Finding the Area of Any Triangle

What you'll learn

How to find a triangle’s area when you know two sides and the angle between them.
Why the formula uses sine.
How to work backwards to find an unknown angle.
How to handle harder questions with algebraic side lengths or ratios.

Start with the area formula you already know

You probably know the basic area formula for a triangle:

A=12×base×heightA=\frac12 \times \text{base} \times \text{height}

The base is the side you choose to measure along the bottom, and the height is the perpendicular distance from the opposite vertex to that base. “Perpendicular” means it meets the base at 90°.

Example

Using base and perpendicular height

A triangle has base 8 cm and perpendicular height 5 cm. Write the usual area formula.
$A=12×base×heightA=\frac12 \times \text{base} \times \text{height}$
Substitute the values.
$A=12×8×5A=\frac12 \times 8 \times 5$
Work it out.
$A = 20$
The area is 20 cm².

The problem is that in many exam questions, the perpendicular height is not given. Instead, you are given two sides and the angle between them. That is where the sine formula comes in.

The included angle

Definition

Included angle

The included angle is the angle between two given sides. If two sides meet at a vertex, the angle at that vertex is the included angle.

For this topic, the key information is usually:

one side length,
another side length,
the angle between those two sides.

The diagram below shows how the height can be created using trigonometry, even when it is not drawn in the original question.

Triangle showing two sides, the included angle, and the perpendicular height used to derive the sine area formula

Why sine appears

In a right-angled triangle, sine connects an angle to the opposite side and the hypotenuse:

sin⁡θ=oppositehypotenuse\sin \theta=\frac{\text{opposite}}{\text{hypotenuse}}

If a sloping side has length $b$ , and it makes an angle $θ\theta$ with the base, then the perpendicular height is:

h=bsin⁡θh=b\sin\theta

So in the usual formula $A=12×base×heightA=\frac12 \times \text{base} \times \text{height}$ , we can replace the height with $bsin⁡θb\sin\theta$ .

Key Idea

Area using two sides and the included angle

If two sides are $a$ and $b$ , and the included angle is $θ\theta$ , then the area is $A=12absin⁡θA=\frac12 ab\sin\theta$ .

Common Mistake

Using a non-included angle

The formula $A=12absin⁡θA=\frac12 ab\sin\theta$ only works when $θ\theta$ is the angle between the two sides $a$ and $b$ . If the angle is somewhere else, you cannot use the formula directly.

Finding the area when two sides and the included angle are known

This is the most common style of question. You simply identify the two sides, identify the included angle, substitute into the formula, then round as requested.

Example

Area with an obtuse included angle

A triangle has two sides of length 13 cm and 11 cm. The angle between them is 120°. Work out the area to 1 decimal place.

Write the formula.
$A=12absin⁡θA=\frac12 ab\sin\theta$
Substitute $a = 13$ , $b = 11$ , and $θ=120∘\theta=120^\circ$ .
$A=12×13×11×sin⁡120∘A=\frac12 \times 13 \times 11 \times \sin 120^\circ$
Evaluate using a calculator in degree mode.
$A=61.9208…A=61.9208\ldots$
Round to 1 decimal place: 61.9 cm².

Tip

Obtuse angles are fine

The formula works for acute, right-angled, and obtuse triangles. For example, $sin⁡120∘\sin 120^\circ$ is positive, so the area still comes out positive.

Rounding your answer

Exam questions often ask for a certain number of decimal places or significant figures.

Decimal places count digits after the decimal point.
Significant figures count important digits from the first non-zero digit.

Keep the full calculator value until the final line, then round once.

Example

Area to 3 significant figures

A triangle has two sides of length 7 m and 12 m. The included angle is 38°. Find the area to 3 significant figures.

Start with the sine area formula.
$A=12absin⁡θA=\frac12 ab\sin\theta$
Substitute the values.
$A=12×7×12×sin⁡38∘A=\frac12 \times 7 \times 12 \times \sin 38^\circ$
Calculate the unrounded value.
$A=25.8577…A=25.8577\ldots$
To 3 significant figures, the area is 25.9 m².

Tip

Check your calculator mode

For IGCSE trigonometry questions, your calculator should usually be in degree mode. If your answer looks wildly too small or too large, check for DEG rather than RAD.

Working backwards to find the angle

Sometimes the area is given, and you need to find the included angle. In that case, use the same formula but rearrange for $sin⁡x\sin x$ .

You will then use the inverse sine button, written as $sin^{-1}$ , to find the angle.

Definition

Inverse sine

The inverse sine, written $sin^{-1}$ , is used to find an angle when you know its sine value.

Example

Finding an unknown included angle

A triangle has sides 12 cm and 17 cm with included angle $x$ . Its area is 80 cm². The diagram shows that $x$ is acute. Find $x$ to 1 decimal place.

Substitute into $A=12absin⁡xA=\frac12 ab\sin x$ .
$80=12×12×17×sin⁡x80=\frac12 \times 12 \times 17 \times \sin x$
Simplify the multiplication.
$80=102sin⁡x80=102\sin x$
Divide by 102 to isolate $sin⁡x\sin x$ .
$sin⁡x=80102\sin x=\frac{80}{102}$
Use inverse sine.
$x=sin⁡−1(80102)x=\sin^{-1}\left(\frac{80}{102}\right)$
Calculate and round: $x=51.6∘x=51.6^\circ$ to 1 decimal place.

Common Mistake

Sine can give two possible angles

For angles inside a triangle, $sin⁡x\sin x$ has the same value for $x$ and $180∘−x180^\circ-x$ . If the diagram or wording shows an obtuse angle, you may need to subtract your calculator angle from 180°.

When the side lengths contain algebra

Some questions give side lengths such as $x$ and $x + 6$ . The method is still the same, but after substituting you may need to expand, factorise, or solve a quadratic equation.

A quadratic equation is an equation involving $x^2$ , such as $x^2+6x-72=0$ .

Example

Algebraic side lengths with a 60° angle

The sides enclosing a 60° angle are $x$ cm and $x + 6$ cm. The area is $18318\sqrt{3}$ cm². Work out $x$ .

Write the area formula with the given sides.
$183=12×x×(x+6)×sin⁡60∘18\sqrt{3}=\frac12 \times x \times (x+6)\times \sin 60^\circ$
Use the exact value $sin⁡60∘=32\sin 60^\circ=\frac{\sqrt{3}}{2}$ .
$183=x(x+6)3418\sqrt{3}=\frac{x(x+6)\sqrt{3}}{4}$
Divide by $3\sqrt{3}$ , then multiply by 4.
$72 = x (x + 6)$
Expand and rearrange into a quadratic equation.
$x^2+6x-72=0$
Factorise.
$(x + 12) (x - 6) = 0$
Solve and reject the negative value because a length cannot be negative: $x = 6$ cm.

Common Mistake

Keeping a negative length

Quadratics can produce two answers, but side lengths must be positive. Always reject any negative length in a geometry question.

Using ratios for side lengths

If two side lengths are given in a ratio, introduce a multiplier. For example, if the ratio is 3:2, the side lengths can be written as $3 k$ and $2 k$ .

Example

Area with side lengths in a ratio

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

Let the two sides be $3 k$ and $2 k$ .
Substitute into the area formula.
$54=12×3k×2k×sin⁡30∘54=\frac12 \times 3k \times 2k \times \sin 30^\circ$
Use $sin⁡30∘=12\sin 30^\circ=\frac12$ and simplify.
$54=32k254=\frac32 k^2$
Solve for $k^2$ .
$k^2=36$
Since $k$ is a length multiplier, take the positive square root: $k = 6$ .
The shorter side is $2 k = 12$ , so the shorter side is 12 cm.

Exam technique

In the exam

Check that the angle given is between the two sides you are using.
Write $A=12absin⁡θA=\frac12 ab\sin\theta$ before substituting values.
Keep full calculator accuracy until the final answer, then round exactly as requested.
If solving for an angle, think about whether the diagram suggests an acute or obtuse angle.
If solving for a length, reject negative answers.

Self review

Check yourself

Can you explain what the included angle means?
If the area and two sides are given, what operation helps you find the angle?
Why might a quadratic equation give an answer you must reject?

Finding the Area of Any Triangle

What you'll learn

How to find a triangle’s area when you know two sides and the angle between them.
Why the formula uses sine.
How to work backwards to find an unknown angle.
How to handle harder questions with algebraic side lengths or ratios.

Start with the area formula you already know

You probably know the basic area formula for a triangle:

A=12×base×heightA=\frac12 \times \text{base} \times \text{height}

Example

Using base and perpendicular height

A triangle has base 8 cm and perpendicular height 5 cm. Write the usual area formula.
$A=12×base×heightA=\frac12 \times \text{base} \times \text{height}$
Substitute the values.
$A=12×8×5A=\frac12 \times 8 \times 5$
Work it out.
$A = 20$
The area is 20 cm².

The problem is that in many exam questions, the perpendicular height is not given. Instead, you are given two sides and the angle between them. That is where the sine formula comes in.

The included angle

Definition

Included angle

The included angle is the angle between two given sides. If two sides meet at a vertex, the angle at that vertex is the included angle.

For this topic, the key information is usually:

one side length,
another side length,
the angle between those two sides.

The diagram below shows how the height can be created using trigonometry, even when it is not drawn in the original question.

Triangle showing two sides, the included angle, and the perpendicular height used to derive the sine area formula

Why sine appears

In a right-angled triangle, sine connects an angle to the opposite side and the hypotenuse:

sin⁡θ=oppositehypotenuse\sin \theta=\frac{\text{opposite}}{\text{hypotenuse}}

If a sloping side has length $b$ , and it makes an angle $θ\theta$ with the base, then the perpendicular height is:

h=bsin⁡θh=b\sin\theta

So in the usual formula $A=12×base×heightA=\frac12 \times \text{base} \times \text{height}$ , we can replace the height with $bsin⁡θb\sin\theta$ .

Key Idea

Area using two sides and the included angle

If two sides are $a$ and $b$ , and the included angle is $θ\theta$ , then the area is $A=12absin⁡θA=\frac12 ab\sin\theta$ .

Common Mistake

Using a non-included angle

The formula $A=12absin⁡θA=\frac12 ab\sin\theta$ only works when $θ\theta$ is the angle between the two sides $a$ and $b$ . If the angle is somewhere else, you cannot use the formula directly.

Finding the area when two sides and the included angle are known

This is the most common style of question. You simply identify the two sides, identify the included angle, substitute into the formula, then round as requested.

Example

Area with an obtuse included angle

A triangle has two sides of length 13 cm and 11 cm. The angle between them is 120°. Work out the area to 1 decimal place.

Write the formula.
$A=12absin⁡θA=\frac12 ab\sin\theta$
Substitute $a = 13$ , $b = 11$ , and $θ=120∘\theta=120^\circ$ .
$A=12×13×11×sin⁡120∘A=\frac12 \times 13 \times 11 \times \sin 120^\circ$
Evaluate using a calculator in degree mode.
$A=61.9208…A=61.9208\ldots$
Round to 1 decimal place: 61.9 cm².

Tip

Obtuse angles are fine

The formula works for acute, right-angled, and obtuse triangles. For example, $sin⁡120∘\sin 120^\circ$ is positive, so the area still comes out positive.

Rounding your answer

Exam questions often ask for a certain number of decimal places or significant figures.

Decimal places count digits after the decimal point.
Significant figures count important digits from the first non-zero digit.

Keep the full calculator value until the final line, then round once.

Example

Area to 3 significant figures

A triangle has two sides of length 7 m and 12 m. The included angle is 38°. Find the area to 3 significant figures.

Start with the sine area formula.
$A=12absin⁡θA=\frac12 ab\sin\theta$
Substitute the values.
$A=12×7×12×sin⁡38∘A=\frac12 \times 7 \times 12 \times \sin 38^\circ$
Calculate the unrounded value.
$A=25.8577…A=25.8577\ldots$
To 3 significant figures, the area is 25.9 m².

Tip

Check your calculator mode

For IGCSE trigonometry questions, your calculator should usually be in degree mode. If your answer looks wildly too small or too large, check for DEG rather than RAD.

Working backwards to find the angle

Sometimes the area is given, and you need to find the included angle. In that case, use the same formula but rearrange for $sin⁡x\sin x$ .

You will then use the inverse sine button, written as $sin^{-1}$ , to find the angle.

Definition

Inverse sine

The inverse sine, written $sin^{-1}$ , is used to find an angle when you know its sine value.

Example

Finding an unknown included angle

A triangle has sides 12 cm and 17 cm with included angle $x$ . Its area is 80 cm². The diagram shows that $x$ is acute. Find $x$ to 1 decimal place.

Substitute into $A=12absin⁡xA=\frac12 ab\sin x$ .
$80=12×12×17×sin⁡x80=\frac12 \times 12 \times 17 \times \sin x$
Simplify the multiplication.
$80=102sin⁡x80=102\sin x$
Divide by 102 to isolate $sin⁡x\sin x$ .
$sin⁡x=80102\sin x=\frac{80}{102}$
Use inverse sine.
$x=sin⁡−1(80102)x=\sin^{-1}\left(\frac{80}{102}\right)$
Calculate and round: $x=51.6∘x=51.6^\circ$ to 1 decimal place.

Common Mistake

Sine can give two possible angles

When the side lengths contain algebra

Some questions give side lengths such as $x$ and $x + 6$ . The method is still the same, but after substituting you may need to expand, factorise, or solve a quadratic equation.

A quadratic equation is an equation involving $x^2$ , such as $x^2+6x-72=0$ .

Example

Algebraic side lengths with a 60° angle

The sides enclosing a 60° angle are $x$ cm and $x + 6$ cm. The area is $18318\sqrt{3}$ cm². Work out $x$ .

Write the area formula with the given sides.
$183=12×x×(x+6)×sin⁡60∘18\sqrt{3}=\frac12 \times x \times (x+6)\times \sin 60^\circ$
Use the exact value $sin⁡60∘=32\sin 60^\circ=\frac{\sqrt{3}}{2}$ .
$183=x(x+6)3418\sqrt{3}=\frac{x(x+6)\sqrt{3}}{4}$
Divide by $3\sqrt{3}$ , then multiply by 4.
$72 = x (x + 6)$
Expand and rearrange into a quadratic equation.
$x^2+6x-72=0$
Factorise.
$(x + 12) (x - 6) = 0$
Solve and reject the negative value because a length cannot be negative: $x = 6$ cm.

Common Mistake

Keeping a negative length

Quadratics can produce two answers, but side lengths must be positive. Always reject any negative length in a geometry question.

Using ratios for side lengths

If two side lengths are given in a ratio, introduce a multiplier. For example, if the ratio is 3:2, the side lengths can be written as $3 k$ and $2 k$ .

Example

Area with side lengths in a ratio

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

Let the two sides be $3 k$ and $2 k$ .
Substitute into the area formula.
$54=12×3k×2k×sin⁡30∘54=\frac12 \times 3k \times 2k \times \sin 30^\circ$
Use $sin⁡30∘=12\sin 30^\circ=\frac12$ and simplify.
$54=32k254=\frac32 k^2$
Solve for $k^2$ .
$k^2=36$
Since $k$ is a length multiplier, take the positive square root: $k = 6$ .
The shorter side is $2 k = 12$ , so the shorter side is 12 cm.

Exam technique

In the exam

Check that the angle given is between the two sides you are using.
Write $A=12absin⁡θA=\frac12 ab\sin\theta$ before substituting values.
Keep full calculator accuracy until the final answer, then round exactly as requested.
If solving for an angle, think about whether the diagram suggests an acute or obtuse angle.
If solving for a length, reject negative answers.

Self review

Check yourself

Can you explain what the included angle means?
If the area and two sides are given, what operation helps you find the angle?
Why might a quadratic equation give an answer you must reject?

Finding the Area of Any Triangle

What you'll learn

Start with the area formula you already know

The included angle

Why sine appears

Finding the area when two sides and the included angle are known

Rounding your answer

Working backwards to find the angle

When the side lengths contain 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 included angle

Why sine appears

Finding the area when two sides and the included angle are known

Rounding your answer

Working backwards to find the angle

When the side lengths contain 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 included angle

Why sine appears

Finding the area when two sides and the included angle are known

Rounding your answer

Working backwards to find the angle

When the side lengths contain 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 included angle

Why sine appears

Finding the area when two sides and the included angle are known

Rounding your answer

Working backwards to find the angle

When the side lengths contain algebra

Using ratios for side lengths

Recap questions

You've reached the end

Finding the Area of Any Triangle Revision Guide