 # Understanding the Sine Rule and Cosine Rule for GCSE Maths The sine rule and cosine rule are trigonometric laws that are used to work out unknown sides and angles in any triangle. Every GCSE Maths student needs a working knowledge of trigonometry, and the sine and cosine rules will be indispensable in your exam.

Every triangle has six measurements: three sides and three angles. To work out any unknown angles or sides you need to know at least three of these measurements in any combination, with the exception of the three angles. You can use either the cosine or sine rule – the choice depends on what you are looking for and what is given. In this quick lesson, we’ll go over the rules and how to use them to solve GCSE-level trigonometry questions.

## 1. The sine rule

Triangles are not always conveniently labelled and it is very important that you perform the following step to avoid confusion. Take any triangle ABC and label the sides a, b, c and the corresponding opposite angles A, B, C like in the diagram below.

The sine rule states that: $\frac{a}{sin\ A} \ =\ \frac{b}{sin\ B} \ =\ \frac{c}{sin\ C}$

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

### Let’s apply this to an actual GCSE-style example

#### Example 1: Work out the value of the side $x$

In this case we use the sine rule with sides on top (the first one). $\mathnormal{\frac{x}{sin\ 84 \textdegree} \ =\ \frac{25}{sin\ 47 \textdegree}}$

Multiplying both sides by $sin\ 84\textdegree$ we have: $x\ =\ \mathnormal{\frac{25\ sin\ 84\textdegree}{sin\ 47\textdegree}} \ =\ 34.0\ cm\ to\ 3\ s.f.$

#### Example 2: Work out the value of the angle $x$

In this example we use the sine rule with the sines on top (the second one). We also need a calculator to work out the inverse sine. $sin\ x=\ \mathnormal{\frac{7\ sin 40 \textdegree}{6}}$

Multiplying both sides by 7 we obtain: $sin\ x\ =\ \mathnormal{\frac{7\ sin 40 \textdegree}{6}}$ $x\ =\ sin^{-1} 0.7499$

Therefore: $x=\ 48.6 \textdegree to\ 3\ s.f.$

## 2. The cosine rule

The cosine rule can be used to find unknown sides, for a triangle like in the diagram below. It can be expressed using the following formulas:

1. $a^{2} \ =\ b^{2} \ +\ c^{2} \ -\ 2bc\ cosA$
2. $b^{2} \ =\ a^{2} \ +\ c^{2} \ -\ 2ac\ cosB$
3. $c^{2} = a^{2} + b^{2} - 2ab cosC$

The above formulas can also be rearranged to work out unknown angles using inverse operations: $cos\ A\ =\ \mathnormal{\frac{b^{2} +\ c^{2} -a^{2}}{ \begin{array}{l} 2bc\ \end{array}}}$ $cos\ B\ =\ \mathnormal{\frac{b^{2} +\ c^{2} -a^{2}}{ \begin{array}{l} 2bc\ \end{array}}}$ $cos\ C\ =\ \mathnormal{\frac{a^{2} +\ b^{2} -c^{2}}{ \begin{array}{l} 2ab\ \end{array}}}$

### Let’s try them out in some more GCSE-level questions.

#### Example 3: Work out the value of the side $x$ in the following triangle

Using the cosine rule 1 we obtain: $x^{2} \ =\ 6^{2} \ +\ 10^{2} \ -\ 2\ \times \ 6\ \times \ 10\ \times \ cos\ 80\textdegree$ $x^{2} =115.16$

Therefore: $x=10.7\ to\ 3\ s.f.$

#### Example 4: Work out the value of the side $x$ in the following triangle $cos\ x\ =\ \mathnormal{\frac{5^{2} +\ 7^{2} -8^{2}}{ \begin{array}{l} 2\ \times \ 5\ \times \ 7\ \end{array}}}$ $cos\ x=0.1428$

Therefore we can work out that: $x=81.8\textdegree to\ 3\ s.f.$

In conclusion, if you are not sure about whether to use the sine rule or the cosine rule, you can take advantage of a handy tactic.

If you are given two sides and the angle between them, like in example 3, you can use the cosine rule to work out the missing side. If you are given one side and the opposite angle is known, together with another angle or side (example 2), you can use the sine rule to work out a missing side. If you are given three sides you can use the cosine rule to work out any angle. ## Let’s get started

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