# Cosine Rule

Here we will learn about the cosine rule including how to use the cosine rule to find missing sides and angles in a non right-angled triangles and when to use the cosine rule instead of using the sine rule, Pythagoras’ Theorem or SOHCAHTOA.

There are also cosine rule worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What is the cosine rule?

The cosine rule (or the law of cosines) is a formula which can be used to calculate the missing sides of a triangle or to find a missing angle. To do this we need to know the two arrangements of the formula and what each variable represents.
Take a look at the triangle ABC below.

This triangle has exactly the same set up as the sine rule, with the sides represented by lower case letters and the opposite angles represented by the same capitalised letters, e.g. side b is opposite the angle at B.

This is the cosine rule:

$\\a^{2}=b^{2}+c^{2}-2bc\cos(A)\\$

## How can I use the cosine rule?

We can use the cosine rule to find missing sides and missing angles in triangles.

1. Label each angle (A, B, C) and each side (a, b, c) of the triangle.

In order to use the cosine rule we need to consider the angle that lies between two known sides.

Take a look at the diagram,

Here, the angle at A lies between the sides of b, and c (a bit like an angle sandwich).

We always label the angle we are going to be using as A, then it doesn’t matter how you label the other vertices (corners). Angle B and angle C can be either vertex with side b and side c being their opposite sides.

2State the cosine rule then substitute the given values into the formula.

Now that we know which sides and angles we have, we need to substitute this information into the cosine rule.
We can then solve this equation to find the missing side or angle.

As these are calculated slightly differently, we can rearrange the cosine rule to suit what we are trying to find.
Here are the two versions:

To find a missing side:

\\a^{2}=b^{2}+c^{2}-2bc\cos(A)\\

To find a missing angle:

\cos(A)=\frac{b^2+c^2-a^2}{2bc}

One equation is a rearrangement of the other.

3Solve the equation.

Once everything is substituted into the cosine rule we can solve the equation to calculate the unknown side or angle.

## How to find the missing side or angle of a triangle using the cosine rule

In order to find the missing side or angle of a triangle using the cosine rule:

1. Label each angle (A, B, C) and each side (a, b, c) of the triangle.
2. State the cosine rule then substitute the given values into the formula.
3. Solve the equation.

## Cosine rule examples (missing side)

### Example 1: find the missing side using the cosine rule

Find the value of x for triangle ABC, correct to 2 decimal places.

1. Label each angle (A, B, C) and each side (a, b, c) of the triangle.

The vertices are already labelled with A located on the angle we are using so we only need to label the opposite sides of a, b, and c.

2State the cosine rule then substitute the given values into the formula.

Here, we need to find the missing side a, therefore we need to state the cosine rule with a2 as the subject:

\begin{aligned} \\a^{2}&=b^{2}+c^{2}-2bc\cos(A)\\ \\x^{2}&=7.1^{2}+6.5^{2}-2\times7.1\times6.5\times\cos(32)\\ \end{aligned}

3Solve the equation.

First we need to simplify the right hand side of the equation, and then square root the solution to find the value for x.

\begin{aligned} \\x^{2}&=50.41+42.25-92.3\times\cos(32)\\ \\x^{2}&=92.66-78.27483928…\\ \\x^{2}&=14.38516072…\\ \\x&=\sqrt{14.38516072…}\\ \\x&=3.79\mathrm{cm}\quad(2dp)\\ \end{aligned}

### Example 2: find the missing side using the cosine rule

Find the length of PQ for triangle PQR, correct to 3 significant figures.

Here, it is important for us to label the angle that we need to use first. For this question, we need to find the side with  length x which is opposite the angle at R.

We need to label the angle R as A.

The other two vertices are then labelled as B and C with sides b and c labelled accordingly.

Here, we need to find the missing side a, therefore we need to state the cosine rule with a2 as the subject:

\begin{aligned} \\a^{2}&=b^{2}+c^{2}-2bc\cos(A)\\ \\x^{2}&=225^{2}+310^{2}-2\times225\times310\times\cos(75)\\ \end{aligned}

\begin{aligned} \\x^{2}&=50625+96100-2\times225\times310\times\cos(75)\\ \\x^{2}&=146725-36105.25679…\\ \\x^{2}&=110619.7432…\\ \\x&=\sqrt{110619.7432…}\\ \\x&=333\mathrm{m}\quad(3sf)\\ \end{aligned}

### Example 3: find the missing side using the cosine rule

Find the length of z for triangle XYZ. Write your answer to a suitable degree of accuracy.

In order to find the length of z, we need to know the opposite angle at Z. As we know the other two angles, 180 − (79 + 62) = 39º so Z = 39º.

We can then label vertex Z as A, the length z as a, and the other angles and sides accordingly.

Here, we need to find the missing side a, therefore we need to state the cosine rule with a2 as the subject:

\begin{aligned} \\a^{2}&=b^{2}+c^{2}-2bc\cos(A)\\ \\z^{2}&=0.8^{2}+0.6^{2}-2\times0.8\times0.6\times\cos(39)\\ \end{aligned}

\begin{aligned} \\z^{2}&=0.64+0.36-2\times0.8\times0.6\times\cos(39)\\ \\z^{2}&=1-0.746060123…\\ \\z^{2}&=0.253939877…\\ \\z&=\sqrt{0.253939877…}\\ \\z&=0.504\mathrm{mm}\quad(3dp)\\ \end{aligned}

## Cosine rule examples (missing angle)

### Example 4: find the missing angle using the cosine rule

Find the size of the angle θ for the isosceles triangle ABC. Write your answer to 2 significant figures.

Here, the vertices are already labelled, and the angle we need to find is already A so we just need to fill in the opposing sides with a, b, and c.

Here, we need to find the missing angle A, therefore we need to state the cosine rule with cos(A) as the subject:

\begin{aligned} \\\cos(A)&=\frac{b^2+c^2-a^2}{2bc}\\ \\\cos(\theta)&=\frac{6^2+12^2-12^2}{2\times6\times12}\\ \end{aligned}

\begin{aligned} \\\cos(\theta)&=\frac{36+144-144}{144}\\ \\\cos(\theta)&=\frac{36}{144}\\ \\\cos(\theta)&=0.25\\ \\\theta&=\cos^{-1}(0.25)\\ \\\theta&=76^{\circ}\quad(2sf)\\ \end{aligned}

### Example 5: find the missing angle using the cosine rule

Find the size of the angle θ for triangle EFG. Write your answer to 3 significant figures.

Here, we need to label each vertex and angle. As we need to know the angle at F, this will be labelled as A and the opposing side is labelled as a. The other vertices are labelled as B and C (it doesn’t matter which) and their opposite side b and c as given below.

Here, we need to find the missing angle A, therefore we need to state the cosine rule with cos(A) as the subject:

\begin{aligned} \\\cos(A)&=\frac{b^2+c^2-a^2}{2bc}\\ \\\cos(\theta)&=\frac{29.7^2+13.8^2-27.5^2}{2\times29.7\times13.8}\\ \end{aligned}

\begin{aligned} \\\cos(\theta)&=\frac{882.09+190.44-756.25}{819.72}\\ \\\cos(\theta)&=\frac{316.28}{819.72}\\ \\\cos(\theta)&=0.385839067…\\ \\\theta&=\cos^{-1}(0.385839067…)\\ \\\theta&=67.3^{\circ}\quad(3sf)\\ \end{aligned}

### Example 6: find the missing obtuse angle using the cosine rule

Find the size of the angle θ for triangle XYZ. Write your answer to 2 decimal places.

Here, we need to label each vertex and angle. As we need to know the angle at Z, this will be labelled as A and the opposing side is labelled as a. The other vertices are labelled as B and C and their opposite side b and c respectively.

Here, we need to find the missing angle A, therefore we need to state the cosine rule with cos(A) as the subject:

\begin{aligned} \\\cos(A)&=\frac{b^2+c^2-a^2}{2bc}\\ \\\cos(\theta)&=\frac{2.7^2+3.8^2-5.1^2}{2\times2.7\times3.8}\\ \end{aligned}

\begin{aligned} \\\cos(\theta)&=\frac{7.29+14.44-26.01}{20.52}\\ \\\cos(\theta)&=\frac{-4.28}{20.52}\\ \\\cos(\theta)&=-0.2085769981…\\ \\\theta&=\cos^{-1}(-0.2085769981…)\\ \\\theta&=102.04^{\circ}\quad(2dp)\\ \end{aligned}

### Common misconceptions

• Pythagoras’ Theorem and trigonometry

A common error is to use Pythagoras’ Theorem instead of trigonometry to find the missing side of the non right angle triangle

• Sine function instead of cosine function

A common error is to use the sine function instead of cosine function

• Incorrect labelling of the vertices and sides

‘A’ is the included angle between the two sides b and c. It is common to use one of the angles in the triangle in place of the angle at A and therefore the calculation will be incorrect.

• Incorrectly using the cosine rule as a2 = b2 + c2 + 2bcCos(A)

Here, 2bcCos(A) has been added to b2 + c2. This is a very common misconception which can easily be avoided.

• Substituting errors

A common error is to incorrectly substitute into the cosine rule using the side length ‘a’ instead of the angle A

• Substituting values into the cosine rule without the correct application of BIDMAS

It is important to follow the order of operations when evaluating the cosine rule.

• Is there an ambiguous case for the cosine rule like there is for the sine rule?

The simple answer is no because of the nature of the cosine function and the link to finding an angle inside a triangle. E.g.
If you take cos(60), this will return the same answer on a calculator as cos(300).

The sum of angles in a triangle must add to 180º so no angle will be greater than 180º. It is important to remember that the inverse cosine of any number between 0 and -1 will return an obtuse angle. For more information, see Trigonometric Graphs.

Cosine rule is part of our series of lessons to support revision on trigonometry. You may find it helpful to start with the main trigonometry lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

### Practice cosine rule questions

1. Use the cosine rule to find the length of x for triangle ABC . Give your answer to 1decimal place.

13.6cm

13.2cm

15.2cm

4.2cm

Label the triangle:

\begin{aligned} a^{2}&=b^{2}+c^{2}-2bc \cos(A)\\\\ x^{2}&=12^{2}+8^{2}-2 \times 12 \times 8 \times \cos(97)\\\\ x^{2}&=208-192 \cos(97)\\\\ x^{2}&=231.3989139\\\\ x&=\sqrt{231.3989139}\\\\ x&=15.2 ~(1\mathrm{dp}) \end{aligned}

2. Find the length of x for triangle PQR . Give your answer to 1 decimal place.

10.2cm

5.8cm

8.3cm

10.5cm

Label the triangle:

\begin{aligned} a^{2}&=b^{2}+c^{2}-2bc \cos(A)\\\\ x^{2}&=6.3^{2}+5.4^{2}-2 \times 6.3 \times 5.4 \times \cos(59)\\\\ x^{2}&=68.85-68.04\cos(59)\\\\ x^{2}&=33.80680938 \\\\ x&=\sqrt{33.80680938}\\\\ x&=5.8 ~(1\mathrm{dp}) \end{aligned}

3. Find the length of XY for triangle XYZ . Write your answer to 1 decimal place.

11.6cm

3.3cm

5.8cm

10.5cm

Label the triangle:

\begin{aligned} a^{2}&=b^{2}+c^{2}-2bc \cos(A)\\\\ x^{2}&=6^{2}+6^{2}-2 \times 6 \times 6 \times \cos(122)\\\\ x^{2}&=72-72 \cos(122)\\\\ x^{2}&=110.154187\\\\ x&=\sqrt{110.145187}\\\\ x&=10.5 ~(1\mathrm{dp}) \end{aligned}

4. Find the size of angle \theta for triangle ABC . Give your answer to 3 significant figures.

43.8^{\circ}

1.00^{\circ}

28.0^{\circ}

134^{\circ}

Label the triangle:

\begin{aligned} \cos(A) &= \frac{b^{2}+c^{2}-a^{2}}{2bc}\\\\ \cos(\theta)&=\frac{5.2^{2}+7.8^{2}-12^{2}}{2 \times 5.2 \times 7.8}\\\\ \cos(\theta)&= -0.6918145957\\\\ \theta&=\cos^{-1}(-0.6918145957)\\\\ \theta&=134^{\circ} \end{aligned}

5. Find the size of angle \theta for triangle XYZ . Give your answer to 3 significant figures.

30.4^{\circ}

59.6^{\circ}

1.00^{\circ}

36.7^{\circ}

Label the triangle:

\begin{aligned} \cos(A) &= \frac{b^{2}+c^{2}-a^{2}}{2bc}\\\\ \cos(\theta)&=\frac{15.1^{2}+9.8^{2}-8.3^{2}}{2 \times 15.1 \times 9.8}\\\\ \cos(\theta)&= 0.8621435329\\\\ \theta&=\cos^{-1}(0.8621435329)\\\\ \theta&=30.4^{\circ} \end{aligned}

6. Find the size of angle \theta for triangle EFG . Write your answer to 2 decimal places.

73.60^{\circ}

16.10^{\circ}

1.00^{\circ}

81.95^{\circ}

Label the triangle:

\begin{aligned} \cos(A) &= \frac{b^{2}+c^{2}-a^{2}}{2bc}\\\\ \cos(\theta)&=\frac{25^{2}+25^{2}-7^{2}}{2 \times 25 \times 25}\\\\ \cos(\theta)&= 0.9608\\\\ \theta&=\cos^{-1}(0.9608)\\\\ \theta&=16.10^{\circ} \end{aligned}

### Cosine rule GCSE exam questions

1.  In triangle ABC, AB = 4.5mm, AC = 8.3mm and BC = 6.1mm.

Work out the size of the angle BAC .

(3 marks)

\cos(A)=\frac{4.5^{2}+8.3^{2}-6.1^{2}}{2\times 4.5 \times 8.3}

(1)

\begin{aligned} \cos(A)&=\frac{51.93}{74.7}\\\\ \cos(A)&=0.695… \end{aligned}

(1)

\begin{aligned} A&=\cos^{-1}(0.695…)\\\\ A&=46.0^{\circ} \end{aligned}

(1)

(5 marks)

\tan(44)=\frac{BD}{76}

(1)

\begin{aligned} BD&=76 \times \tan(44)\\\\ BD&=73.39 \mathrm{cm} \end{aligned}

(1)

(AD)^{2}=63^{2}+73.39^{2}-2 \times 63 \times 73.39 \times \cos(79)

(1)

(1)

(1)

3.  A satellite takes measurements to some triangulation stations.

Calculate the distance between the triangulation stations.

(3 marks)

a^{2}=230^{2}+201^{2}-2 \times 230 \times 201 \times \cos(4)

(1)

\begin{aligned} a^{2}&=93310-92460 \times \cos(4)\\\\ a^{2}&=1066.228 \end{aligned}

(1)

\begin{aligned} a&=\sqrt{1066.228}\\\\ a&=32.7\mathrm{km} \end{aligned}

(1)

### Did you know?

The cosine rule is derived from the use of Pythagoras’ theorem c2 = a2 + b2 and the cosine function:

$\cos(\theta)=\frac{A}{H}$

To do this, the triangle is split into two right-angled triangles. We can then use Pythagoras’ theorem to work out c2 and a2 separately, and then we use the cosine function to calculate the width of one of the two triangles.

Note: You are not required to know how to derive the cosine rule, however the derivation only requires knowledge of the GCSE curriculum, similar to the derivation of the quadratic formula.

## Learning checklist

You have now learned how to:

• know and apply the cosine rule to find unknown lengths and angles

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