GCSE Maths · Geometry and measures

Trigonometry GCSE Questions, Worked Examples and Answers

Trigonometry links the angles and side lengths of a triangle. In a right-angled triangle, SOHCAHTOA helps you choose sine, cosine or tangent. Opposite and adjacent are named relative to the angle you select, while the hypotenuse is always opposite the right angle. GCSE trigonometry questions usually ask you to find a missing side, find an angle or solve a contextual problem.

Edexcel · AQA · OCRFoundation & Higher12 original questions

Start the questions Learn the method

Core knowledge

What you need to know

Start by identifying the triangle and naming its sides. The side labels determine the ratio.

Select angle αThe horizontal side is adjacent; the vertical side is opposite.

Now select angle βOpposite and adjacent swap; the hypotenuse does not.

Right angle and hypotenuse

A right angle is $90^\circ$ . The hypotenuse is opposite it and is the longest side.

Opposite and adjacent

Opposite is across from the selected acute angle. Adjacent touches that angle but is not the hypotenuse.

Labels can change

Select the other acute angle and opposite and adjacent swap. The hypotenuse stays fixed.

Calculator setup

Use degree mode. Keep the full calculator value through the working and round only the final answer.

SOH

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

CAH

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

TOA

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

Finding a side

Use $\sin$ , $\cos$ or $\tan$ , then rearrange.

Finding an angle

Use $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ after forming the side ratio.

Use this every time

The five-step SOHCAHTOA method

Keep the same order for side, angle and contextual questions. It makes the ratio choice visible and protects method marks.

Label the hypotenuse, opposite and adjacent sides for the given angle.
Choose sin, cos or tan using the two sides involved.
Substitute the known values and the unknown into the ratio.
Rearrange for a side, or use the inverse function for an angle.
Calculate in degree mode, then check and round only at the end.

Fully worked

GCSE trigonometry worked examples

The examples progress from direct calculations to contextual and 3D reasoning.

Example 1

Find an opposite side using tangent

2 marks

Question

The side adjacent to a $38^\circ$ angle is $7$ cm. Find the opposite side, $x$ , to 3 significant figures.

Mark the right angle first, then label sides relative to the shown angle.

Method choice: opposite and adjacent appear in TOA, so use tangent.

$\tan 38^\circ=\frac{x}{7}$

$x=7\tan 38^\circ$

$x=5.468999\ldots$

$\boxed{x=5.47\text{ cm}}$

Exam tip: write the ratio equation before pressing calculator buttons.

Example 2

Find the hypotenuse using cosine

3 marks

Question

A right-angled triangle has an angle of $54^\circ$ and an adjacent side of $8.2$ m. Find the hypotenuse, $h$ , to 3 significant figures.

Method choice: adjacent and hypotenuse appear in CAH, so use cosine. Because the unknown is in the denominator, rearrange carefully.

$\cos 54^\circ=\frac{8.2}{h}$

$h\cos 54^\circ=8.2$

$h=\frac{8.2}{\cos 54^\circ}$

$h=13.950673\ldots$

$\boxed{h=14.0\text{ m}}$

Exam tip: the hypotenuse must be longer than $8.2$ m; use that to check the result.

Example 3

Find an angle using inverse tangent

3 marks

Question

The opposite side of a right-angled triangle is $9.4$ cm and the adjacent side is $12.7$ cm. Find the angle $\theta$ to 1 decimal place.

Method choice: form the opposite-to-adjacent ratio, then use inverse tangent because the angle is unknown.

$\tan\theta=\frac{9.4}{12.7}$

$\theta=\tan^{-1}\left(\frac{9.4}{12.7}\right)$

$\theta=36.507270\ldots^\circ$

$\boxed{\theta=36.5^\circ}$

Exam tip: $\tan^{-1}$ means inverse tangent here, not $1\div\tan$ .

Example 4 · Context

Find a tree height

4 marks

Question

A learner stands $14.5$ m from a tree on level ground. Her eye level is $1.6$ m above the ground. The angle of elevation to the top is $32^\circ$ . Find the total height of the tree to 3 significant figures.

The tangent calculation gives only the height above eye level. Add

1.6

m afterwards.

Method choice: tangent finds the vertical height above eye level. The final step must add the learner's eye height.

$\tan 32^\circ=\frac{x}{14.5}$

$x=14.5\tan 32^\circ$

$x=9.060605\ldots$

$\text{total height}=9.060605\ldots+1.6$

$\text{total height}=10.660605\ldots$

$\boxed{\text{total height}=10.7\text{ m}}$

Exam tip: translate the context into a triangle before choosing the ratio, and check whether the calculated side is the final answer.

Example 5 · 3D

Find an angle in a cuboid

Higher only4 marks

Question

A cuboid has a base measuring $8$ cm by $6$ cm and a height of $7.5$ cm. Find the angle between the space diagonal and the base, to 1 decimal place.

Find the base diagonal with Pythagoras, then use it as the adjacent side in the right-angled vertical cross-section.

Method choice: first use Pythagoras to expose a right-angled vertical cross-section. Then use inverse tangent.

$d^2=8^2+6^2$

$d^2=100$

$d=10\text{ cm}$

$\tan\theta=\frac{7.5}{10}$

$\theta=\tan^{-1}(0.75)$

$\theta=36.869897\ldots^\circ$

$\boxed{\theta=36.9^\circ}$

Exam tip: in 3D, draw or highlight the right-angled cross-section containing the required angle.

Original practice

Trigonometry GCSE exam questions

Try all 12 questions before opening the worked answers. The set moves from side labels to multi-step applications.

How to use this setAllow about 40 minutes · show each ratio and rearrangement · total 32 marks

Name the sides

2 marks

A right-angled triangle has a selected acute angle $\theta$ . The side across from $\theta$ is labelled $p$ , the side beside $\theta$ that is not the hypotenuse is labelled $q$ , and the longest side is labelled $r$ . Name $p$ , $q$ and $r$ .

Show worked answer

The names are always taken relative to the selected angle.

$p=\text{opposite}$

$q=\text{adjacent}$

$r=\text{hypotenuse}$

Find an opposite side

2 marks

In a right-angled triangle, the side adjacent to an angle of $42^\circ$ is $6.4$ cm. Find the opposite side, $x$ . Give your answer to 3 significant figures.

Show worked answer

Opposite and adjacent mean use tangent.

$\tan 42^\circ=\frac{x}{6.4}$

$x=6.4\tan 42^\circ$

$x=5.762585\ldots$

$\boxed{x=5.76\text{ cm}}$

Find an adjacent side

2 marks

The side opposite a $35^\circ$ angle is $8.1$ m. Find the adjacent side, $a$ . Give your answer to 3 significant figures.

Show worked answer

Opposite and adjacent mean use tangent.

$\tan 35^\circ=\frac{8.1}{a}$

$a\tan 35^\circ=8.1$

$a=\frac{8.1}{\tan 35^\circ}$

$a=11.567998\ldots$

$\boxed{a=11.6\text{ m}}$

Find a hypotenuse with sine

2 marks

A right-angled triangle has an angle of $28^\circ$ and an opposite side of $7.3$ cm. Find the hypotenuse, $h$ , to 3 significant figures.

Show worked answer

Opposite and hypotenuse mean use sine.

$\sin 28^\circ=\frac{7.3}{h}$

$h\sin 28^\circ=7.3$

$h=\frac{7.3}{\sin 28^\circ}$

$h=15.549397\ldots$

$\boxed{h=15.5\text{ cm}}$

Find an adjacent side with cosine

2 marks

The hypotenuse of a right-angled triangle is $12.8$ cm. An acute angle is $63^\circ$ . Find the adjacent side, $y$ , to 3 significant figures.

Show worked answer

Adjacent and hypotenuse mean use cosine.

$\cos 63^\circ=\frac{y}{12.8}$

$y=12.8\cos 63^\circ$

$y=5.811078\ldots$

$\boxed{y=5.81\text{ cm}}$

Find an angle with inverse sine

2 marks

In a right-angled triangle, the opposite side is $6.2$ cm and the hypotenuse is $9.5$ cm. Find the angle $\theta$ to 1 decimal place.

Show worked answer

Opposite and hypotenuse mean use sine, then inverse sine.

$\sin\theta=\frac{6.2}{9.5}$

$\theta=\sin^{-1}\left(\frac{6.2}{9.5}\right)$

$\theta=40.740306\ldots^\circ$

$\boxed{\theta=40.7^\circ}$

Find an angle with inverse cosine

2 marks

The adjacent side of a right-angled triangle is $11.5$ mm and the hypotenuse is $14.2$ mm. Find the angle $\alpha$ to 1 decimal place.

Show worked answer

Adjacent and hypotenuse mean use cosine, then inverse cosine.

$\cos\alpha=\frac{11.5}{14.2}$

$\alpha=\cos^{-1}\left(\frac{11.5}{14.2}\right)$

$\alpha=35.917827\ldots^\circ$

$\boxed{\alpha=35.9^\circ}$

Ramp angle

3 marks

A straight ramp rises $0.72$ m over a horizontal distance of $5.8$ m. Calculate the angle the ramp makes with the horizontal. Give your answer to 1 decimal place.

Show worked answer

The rise is opposite and the horizontal distance is adjacent.

$\tan\theta=\frac{0.72}{5.8}$

$\theta=\tan^{-1}\left(\frac{0.72}{5.8}\right)$

$\theta=7.076378\ldots^\circ$

$\boxed{\theta=7.1^\circ}$

Support cable

3 marks

A vertical mast is $9.6$ m high. A straight support cable runs from its top to level ground and makes an angle of $68^\circ$ with the ground. Find the cable length to 3 significant figures.

Show worked answer

The mast is opposite the ground angle and the cable is the hypotenuse.

$\sin 68^\circ=\frac{9.6}{c}$

$c\sin 68^\circ=9.6$

$c=\frac{9.6}{\sin 68^\circ}$

$c=10.353933\ldots$

$\boxed{c=10.4\text{ m}}$

Height from an angle of elevation

4 marks

Priya stands $18$ m from a vertical sculpture on level ground. Her eye level is $1.55$ m above the ground. The angle of elevation from her eyes to the top is $27^\circ$ . Calculate the sculpture's height to 3 significant figures.

Show worked answer

First find the height above Priya's eye level using tangent.

$\tan 27^\circ=\frac{x}{18}$

$x=18\tan 27^\circ$

$x=9.171458\ldots$

Now add the eye height.

$h=9.171458\ldots+1.55$

$h=10.721458\ldots$

$\boxed{h=10.7\text{ m}}$

Angle in a cuboid

Higher only4 marks

A cuboid has a rectangular base $8$ cm by $6$ cm and a height of $7.5$ cm. A line joins one bottom corner to the opposite top corner. Find the angle between this line and the base, to 1 decimal place.

Show worked answer

First find the diagonal across the rectangular base.

$d^2=8^2+6^2$

$d^2=100$

$d=10\text{ cm}$

The height is opposite the required angle and the base diagonal is adjacent.

$\tan\theta=\frac{7.5}{10}$

$\theta=\tan^{-1}(0.75)$

$\theta=36.869897\ldots^\circ$

$\boxed{\theta=36.9^\circ}$

Roof cross-section

4 marks

A symmetrical roof has a horizontal half-width of $4.8$ m. Each sloping side makes an angle of $37^\circ$ with the horizontal. Find (a) the vertical rise and (b) the length of one sloping side. Give both answers to 3 significant figures.

Show worked answer

For the rise $r$ , use opposite and adjacent.

$\tan 37^\circ=\frac{r}{4.8}$

$r=4.8\tan 37^\circ$

$r=3.617059\ldots$

$\boxed{r=3.62\text{ m}}$

For the sloping length $s$ , use adjacent and hypotenuse.

$\cos 37^\circ=\frac{4.8}{s}$

$s=\frac{4.8}{\cos 37^\circ}$

$s=6.010251\ldots$

$\boxed{s=6.01\text{ m}}$

Examiner-style feedback

Common trigonometry mistakes

Labels taken from the wrong angle

Point to the selected angle before writing O, A and H. Opposite and adjacent can swap when the angle changes.

Calculator in radian mode

Check for DEG on the display. An acute angle in a right-angled triangle must be between $0^\circ$ and $90^\circ$ .

Inverse function used for a side

Use an inverse function only when an angle is unknown. For a missing side, rearrange the ordinary sine, cosine or tangent equation.

Rounded too early

Keep unrounded values in multi-step calculations. Early rounding can move the final answer outside the accepted accuracy.

30-second recap

For every right-angled triangle

Let the selected angle determine the side labels, then let the two relevant sides determine the ratio.

Label the hypotenuse, opposite and adjacent sides for the given angle.
Choose sin, cos or tan using the two sides involved.
Substitute the known values and the unknown into the ratio.
Rearrange for a side, or use the inverse function for an angle.
Calculate in degree mode, then check and round only at the end.

Quick answers

GCSE trigonometry FAQ

How do I know which trigonometric ratio to use?

Label opposite, adjacent and hypotenuse relative to the selected angle. Then choose the SOHCAHTOA ratio containing the known side and the side you need.

When do I use inverse sin, cos or tan?

Use an inverse trigonometric function when the unknown is an angle. Form the side ratio first, then apply the matching inverse function.

Why is my calculator giving a different angle?

Check that the calculator is in degree mode. GCSE triangle angles are normally measured in degrees, not radians.

Should I round during a trigonometry calculation?

Keep the full calculator value through intermediate steps and round only the final answer to the accuracy requested.

Is SOHCAHTOA only for right-angled triangles?

Yes. For non-right-angled triangles, Higher-tier questions may require the sine rule, cosine rule or trigonometric area formula instead.

Build the skill

Prerequisites and related Higher topics

Revise first

Angles, algebra and Pythagoras

Be confident with angle facts, rearranging equations, square roots and finding a missing side with Pythagoras.

Higher topics

Extend beyond right triangles

Exact trigonometric values, the sine rule, cosine rule and trigonometric area formula are related Higher topics and need their own methods.

Content standards

Curriculum and rights review

Curriculum references checked 1 September 2026. All questions, values, contexts, solution wording and diagrams on this page are original Pass an Exam content; no past-paper question text has been reproduced.

Official specification references

Trigonometry GCSE Questions, Worked Examples and Answers

What you need to know

Right angle and hypotenuse

Opposite and adjacent

Labels can change

Calculator setup

The five-step SOHCAHTOA method

GCSE trigonometry worked examples

Trigonometry GCSE exam questions

Common trigonometry mistakes

For every right-angled triangle

GCSE trigonometry FAQ

Prerequisites and related Higher topics

Angles, algebra and Pythagoras

Extend beyond right triangles

Turn this method into exam practice

Curriculum and rights review