    • more complicated 2D shapes that have parts or sections that can be considered as a triangle

      • e.g. a trapezium

    • 3D shapes that have triangular parts or features

      • e.g. a triangular prism, the cross section of a cone

  • Many problems will involve algebraic answers

    • sides may be given in terms of a variable such as x

    • you may be asked to find an expression for a length or area "in terms of x"

  • Problems may involve surds, arising from exact values to common trigonometric ratios

    • e.g. sin space 60 degree equals fraction numerator square root of 3 over denominator 2 end fraction comma space space cos space 45 degree equals fraction numerator 1 over denominator square root of 2 end fraction open parentheses equals fraction numerator square root of 2 over denominator 2 end fraction close parentheses

    • You need to learn these exact values

Worked Example

The diagram below shows a cone with slant height of 2 x cm.
Given that angle ABC is 30°, show that the diameter of the base, d cm, is given by the expression  d squared equals 4 x squared open parentheses 2 minus square root of 3 close parentheses.

picture-2

BA = BC by symmetry
Draw a cross section through the centre of the cone and include any angles and lengths

sKerm2fD_cone-cross-section

Use the cosine rule with a = d, b = 2x, c = 2and A = 30°

 d squared equals open parentheses 2 x close parentheses squared plus open parentheses 2 x close parentheses squared minus 2 open parentheses 2 x close parentheses open parentheses 2 x close parentheses cos space 30

Simplify the powers and collect like terms 

table attributes columnalign right center left columnspacing 0px end attributes row cell d squared end cell equals cell 4 x squared plus 4 x squared minus 8 x squared space cos space 30 end cell row cell d squared end cell equals cell 8 x squared minus 8 x squared space cos space 30 end cell end table

Use the exact value of cos 30 and simplify
 d squared equals 8 x squared minus 8 x squared cross times fraction numerator square root of 3 over denominator 2 end fraction
d squared equals 8 x squared minus 4 x squared square root of 3
 

Factorise out 4x2 on the right-hand side

bold italic d to the power of bold 2 bold equals bold 4 bold italic x to the power of bold 2 stretchy left parenthesis 2 minus square root of 3 stretchy right parenthesis

Splitting the triangle into two right-angled triangles doesn't help with this question in a non-calculator paper (you do not know exact trig values for 15° or 75°)

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Paul

Author: Paul

Expertise: Maths Content Creator (Previous)

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.