The Sine Rule (Cambridge International IGCSE Mathematics)

Sine Rule

What is the sine rule?

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

  • It allows us to find missing side lengths or angles

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

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

• Where

  • $a$* *is the side opposite angle A
  • $b$* *is the side opposite angle B
  • $c$* *is the side opposite angle C

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

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

    • Then solve to find the side you need

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

  • 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 calculation, $\sin ^{- 1} (... )$

Exam Hint You are given the formula $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}$ in your exam.

You are not given the formula $\frac{\mathrm{sin}A}{a}=\frac{\mathrm{sin}B}{b}=\frac{\mathrm{sin}C}{c}$. However, you can derive it by flipping each fraction in the formula you are given.

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, $θ$, 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 $θ$

• The other base angle could either be obtuse or acute

  • The sine rule only gives the acute answer on your calculator

    • You need to check the diagram to see if the angle you need is actually obtuse
    • If it is, use this rule: obtuse angle = 180 - acute angle

Worked Example The following diagram shows triangle ABC.

$\mathrm{AB}=8.1\mathrm{cm}$, $\mathrm{BC}=12.3\mathrm{cm}$ and angle $\mathrm{BCA}=27^{\circ}$.

(a) Calculate the value of $x$.

Answer:

Label the sides of the diagram

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

$\frac{\mathrm{sin}A}{a}=\frac{\mathrm{sin}B}{b}=\frac{\mathrm{sin}C}{c}$ In practice, you only need to equate two of these three parts

$\begin{gathered} \frac{\sin x}{12.3} & = \frac{\sin 27}{8.1} \\ \sin x & = \frac{12.3 \sin 27}{8.1} \\ x = \sin ^{- 1} ( \frac{12.3 \sin 27}{8.1} ) \\ x = 43.58207... \end{gathered}$

Final answer: $43.6^{\circ}$ (to 1 d.p)

(b) Calculate the value of $y$.

Answer:

To find y you need to know the angle opposite (angle ABC) You know 27 and x from above, so subtract these from 180

$\begin{gathered} \text{Angle} A B C & = 180 - 27 - 43.58207... \\ = 109.41792... \end{gathered}$

y* is a length so use the sin rule with the sides on the top* $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}$

$\begin{gathered} \frac{y}{\sin ( 109.41792... )} & = \frac{8.1}{\sin 27} \\ y = \frac{8.1 \sin ( 109.41792... )}{\sin 27} \\ y = 16.82691... \end{gathered}$

Final answer: $y=16.8\mathrm{cm}$(to 3 s.f.)