Finding the Area of Any Triangle

What you'll learn

How the usual triangle area formula links to sine.
How to find an area when you know two sides and the angle between them.
How to work backwards to find a missing angle.
How to handle algebraic side lengths and ratios.

Start with the area formula you already know

For any triangle, if you know a base and its perpendicular height, you can find the area using:

\frac{1}{2}\times \text{base}\times \text{height}

Definition

Perpendicular height

The perpendicular height is the shortest distance from a chosen base to the opposite vertex. It meets the base at 90°.

The perpendicular height drops from the opposite vertex to the chosen base at a right angle.

Example

Using base and perpendicular height

A triangle has base 10 cm and perpendicular height 7 cm.

Triangle with base 10 cm and perpendicular height 7 cm for using A = 1/2 × base × height.

Substitute into the formula.
$\frac{1}{2}\times 10\times 7$
Calculate the area.
$A = 35$
The area is 35 cm².

The problem is that in many triangles, the perpendicular height is not labelled. That is where sine helps.

The sine area formula

Definition

Included angle

The included angle is the angle formed by two given sides — the angle directly between them.

The included angle is the angle directly between the two known sides.

If you know two sides and the included angle, you can use:

\frac{1}{2}ab\sin C

Here, $a$ and $b$ are the two known side lengths, and $C$ is the included angle.

Key Idea

Area of any triangle

To find the area from two sides and the angle between them, use $A=12absin⁡CA=\frac{1}{2}ab\sin C$ .

Common Mistake

Using the wrong angle

The angle must be between the two sides you are using. If the angle is not included, this formula does not directly apply.

Tip

Calculator and rounding

Make sure your calculator is in degrees mode, often shown as DEG. Keep the full calculator value until the final answer, then round as requested.

Example

Finding an area from two sides and an included angle

A triangle has sides 14 cm and 11 cm with an included angle of 115°. Find its area to 1 decimal place.

The 115° angle is included between the sides 14 cm and 11 cm.

Use the sine area formula.
$\frac{1}{2}ab\sin C$
Substitute the values.
$\frac{1}{2}\times 14\times 11\times \sin 115^\circ$
Calculate.
$69.785\ldots$
Round to 1 decimal place: the area is 69.8 cm².

Exact angles can make the calculation shorter

Some angles have exact sine values. The most common one in this topic is:

sin⁡30∘=12\sin 30^\circ = \frac{1}{2}

This can make the area calculation very quick.

Example

Using a 30° included angle

A triangle has sides 12 m and 9 m with an included angle of 30°.

The exact value sin 30° can be used because the 30° angle is between the two given sides.

Substitute into the formula.
$\frac{1}{2}\times 12\times 9\times \sin 30^\circ$
Use $sin⁡30∘=12\sin 30^\circ=\frac{1}{2}$ .
$\frac{1}{2}\times 12\times 9\times \frac{1}{2}$
Calculate.
$A = 27$
The area is 27 m².

Working backwards to find a missing angle

Sometimes you are given the area and the two sides, and you need to find the included angle.

Definition

Inverse sine

Inverse sine, written $sin^{-1}$ , is the calculator operation that finds an angle when you know its sine.

Start with:

\frac{1}{2}ab\sin x

Then rearrange to make $sin⁡x\sin x$ the subject:

sin⁡x=2Aab\sin x = \frac{2A}{ab}

Common Mistake

Two possible angles

If $sin⁡x\sin x$ is positive, there may be two possible triangle angles: the calculator angle and $180∘180^\circ$ minus that angle. Use the diagram to decide whether the angle is acute or obtuse.

Example

Finding an obtuse missing angle

A triangle has sides 12 cm and 15 cm. The included angle is $x∘x^\circ$ , the area is 72 cm², and the angle shown is obtuse.

The diagram shows the included angle x° is obtuse, so the obtuse sine solution is needed.

Substitute into the formula.
$\frac{1}{2}\times 12\times 15\times \sin x$
Simplify.
$90\sin x$
Divide by 90.
$sin⁡x=0.8\sin x = 0.8$
Use inverse sine to find the calculator angle.
$\sin^{-1}(0.8)=53.130\ldots^\circ$
Because the diagram shows an obtuse angle, subtract from 180°.
$180∘−53.130…∘=126.869…∘180^\circ - 53.130\ldots^\circ = 126.869\ldots^\circ$
To 1 decimal place, $x=126.9∘x=126.9^\circ$ .

When the side lengths include algebra

If the sides are written using $x$ , you still use the same formula. This time, the formula creates an equation that you solve.

A useful exact value is:

sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}

Example

Finding an unknown side expression

A triangle has side lengths $x$ cm and $(x + 2)$ cm, with included angle 60°. Its area is $12312\sqrt{3}$ cm².

Algebraic side lengths can be substituted into the same sine area formula.

Substitute into the sine area formula.
$123=12×x(x+2)×sin⁡60∘12\sqrt{3}=\frac{1}{2}\times x(x+2)\times \sin 60^\circ$
Replace $sin⁡60∘\sin 60^\circ$ with its exact value.
$123=12×x(x+2)×3212\sqrt{3}=\frac{1}{2}\times x(x+2)\times \frac{\sqrt{3}}{2}$
Simplify.
$123=x(x+2)3412\sqrt{3}=\frac{x(x+2)\sqrt{3}}{4}$
Divide by $3\sqrt{3}$ and multiply by 4.
$x (x + 2) = 48$
Expand and rearrange.
$x^2+2x-48=0$
Factorise.
$(x + 8) (x - 6) = 0$
The solutions are $x = - 8$ and $x = 6$ . A length cannot be negative, so $x = 6$ .

Using ratios for side lengths

Definition

Ratio

A ratio compares quantities in parts. If two lengths are in the ratio 3:2, they can be written as $3 k$ and $2 k$ , where $k$ is the value of one part.

Example

Using a side ratio

Two sides of a triangle meet at 30°. Their lengths are in the ratio 3:2, and the area is 54 cm². Find the shorter side.

Writing the ratio as 3k and 2k lets the two given sides be used in the sine area formula.

Let the two sides be $3 k$ cm and $2 k$ cm.
Substitute into the formula.
$54=12(3k)(2k)sin⁡30∘54=\frac{1}{2}(3k)(2k)\sin 30^\circ$
Use $sin⁡30∘=12\sin 30^\circ=\frac{1}{2}$ .
$54=32k254=\frac{3}{2}k^2$
Solve for $k^2$ .
$k^2=36$
Since lengths are positive, $k = 6$ . The shorter side is $2 k = 12$ cm.

Exam technique

In the exam

Check that the angle is between the two sides before using $A=12absin⁡CA=\frac{1}{2}ab\sin C$ .
If finding an angle, isolate $sin⁡x\sin x$ first, then use $sin^{-1}$ .
Look carefully at the diagram: if the angle is obtuse, use $180∘180^\circ$ minus the calculator angle.

Self review

Check yourself

Can you identify the included angle from a triangle diagram?
When would you use $A=12absin⁡CA=\frac{1}{2}ab\sin C$ instead of base times height?
If your calculator gives an acute angle, how do you find the possible obtuse angle?

Finding the Area of Any Triangle

What you'll learn

How the usual triangle area formula links to sine.
How to find an area when you know two sides and the angle between them.
How to work backwards to find a missing angle.
How to handle algebraic side lengths and ratios.

Start with the area formula you already know

For any triangle, if you know a base and its perpendicular height, you can find the area using:

\frac{1}{2}\times \text{base}\times \text{height}

Definition

Perpendicular height

The perpendicular height is the shortest distance from a chosen base to the opposite vertex. It meets the base at 90°.

The perpendicular height drops from the opposite vertex to the chosen base at a right angle.

Example

Using base and perpendicular height

A triangle has base 10 cm and perpendicular height 7 cm.

Triangle with base 10 cm and perpendicular height 7 cm for using A = 1/2 × base × height.

Substitute into the formula.
$\frac{1}{2}\times 10\times 7$
Calculate the area.
$A = 35$
The area is 35 cm².

The problem is that in many triangles, the perpendicular height is not labelled. That is where sine helps.

The sine area formula

Definition

Included angle

The included angle is the angle formed by two given sides — the angle directly between them.

The included angle is the angle directly between the two known sides.

If you know two sides and the included angle, you can use:

\frac{1}{2}ab\sin C

Here, $a$ and $b$ are the two known side lengths, and $C$ is the included angle.

Key Idea

Area of any triangle

To find the area from two sides and the angle between them, use $A=12absin⁡CA=\frac{1}{2}ab\sin C$ .

Common Mistake

Using the wrong angle

The angle must be between the two sides you are using. If the angle is not included, this formula does not directly apply.

Tip

Calculator and rounding

Make sure your calculator is in degrees mode, often shown as DEG. Keep the full calculator value until the final answer, then round as requested.

Example

Finding an area from two sides and an included angle

A triangle has sides 14 cm and 11 cm with an included angle of 115°. Find its area to 1 decimal place.

The 115° angle is included between the sides 14 cm and 11 cm.

Use the sine area formula.
$\frac{1}{2}ab\sin C$
Substitute the values.
$\frac{1}{2}\times 14\times 11\times \sin 115^\circ$
Calculate.
$69.785\ldots$
Round to 1 decimal place: the area is 69.8 cm².

Exact angles can make the calculation shorter

Some angles have exact sine values. The most common one in this topic is:

sin⁡30∘=12\sin 30^\circ = \frac{1}{2}

This can make the area calculation very quick.

Example

Using a 30° included angle

A triangle has sides 12 m and 9 m with an included angle of 30°.

The exact value sin 30° can be used because the 30° angle is between the two given sides.

Substitute into the formula.
$\frac{1}{2}\times 12\times 9\times \sin 30^\circ$
Use $sin⁡30∘=12\sin 30^\circ=\frac{1}{2}$ .
$\frac{1}{2}\times 12\times 9\times \frac{1}{2}$
Calculate.
$A = 27$
The area is 27 m².

Working backwards to find a missing angle

Sometimes you are given the area and the two sides, and you need to find the included angle.

Definition

Inverse sine

Inverse sine, written $sin^{-1}$ , is the calculator operation that finds an angle when you know its sine.

Start with:

\frac{1}{2}ab\sin x

Then rearrange to make $sin⁡x\sin x$ the subject:

sin⁡x=2Aab\sin x = \frac{2A}{ab}

Common Mistake

Two possible angles

If $sin⁡x\sin x$ is positive, there may be two possible triangle angles: the calculator angle and $180∘180^\circ$ minus that angle. Use the diagram to decide whether the angle is acute or obtuse.

Example

Finding an obtuse missing angle

A triangle has sides 12 cm and 15 cm. The included angle is $x∘x^\circ$ , the area is 72 cm², and the angle shown is obtuse.

The diagram shows the included angle x° is obtuse, so the obtuse sine solution is needed.

Substitute into the formula.
$\frac{1}{2}\times 12\times 15\times \sin x$
Simplify.
$90\sin x$
Divide by 90.
$sin⁡x=0.8\sin x = 0.8$
Use inverse sine to find the calculator angle.
$\sin^{-1}(0.8)=53.130\ldots^\circ$
Because the diagram shows an obtuse angle, subtract from 180°.
$180∘−53.130…∘=126.869…∘180^\circ - 53.130\ldots^\circ = 126.869\ldots^\circ$
To 1 decimal place, $x=126.9∘x=126.9^\circ$ .

When the side lengths include algebra

If the sides are written using $x$ , you still use the same formula. This time, the formula creates an equation that you solve.

A useful exact value is:

sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}

Example

Finding an unknown side expression

A triangle has side lengths $x$ cm and $(x + 2)$ cm, with included angle 60°. Its area is $12312\sqrt{3}$ cm².

Algebraic side lengths can be substituted into the same sine area formula.

Substitute into the sine area formula.
$123=12×x(x+2)×sin⁡60∘12\sqrt{3}=\frac{1}{2}\times x(x+2)\times \sin 60^\circ$
Replace $sin⁡60∘\sin 60^\circ$ with its exact value.
$123=12×x(x+2)×3212\sqrt{3}=\frac{1}{2}\times x(x+2)\times \frac{\sqrt{3}}{2}$
Simplify.
$123=x(x+2)3412\sqrt{3}=\frac{x(x+2)\sqrt{3}}{4}$
Divide by $3\sqrt{3}$ and multiply by 4.
$x (x + 2) = 48$
Expand and rearrange.
$x^2+2x-48=0$
Factorise.
$(x + 8) (x - 6) = 0$
The solutions are $x = - 8$ and $x = 6$ . A length cannot be negative, so $x = 6$ .

Using ratios for side lengths

Definition

Ratio

A ratio compares quantities in parts. If two lengths are in the ratio 3:2, they can be written as $3 k$ and $2 k$ , where $k$ is the value of one part.

Example

Using a side ratio

Two sides of a triangle meet at 30°. Their lengths are in the ratio 3:2, and the area is 54 cm². Find the shorter side.

Writing the ratio as 3k and 2k lets the two given sides be used in the sine area formula.

Let the two sides be $3 k$ cm and $2 k$ cm.
Substitute into the formula.
$54=12(3k)(2k)sin⁡30∘54=\frac{1}{2}(3k)(2k)\sin 30^\circ$
Use $sin⁡30∘=12\sin 30^\circ=\frac{1}{2}$ .
$54=32k254=\frac{3}{2}k^2$
Solve for $k^2$ .
$k^2=36$
Since lengths are positive, $k = 6$ . The shorter side is $2 k = 12$ cm.

Exam technique

In the exam

Check that the angle is between the two sides before using $A=12absin⁡CA=\frac{1}{2}ab\sin C$ .
If finding an angle, isolate $sin⁡x\sin x$ first, then use $sin^{-1}$ .
Look carefully at the diagram: if the angle is obtuse, use $180∘180^\circ$ minus the calculator angle.

Self review

Check yourself

Can you identify the included angle from a triangle diagram?
When would you use $A=12absin⁡CA=\frac{1}{2}ab\sin C$ instead of base times height?
If your calculator gives an acute angle, how do you find the possible obtuse angle?

Finding the Area of Any Triangle

What you'll learn

Start with the area formula you already know

The sine area formula

Exact angles can make the calculation shorter

Working backwards to find a missing angle

When the side lengths include algebra

Using ratios for side lengths

Recap questions

You've reached the end

Finding the Area of Any Triangle

What you'll learn

Start with the area formula you already know

The sine area formula

Exact angles can make the calculation shorter

Working backwards to find a missing angle

When the side lengths include algebra

Using ratios for side lengths

Recap questions

You've reached the end

Finding the Area of Any Triangle

What you'll learn

Start with the area formula you already know

The sine area formula

Exact angles can make the calculation shorter

Working backwards to find a missing angle

When the side lengths include algebra

Using ratios for side lengths

Recap questions

You've reached the end

Finding the Area of Any Triangle Revision Guide

Finding the Area of Any Triangle

What you'll learn

Start with the area formula you already know

The sine area formula

Exact angles can make the calculation shorter

Working backwards to find a missing angle

When the side lengths include algebra

Using ratios for side lengths

Recap questions

You've reached the end

Finding the Area of Any Triangle Revision Guide