National 5MathsSQARevision NotesTrigonometric SkillsTrigonometry with TrianglesSine Rule

Sine Rule (SQA National 5 Maths): Revision Note

Exam code: X847 75

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Using sine rule to find a length

What is the sine rule?

  • The sine rule is used in non right-angled triangles

    • It allows you to find missing side lengths or angles

  • It states that for any triangle with angles A, B and C

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

  • Where

    • a is the side opposite angle A

    • b is the side opposite angle B

    • c is the side opposite angle C

Non Right-Angled Triangle labelled with angles A, B and C and opposite corresponding sides a, b and c.

Examiner Tips and Tricks

The Sine Rule formula is given to you on the Formulae List in the exam paper.

How do I use the sine rule to find missing lengths?

  • Use the sine rule

    • when you have opposite pairs of sides and angles in the question

      • a and A, or b and B, or c and C

  • Start by labelling your triangle with the angles and sides

    • Angles have upper case letters

    • Sides opposite the angles have the equivalent lower case letter

  • To find a missing length, substitute numbers into the formula

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

    • You only need to have two parts equal to each other (not all three)

      • Then solve to find the side you need

Using sine rule to find an angle

How do I use the sine rule to find missing angles?

  • To find a missing angle, it is easier to rearrange the formula first by flipping each part

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

    • The angles are now in the numerators of the fractions

    • Substitute the values you have into the formula and solve

      • You will need to use inverse sine in your calculator, sin to the power of negative 1 end exponent open parentheses... close parentheses

Worked Example

The diagram shows triangle ABC. 

  • Angle ACB = 25°

  • Angle BAC = x°

  • AB = 8.1 cm

  • BC = 12.3 cm

  • AC = space y cm

Triangle ABC with AB = 8.1 cm, BC = 12.3 cm, AC = y cm, angle BAC = xº and angle BCA = 27º.

a) Calculate the value of x.

b) Calculate the value of space y.

Answer:

Part (a)

Label the sides of the triangle

Triangle ABC with sides opposite angles labelled with corresponding lowercase letters.

x is an angle so use the sine rule with the angles on top

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

In questions, you only need to equate two of these three parts

table row cell fraction numerator sin space x over denominator 12.3 end fraction end cell equals cell fraction numerator sin 27 over denominator 8.1 end fraction end cell row cell sin space x end cell equals cell 12.3 cross times fraction numerator sin 27 over denominator 8.1 end fraction end cell row x equals cell sin to the power of negative 1 end exponent open parentheses 12.3 cross times fraction numerator sin 27 over denominator 8.1 end fraction close parentheses end cell row x equals cell 43.582077... end cell end table

Round to a sensible degree of accuracy

  • Unless a question tells you otherwise, 1 decimal place is usually a good choice for angles

x equals 43.6 degree space space open parentheses 1 space straight d. straight p. close parentheses

Part (b)

To find y you need to know the angle opposite (angle ABC)

  • You know 27 and x from above, so subtract these from 180

table row cell Angle space ABC end cell equals cell 180 minus 27 minus 43.582077... end cell row blank equals cell 109.417922... end cell end table

y is a length so use the sin rule with the sides on the top

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

(You could also use the Cosine Rule to find y, but the Sine Rule is a much simpler formula!)

table row cell fraction numerator y over denominator sin open parentheses 109.417922... close parentheses end fraction end cell equals cell fraction numerator 8.1 over denominator sin 27 end fraction end cell row y equals cell fraction numerator 8.1 over denominator sin space 27 end fraction cross times sin open parentheses 109.417922... close parentheses end cell row y equals cell 16.826919... end cell end table

Round to a sensible degree of accuracy

  • Unless a question tells you otherwise, 3 significant figures is usually a good choice

y equals 16.8 space space open parentheses 3 space straight s. straight f. close parentheses

What is the ambiguous case of the sine rule?

  • Given information about a triangle, there may be two different ways to draw it

  • In the diagram below, the lengths of two sides are given, a and b

    • A base angle is also given, theta, but no angle near b is given

    • It turns out that there are two possible ways to arrange b to complete the triangle!

      • Both triangles have the correct values of a, b and theta

  • The other base angle could either be obtuse or acute

    • After using the sine rule, using sin-1 on your calculator only gives the acute angle

      • You need to check the diagram or the other information given in the question to see if the angle you need is actually obtuse

      • If it is, use this rule: obtuse angle = 180 - acute angle

aa-sl-3-3-2-ambiguous-sine-rule-diagram-1

Examiner Tips and Tricks

A case of the ambiguous sine rule is unlikely to appear on your exam.

However knowing that the sine of an acute angle is equal to the sine of 180° minus that acute angle is an important skill for solving trigonometric equations.

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths Content Creator

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject 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.