The Cosine Rule

What you'll learn

When to use the Cosine Rule instead of Pythagoras or SOHCAHTOA.
How to find a missing side when you know two sides and the included angle.
How to find a missing angle when you know all three sides.
How the Cosine Rule can appear in algebra and perimeter questions.

1. Naming the parts of a triangle

A vertex is a corner of a triangle. In triangle $A B C$ , the vertices are $A$ , $B$ and $C$ .

The side opposite an angle is the side across from it. For example, the side opposite angle $A$ is $B C$ .

Definition

Included angle

The included angle is the angle between two known sides. If you know sides $A B$ and $A C$ , the included angle is angle $B A C$ , because both sides meet at $A$ .

The included angle is the angle formed where the two known sides meet.

Example

Spotting the opposite side

A triangle has $A B = 6$ m, $A C = 8$ m, and angle $50^\circ$ . Which side is opposite the 50° angle?

Angle BAC is at A, so the opposite side is BC across the triangle.

The angle $B A C$ is at vertex $A$ .
The side opposite vertex $A$ is the side joining the other two vertices.
So the opposite side is $B C$ .

Key Idea

What to look for

The Cosine Rule connects an angle with the side directly opposite it. Always pair the angle with its opposite side before substituting numbers.

2. The Cosine Rule for finding a missing side

Use this when you know two sides and the included angle, and you want the third side.

Definition

The Cosine Rule

If side $a$ is opposite angle $A$ , then:

In the Cosine Rule, side a is directly opposite angle A, while sides b and c meet at angle A.

a^2 = b^2 + c^2 - 2bc\cos A

Here, $b$ and $c$ are the two sides that meet at angle $A$ .

This looks a bit like Pythagoras, but the extra part involving $cos⁡A\cos A$ adjusts for triangles that are not right-angled.

Example

Finding a missing side

A triangle has two sides 12 cm and 9 cm with an included angle of 100°. Find the third side, giving your answer to 1 decimal place.

The unknown third side is opposite the included 100° angle between the 12 cm and 9 cm sides.

The unknown side is opposite the 100° angle, so call it $x$ .
Substitute into the Cosine Rule using $b = 12$ , $c = 9$ and $100^\circ$ :
$x2=122+92−2×12×9×cos⁡100∘x2=144+81−216cos⁡100∘x2=262.508…\begin{aligned} x^2 &= 12^2 + 9^2 - 2 \times 12 \times 9 \times \cos 100^\circ \\ x^2 &= 144 + 81 - 216\cos 100^\circ \\ x^2 &= 262.508\ldots \end{aligned}$
Square root the answer:
$\sqrt{262.508\ldots} = 16.202\ldots$
The third side is 16.2 cm to 1 decimal place.

Common Mistake

Forgetting the square root

The formula gives $x^2$ , not $x$ . After using the Cosine Rule to find a side, you must square root before rounding.

Tip

Calculator mode

For GCSE triangle questions, your calculator should normally be in degrees mode. If your answer looks wildly wrong, check this first.

3. The Cosine Rule for finding a missing angle

Use this when you know all three sides and want an angle.

Rearrange the Cosine Rule:

cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Then use inverse cosine, written $cos^{-1}$ , to get the angle.

Definition

Inverse cosine

$cos^{-1}$ is the calculator function that finds the angle when you know its cosine value.

Example

Finding a missing angle

A triangle has sides 14 cm, 15 cm and 22 cm. Find the angle between the 14 cm and 15 cm sides, giving your answer to 3 significant figures.

The required angle is between the 14 cm and 15 cm sides, with the 22 cm side opposite it.

The angle we want is between the 14 cm and 15 cm sides, so the opposite side is 22 cm.
Use $a = 22$ , $b = 14$ and $c = 15$ :
$cos⁡A=142+152−2222×14×15\cos A = \frac{14^2 + 15^2 - 22^2}{2 \times 14 \times 15}$
Simplify the fraction:
$cos⁡A=196+225−484420=−63420\cos A = \frac{196 + 225 - 484}{420} = \frac{-63}{420}$
Use inverse cosine:
$\cos^{-1}\left(\frac{-63}{420}\right) = 98.626\ldots^\circ$
The angle is 98.6° to 3 significant figures.

Common Mistake

Using the wrong opposite side

When finding an angle, the side being subtracted in the numerator is the side opposite the angle. In $b2+c2−a22bc\frac{b^2+c^2-a^2}{2bc}$ , the $a^2$ side is opposite angle $A$ .

4. Algebraic side lengths

Sometimes the side lengths contain expressions like $x + 1$ or $2 x + 1$ . You still use the Cosine Rule, but you may need to expand brackets and solve a quadratic.

Example

Finding $x$ from algebraic lengths

A triangle has sides $x + 1$ and $2 x + 1$ with an included angle of 60°. The opposite side is $91\sqrt{91}$ . Find $x$ .

The algebraic sides meet at the 60° included angle, and the opposite side is √91.

Write the Cosine Rule. Since the opposite side is $91\sqrt{91}$ :
$(91)2=(x+1)2+(2x+1)2−2(x+1)(2x+1)cos⁡60∘\left(\sqrt{91}\right)^2 = (x+1)^2 + (2x+1)^2 - 2(x+1)(2x+1)\cos 60^\circ$
Use $(91)2=91\left(\sqrt{91}\right)^2 = 91$ and $cos⁡60∘=12\cos 60^\circ = \frac{1}{2}$ :
$91 = (x+1)^2 + (2x+1)^2 - (x+1)(2x+1)$
Expand and simplify:
$91=x2+2x+1+4x2+4x+1−(2x2+3x+1)91=3x2+3x+1\begin{aligned} 91 &= x^2 + 2x + 1 + 4x^2 + 4x + 1 - (2x^2 + 3x + 1) \\ 91 &= 3x^2 + 3x + 1 \end{aligned}$
Rearrange and solve:
$3x2+3x−90=0x2+x−30=0(x+6)(x−5)=0\begin{aligned} 3x^2 + 3x - 90 &= 0 \\ x^2 + x - 30 &= 0 \\ (x+6)(x-5) &= 0 \end{aligned}$
The possible values are $x = - 6$ or $x = 5$ .
Reject $x = - 6$ because it would make side lengths negative. So $x = 5$ .

5. Combining area and the Cosine Rule

A 5-mark question may ask for a perimeter but give you two sides and the area first.

Definition

Area using two sides and an included angle

For two sides $a$ and $b$ with included angle $C$ :

The area formula uses two sides and the included angle between them.

Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

Once you find the included angle, you can use the Cosine Rule to find the missing side. The perimeter is the total distance around the triangle.

Example

Area first, then perimeter

A triangle has sides 15 m and 13 m meeting at an acute angle. Its area is 78 m². Find the perimeter to 3 significant figures.

First find the acute included angle between the 15 m and 13 m sides, then find the missing side for the perimeter.

Use the area formula to find the included angle $C$ :
$\frac{1}{2} \times 15 \times 13 \times \sin C$
Simplify:
$97.5\sin C$
Divide by 97.5:
$sin⁡C=0.8\sin C = 0.8$
Because the angle is acute:
$\sin^{-1}(0.8) = 53.130\ldots^\circ$
Use the Cosine Rule to find the missing side $x$ :
$x2=152+132−2×15×13×cos⁡53.130…∘x2=160\begin{aligned} x^2 &= 15^2 + 13^2 - 2 \times 15 \times 13 \times \cos 53.130\ldots^\circ \\ x^2 &= 160 \end{aligned}$
Square root:
$\sqrt{160} = 12.649\ldots$
Add the three sides:
$12.649\ldots = 40.649\ldots$
The perimeter is 40.6 m to 3 significant figures.

Common Mistake

Area can give two angles

If you use $sin^{-1}$ , there may be an acute and an obtuse angle with the same sine value. Use the diagram or wording carefully to decide which one is intended.

Exam technique

In the exam

Mark the angle and its opposite side before writing the formula.
If finding a side, calculate $a^2$ first, then square root.
If finding an angle, use $cos^{-1}$ only after you have found the cosine value.
Keep full calculator values until the final line, then round to the accuracy asked for.

Self review

Check yourself

In $a^2 = b^2 + c^2 - 2bc\cos A$ , which side is opposite angle $A$ ?
When would you use the Cosine Rule instead of SOHCAHTOA?
Why might an area question lead to two possible included angles?

The Cosine Rule

What you'll learn

When to use the Cosine Rule instead of Pythagoras or SOHCAHTOA.
How to find a missing side when you know two sides and the included angle.
How to find a missing angle when you know all three sides.
How the Cosine Rule can appear in algebra and perimeter questions.

1. Naming the parts of a triangle

A vertex is a corner of a triangle. In triangle $A B C$ , the vertices are $A$ , $B$ and $C$ .

The side opposite an angle is the side across from it. For example, the side opposite angle $A$ is $B C$ .

Definition

Included angle

The included angle is the angle between two known sides. If you know sides $A B$ and $A C$ , the included angle is angle $B A C$ , because both sides meet at $A$ .

The included angle is the angle formed where the two known sides meet.

Example

Spotting the opposite side

A triangle has $A B = 6$ m, $A C = 8$ m, and angle $50^\circ$ . Which side is opposite the 50° angle?

Angle BAC is at A, so the opposite side is BC across the triangle.

The angle $B A C$ is at vertex $A$ .
The side opposite vertex $A$ is the side joining the other two vertices.
So the opposite side is $B C$ .

Key Idea

What to look for

The Cosine Rule connects an angle with the side directly opposite it. Always pair the angle with its opposite side before substituting numbers.

2. The Cosine Rule for finding a missing side

Use this when you know two sides and the included angle, and you want the third side.

Definition

The Cosine Rule

If side $a$ is opposite angle $A$ , then:

In the Cosine Rule, side a is directly opposite angle A, while sides b and c meet at angle A.

a^2 = b^2 + c^2 - 2bc\cos A

Here, $b$ and $c$ are the two sides that meet at angle $A$ .

This looks a bit like Pythagoras, but the extra part involving $cos⁡A\cos A$ adjusts for triangles that are not right-angled.

Example

Finding a missing side

A triangle has two sides 12 cm and 9 cm with an included angle of 100°. Find the third side, giving your answer to 1 decimal place.

The unknown third side is opposite the included 100° angle between the 12 cm and 9 cm sides.

The unknown side is opposite the 100° angle, so call it $x$ .
Substitute into the Cosine Rule using $b = 12$ , $c = 9$ and $100^\circ$ :
$x2=122+92−2×12×9×cos⁡100∘x2=144+81−216cos⁡100∘x2=262.508…\begin{aligned} x^2 &= 12^2 + 9^2 - 2 \times 12 \times 9 \times \cos 100^\circ \\ x^2 &= 144 + 81 - 216\cos 100^\circ \\ x^2 &= 262.508\ldots \end{aligned}$
Square root the answer:
$\sqrt{262.508\ldots} = 16.202\ldots$
The third side is 16.2 cm to 1 decimal place.

Common Mistake

Forgetting the square root

The formula gives $x^2$ , not $x$ . After using the Cosine Rule to find a side, you must square root before rounding.

Tip

Calculator mode

For GCSE triangle questions, your calculator should normally be in degrees mode. If your answer looks wildly wrong, check this first.

3. The Cosine Rule for finding a missing angle

Use this when you know all three sides and want an angle.

Rearrange the Cosine Rule:

cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Then use inverse cosine, written $cos^{-1}$ , to get the angle.

Definition

Inverse cosine

$cos^{-1}$ is the calculator function that finds the angle when you know its cosine value.

Example

Finding a missing angle

A triangle has sides 14 cm, 15 cm and 22 cm. Find the angle between the 14 cm and 15 cm sides, giving your answer to 3 significant figures.

The required angle is between the 14 cm and 15 cm sides, with the 22 cm side opposite it.

The angle we want is between the 14 cm and 15 cm sides, so the opposite side is 22 cm.
Use $a = 22$ , $b = 14$ and $c = 15$ :
$cos⁡A=142+152−2222×14×15\cos A = \frac{14^2 + 15^2 - 22^2}{2 \times 14 \times 15}$
Simplify the fraction:
$cos⁡A=196+225−484420=−63420\cos A = \frac{196 + 225 - 484}{420} = \frac{-63}{420}$
Use inverse cosine:
$\cos^{-1}\left(\frac{-63}{420}\right) = 98.626\ldots^\circ$
The angle is 98.6° to 3 significant figures.

Common Mistake

Using the wrong opposite side

When finding an angle, the side being subtracted in the numerator is the side opposite the angle. In $b2+c2−a22bc\frac{b^2+c^2-a^2}{2bc}$ , the $a^2$ side is opposite angle $A$ .

4. Algebraic side lengths

Sometimes the side lengths contain expressions like $x + 1$ or $2 x + 1$ . You still use the Cosine Rule, but you may need to expand brackets and solve a quadratic.

Example

Finding $x$ from algebraic lengths

A triangle has sides $x + 1$ and $2 x + 1$ with an included angle of 60°. The opposite side is $91\sqrt{91}$ . Find $x$ .

The algebraic sides meet at the 60° included angle, and the opposite side is √91.

Write the Cosine Rule. Since the opposite side is $91\sqrt{91}$ :
$(91)2=(x+1)2+(2x+1)2−2(x+1)(2x+1)cos⁡60∘\left(\sqrt{91}\right)^2 = (x+1)^2 + (2x+1)^2 - 2(x+1)(2x+1)\cos 60^\circ$
Use $(91)2=91\left(\sqrt{91}\right)^2 = 91$ and $cos⁡60∘=12\cos 60^\circ = \frac{1}{2}$ :
$91 = (x+1)^2 + (2x+1)^2 - (x+1)(2x+1)$
Expand and simplify:
$91=x2+2x+1+4x2+4x+1−(2x2+3x+1)91=3x2+3x+1\begin{aligned} 91 &= x^2 + 2x + 1 + 4x^2 + 4x + 1 - (2x^2 + 3x + 1) \\ 91 &= 3x^2 + 3x + 1 \end{aligned}$
Rearrange and solve:
$3x2+3x−90=0x2+x−30=0(x+6)(x−5)=0\begin{aligned} 3x^2 + 3x - 90 &= 0 \\ x^2 + x - 30 &= 0 \\ (x+6)(x-5) &= 0 \end{aligned}$
The possible values are $x = - 6$ or $x = 5$ .
Reject $x = - 6$ because it would make side lengths negative. So $x = 5$ .

5. Combining area and the Cosine Rule

A 5-mark question may ask for a perimeter but give you two sides and the area first.

Definition

Area using two sides and an included angle

For two sides $a$ and $b$ with included angle $C$ :

The area formula uses two sides and the included angle between them.

Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin C

Once you find the included angle, you can use the Cosine Rule to find the missing side. The perimeter is the total distance around the triangle.

Example

Area first, then perimeter

A triangle has sides 15 m and 13 m meeting at an acute angle. Its area is 78 m². Find the perimeter to 3 significant figures.

First find the acute included angle between the 15 m and 13 m sides, then find the missing side for the perimeter.

Use the area formula to find the included angle $C$ :
$\frac{1}{2} \times 15 \times 13 \times \sin C$
Simplify:
$97.5\sin C$
Divide by 97.5:
$sin⁡C=0.8\sin C = 0.8$
Because the angle is acute:
$\sin^{-1}(0.8) = 53.130\ldots^\circ$
Use the Cosine Rule to find the missing side $x$ :
$x2=152+132−2×15×13×cos⁡53.130…∘x2=160\begin{aligned} x^2 &= 15^2 + 13^2 - 2 \times 15 \times 13 \times \cos 53.130\ldots^\circ \\ x^2 &= 160 \end{aligned}$
Square root:
$\sqrt{160} = 12.649\ldots$
Add the three sides:
$12.649\ldots = 40.649\ldots$
The perimeter is 40.6 m to 3 significant figures.

Common Mistake

Area can give two angles

If you use $sin^{-1}$ , there may be an acute and an obtuse angle with the same sine value. Use the diagram or wording carefully to decide which one is intended.

Exam technique

In the exam

Mark the angle and its opposite side before writing the formula.
If finding a side, calculate $a^2$ first, then square root.
If finding an angle, use $cos^{-1}$ only after you have found the cosine value.
Keep full calculator values until the final line, then round to the accuracy asked for.

Self review

Check yourself

In $a^2 = b^2 + c^2 - 2bc\cos A$ , which side is opposite angle $A$ ?
When would you use the Cosine Rule instead of SOHCAHTOA?
Why might an area question lead to two possible included angles?

The Cosine Rule

What you'll learn

1. Naming the parts of a triangle

2. The Cosine Rule for finding a missing side

3. The Cosine Rule for finding a missing angle

4. Algebraic side lengths

5. Combining area and the Cosine Rule

Recap questions

You've reached the end

The Cosine Rule

What you'll learn

1. Naming the parts of a triangle

2. The Cosine Rule for finding a missing side

3. The Cosine Rule for finding a missing angle

4. Algebraic side lengths

5. Combining area and the Cosine Rule

Recap questions

You've reached the end

The Cosine Rule

What you'll learn

1. Naming the parts of a triangle

2. The Cosine Rule for finding a missing side

3. The Cosine Rule for finding a missing angle

4. Algebraic side lengths

5. Combining area and the Cosine Rule

Recap questions

You've reached the end

The Cosine Rule Revision Guide

The Cosine Rule

What you'll learn

1. Naming the parts of a triangle

2. The Cosine Rule for finding a missing side

3. The Cosine Rule for finding a missing angle

4. Algebraic side lengths

5. Combining area and the Cosine Rule

Recap questions

You've reached the end

The Cosine Rule Revision Guide