What does SOHCAHTOA stand for?

SOHCAHTOA summarises the three right-angled trigonometric ratios: $\sin\theta=O/H$, $\cos\theta=A/H$ and $\tan\theta=O/A$. Here, $O$ is opposite, $A$ is adjacent and $H$ is the hypotenuse.

How do I know which trigonometric ratio to use?

Label the sides relative to the chosen angle, then select the ratio containing the known side and the required side. Use sine for opposite and hypotenuse, cosine for adjacent and hypotenuse, and tangent for opposite and adjacent.

How do I find an angle using SOHCAHTOA?

Form the appropriate side ratio, then use an inverse function. For example, if the opposite and hypotenuse are known, use $\theta=\sin^{-1}(O/H)$. Ensure the calculator is in degree mode.

Does SOHCAHTOA work for every triangle?

No. SOHCAHTOA applies directly to right-angled triangles. For a non-right-angled triangle, a Higher-tier question may require the sine rule, cosine rule or area formula instead.

What calculator mode should I use for GCSE trigonometry?

Use **degree mode**, usually shown by “DEG”, because GCSE angles in these questions are measured in degrees. Radian mode will generally produce an incorrect answer.

Is SOHCAHTOA on Foundation or Higher GCSE Maths?

Right-angled trigonometry is studied at both Foundation and Higher tier. Higher-tier assessments may place it in less familiar diagrams or combine it with other methods, so always check your exam board’s current specification and tier requirements.

Trigonometry (SOHCAHTOA) GCSE Maths Guide

Trigonometry (SOHCAHTOA) GCSE Maths revision guide

Trigonometry (SOHCAHTOA) GCSE Maths questions use the sine, cosine and tangent ratios to connect angles and side lengths in right-angled triangles. You may need to find a missing side, calculate an angle or apply trigonometry in a geometrical problem.

This guide explains the method step by step, including trigonometry (SOHCAHTOA) worked examples, common mistakes and exam-style questions. These skills are relevant to both Foundation and Higher tier, although Higher questions may involve more complex diagrams and multi-step reasoning. You can compare this topic with other content in the GCSE Maths topics by exam board and tier.

What is right-angled trigonometry?

Right-angled trigonometry describes relationships between the sides and angles of a right-angled triangle. The three relationships used at GCSE are:

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

cos⁡θ=adjacenthypotenuse\cos \theta=\frac{\text{adjacent}}{\text{hypotenuse}}

tan⁡θ=oppositeadjacent\tan \theta=\frac{\text{opposite}}{\text{adjacent}}

Here, $θ\theta$ represents the angle being used. The Greek letter $θ\theta$ , pronounced “theta”, is commonly used for an unknown or general angle.

The mnemonic SOHCAHTOA helps you remember the three ratios:

SOH means $sin⁡θ=OH\sin \theta=\dfrac{O}{H}$ .
CAH means $cos⁡θ=AH\cos \theta=\dfrac{A}{H}$ .
TOA means $tan⁡θ=OA\tan \theta=\dfrac{O}{A}$ .

SOHCAHTOA applies directly only to right-angled triangles. You should therefore check for a right-angle symbol before choosing this method.

Identifying the sides of a right-angled triangle

The side names depend on the position of the marked angle $θ\theta$ .

The hypotenuse

The hypotenuse is always opposite the right angle. It is the longest side of a right-angled triangle, and its position does not change when a different acute angle is selected.

The opposite side

The opposite side is directly across from the angle $θ\theta$ . It does not touch that angle.

The adjacent side

The adjacent side is next to the angle $θ\theta$ , but it is not the hypotenuse. The word adjacent means “next to”.

If the selected acute angle changes, the opposite and adjacent sides exchange roles. This is why you should label the sides for the particular angle in the question instead of relying on the triangle’s orientation.

How to choose sine, cosine or tangent

Once you have labelled the relevant sides $O$ , $A$ and $H$ , identify which two sides are involved. Choose the ratio containing both of them.

Sides involved	Ratio to use	Relationship
Opposite and hypotenuse	Sine	$sin⁡θ=OH\sin\theta=\dfrac{O}{H}$
Adjacent and hypotenuse	Cosine	$cos⁡θ=AH\cos\theta=\dfrac{A}{H}$
Opposite and adjacent	Tangent	$tan⁡θ=OA\tan\theta=\dfrac{O}{A}$

For example, if the question gives an angle and the hypotenuse, then asks for the opposite side, use sine. If it gives the opposite and adjacent sides and asks for an angle, use tangent.

A reliable exam method is:

Confirm that the triangle is right-angled.
Mark the angle being used.
Label the relevant sides $O$ , $A$ and $H$ .
Select the ratio containing the known side and the required side.
Substitute values before rearranging where possible.
Calculate using degree mode.
Round only at the end and include units when finding a length.

Trigonometry (SOHCAHTOA) worked examples

Worked example 1: finding the opposite side

A right-angled triangle has an angle of $38∘38^\circ$ and a hypotenuse of $12$ cm. Find the side opposite the $38∘38^\circ$ angle.

The known and required sides are the hypotenuse and opposite, so use sine:

sin⁡θ=OH\sin\theta=\frac{O}{H}

Substitute $θ=38∘\theta=38^\circ$ and $H = 12$ :

sin⁡38∘=x12\sin 38^\circ=\frac{x}{12}

Multiply both sides by $12$ :

x=12sin⁡38∘x=12\sin 38^\circ

x=7.387…x=7.387\ldots

Therefore, to $3$ significant figures,

cm\boxed{x=7.39\text{ cm}}

The answer is shorter than the hypotenuse, which is sensible.

Worked example 2: finding the adjacent side

A right-angled triangle has an angle of $54∘54^\circ$ and a hypotenuse of $9.5$ m. Calculate the side adjacent to the $54∘54^\circ$ angle.

The adjacent side and hypotenuse indicate cosine:

cos⁡θ=AH\cos\theta=\frac{A}{H}

Substitute the information:

cos⁡54∘=x9.5\cos54^\circ=\frac{x}{9.5}

Multiply by $9.5$ :

x=9.5cos⁡54∘x=9.5\cos54^\circ

x=5.584…x=5.584\ldots

To $3$ significant figures,

m\boxed{x=5.58\text{ m}}

Worked example 3: finding the hypotenuse

The side adjacent to an angle of $41∘41^\circ$ is $8$ cm. Find the hypotenuse.

Cosine connects the adjacent side and hypotenuse:

cos⁡41∘=8x\cos41^\circ=\frac{8}{x}

Since $x$ is in the denominator, rearrange carefully:

xcos⁡41∘=8x\cos41^\circ=8

x=8cos⁡41∘x=\frac{8}{\cos41^\circ}

x=10.600…x=10.600\ldots

Therefore,

cm\boxed{x=10.6\text{ cm}}

Dividing by the trigonometric value produces a length greater than $8$ cm. This is expected because the unknown is the hypotenuse.

Worked example 4: finding an angle

A right-angled triangle has an opposite side of $7$ cm and an adjacent side of $11$ cm. Find the angle $θ\theta$ .

Opposite and adjacent indicate tangent:

tan⁡θ=OA\tan\theta=\frac{O}{A}

Substitute the side lengths:

tan⁡θ=711\tan\theta=\frac{7}{11}

To undo tangent, use the inverse tangent function:

θ=tan⁡−1(711)\theta=\tan^{-1}\left(\frac{7}{11}\right)

θ=32.471…∘\theta=32.471\ldots^\circ

To $1$ decimal place,

θ=32.5∘\boxed{\theta=32.5^\circ}

The notation $tan^{-1}$ means inverse tangent here. It does not mean $1÷tan⁡θ1\div\tan\theta$ .

Finding missing sides and angles

The algebra depends on what the question asks you to find.

Finding a side in the numerator

If the required side is on top of the fraction, multiply by the denominator. For example:

sin⁡35∘=x14\sin35^\circ=\frac{x}{14}

becomes

x=14sin⁡35∘.x=14\sin35^\circ.

Finding a side in the denominator

If the required side is on the bottom of the fraction, divide the known length by the trigonometric value. For example:

tan⁡28∘=6x\tan28^\circ=\frac{6}{x}

becomes

x=6tan⁡28∘.x=\frac{6}{\tan28^\circ}.

This pattern is particularly important when finding a hypotenuse with sine or cosine.

Finding an angle

When the angle is unknown, use an inverse trigonometric function:

θ=sin⁡−1(OH),\theta=\sin^{-1}\left(\frac{O}{H}\right),

θ=cos⁡−1(AH),\theta=\cos^{-1}\left(\frac{A}{H}\right),

θ=tan⁡−1(OA).\theta=\tan^{-1}\left(\frac{O}{A}\right).

On many scientific calculators, inverse functions are accessed using the shift or second-function key. The display should show $sin^{-1}$ , $cos^{-1}$ or $tan^{-1}$ .

Calculator settings and rounding

Your calculator must be in degree mode for GCSE questions in which angles are measured in degrees. Look for a degree indicator such as “DEG”. Radian mode gives a different answer and is a common source of unexplained calculator errors.

Keep the full calculator value during intermediate steps. Round only the final answer to the accuracy requested, such as $1$ decimal place or $3$ significant figures. If no accuracy is specified, give a sensible degree of accuracy and show enough working for the method to be clear.

Lengths require units, such as cm, m or km. Angles require the degree symbol, $∘^\circ$ .

Multi-step trigonometry questions

Some trigonometry (SOHCAHTOA) exam questions do not present a single isolated triangle. You may need to:

split a shape into right-angled triangles;
calculate one length before using it in a second triangle;
combine trigonometry with Pythagoras’ theorem;
identify an angle of elevation or depression;
use a diagram that is not drawn to scale.

Write down the ratio for each triangle separately. Do not use a side found in an earlier step until you have kept sufficient unrounded digits, because premature rounding can make the final answer inaccurate.

An angle of elevation is measured upwards from a horizontal line. An angle of depression is measured downwards from a horizontal line. Marking the horizontal clearly often makes the correct right-angled triangle easier to identify.

Common mistakes in SOHCAHTOA questions

Labelling sides before selecting the angle

Opposite and adjacent are defined relative to the chosen angle. Identify $θ\theta$ first, then label the sides.

Calling any long-looking side the hypotenuse

The hypotenuse is not identified by the way the diagram is drawn. It must be directly opposite the right angle.

Using SOHCAHTOA on a non-right-angled triangle

SOHCAHTOA requires a right angle. A general triangle may require a different method, such as the sine rule or cosine rule at Higher tier.

Reversing a fraction

For tangent, the relationship is $O / A$ , not $A / O$ . Writing the formula before substituting values reduces this risk.

Using a normal ratio when finding an angle

To find an unknown angle, use the relevant inverse function. For example, if $sin⁡θ=0.6\sin\theta=0.6$ , then $θ=sin⁡−1(0.6)\theta=\sin^{-1}(0.6)$ .

Rounding too early

Store the calculator value or copy several decimal places into the next step. Only round the final result.

Giving an impossible length

The hypotenuse must be the longest side. If your calculated hypotenuse is shorter than another side, check your rearrangement and side labels.

Trigonometry (SOHCAHTOA) exam question

A ladder $6.4$ m long rests against a vertical wall. The ladder makes an angle of $68∘68^\circ$ with the horizontal ground. Calculate the vertical height reached by the ladder. Give your answer to $2$ decimal places.

The ladder is the hypotenuse, and the vertical height is opposite the $68∘68^\circ$ angle. Therefore:

sin⁡68∘=h6.4\sin68^\circ=\frac{h}{6.4}

h=6.4sin⁡68∘h=6.4\sin68^\circ

h=5.933…h=5.933\ldots

So the ladder reaches a vertical height of

m.\boxed{5.93\text{ m}}.

Try 4 more trigonometry practice questions.

Further exam-style practice questions

Question 1

A right-angled triangle has an angle of $27∘27^\circ$ and a hypotenuse of $15$ cm. Find the side adjacent to the angle. Give your answer to $3$ significant figures.

Question 2

In a right-angled triangle, the side opposite $θ\theta$ is $9$ cm and the adjacent side is $13$ cm. Find $θ\theta$ to $1$ decimal place.

Question 3

The side opposite an angle of $36∘36^\circ$ is $7.2$ m. Calculate the hypotenuse to $2$ decimal places.

Question 4

A tree casts a horizontal shadow $18$ m long. The angle of elevation from the end of the shadow to the top of the tree is $31∘31^\circ$ . Calculate the height of the tree to $1$ decimal place.

Answers

$15cos⁡27∘=13.365…15\cos27^\circ=13.365\ldots$ , so the adjacent side is $cm\boxed{13.4\text{ cm}}$ .
$θ=tan⁡−1(913)=34.695…∘\theta=\tan^{-1}\left(\dfrac{9}{13}\right)=34.695\ldots^\circ$ , so $34.7∘\boxed{34.7^\circ}$ .
$sin⁡36∘=7.2H\sin36^\circ=\dfrac{7.2}{H}$ , so $mH=\dfrac{7.2}{\sin36^\circ}=\boxed{12.25\text{ m}}$ .
$tan⁡31∘=h18\tan31^\circ=\dfrac{h}{18}$ , so $h=18tan⁡31∘=10.815…h=18\tan31^\circ=10.815\ldots$ , giving $m\boxed{10.8\text{ m}}$ .

How to revise trigonometry effectively

Start by learning the meaning of opposite, adjacent and hypotenuse rather than memorising calculations. Then practise choosing the ratio without completing the arithmetic. This isolates the most important decision in each question.

Next, practise the three main question types: finding a side in the numerator, finding a side in the denominator and finding an angle. Finish with mixed exam questions so that the required ratio is not signposted.

Always show the trigonometric equation before entering values into your calculator. This can make your reasoning clear and may earn method marks even if a later arithmetic error occurs. MathsGenie practice questions, mark schemes and past-paper practice can then help you apply the method in unfamiliar diagrams.

Conclusion

SOHCAHTOA links the sides and angles of right-angled triangles through sine, cosine and tangent. The essential process is to identify the right angle, label the sides relative to $θ\theta$ , choose the ratio containing the relevant sides and rearrange accurately.

Use inverse functions when finding angles, keep your calculator in degree mode and round only at the end. After reviewing the worked examples, use the MathsGenie trigonometry practice questions and mark schemes to strengthen your exam technique.

Sources and referenced URLs