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.

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Trigonometry GCSE basics video

Start here for H, O and A, the meaning of SOH, CAH and TOA, typical GCSE question types and worked examples for a missing side and angle.

Trigonometry GCSE Basics: SOHCAHTOA, Sides and Angles2 minutes 47 seconds · British English · Right-angled trigonometry

Core idea, first example and exam skills

What you need to know about GCSE trigonometry

Imagine a ramp: its sloping length, horizontal run and vertical rise form a triangle with a 90° corner — a right-angled triangle. Trigonometry connects its side lengths to its angles. The longest side, opposite the 90° corner, is the hypotenuse. First choose one of the other angles: “opposite” means the side across from it, and “adjacent” means the shorter side beside it. In the diagrams, α (alpha) and β (beta) simply name the two angles; θ (theta) later names whichever angle we use.

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.

Sine, cosine and tangent compare pairs of sides. Right-angled triangles with the same acute angle have the same shape, so their matching side lengths stay in the same proportion even when the triangle is larger or smaller. SOHCAHTOA records the three ratios.

SOH

S means sine, O means opposite and H means hypotenuse.

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

CAH

C means cosine, A means adjacent and H means hypotenuse.

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

TOA

T means tangent, O means opposite and A means adjacent.

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

Task: find a missing side

For a $35^\circ$ angle, hypotenuse $12$ cm and opposite side $x$ , sine links the two relevant sides: $\sin35^\circ=x/12$ . Multiplying both sides by 12 gives $x=12\sin35^\circ$ .

Task: find a missing angle

If the opposite side is 4 and the hypotenuse is 6, then $\sin\theta=4/6$ . Use inverse sine to undo sine: $\theta=\sin^{-1}(4/6)$ .

Exam skill checklist

What you need to be able to do in GCSE trigonometry

First confirm that the triangle is right-angled. Then identify what the question gives you and what it asks you to find.

Find a missing side

What the problem asks: Calculate the length of another side.
Information given: A right-angled triangle gives one acute angle and one side length.
How to solve it: Label H, O and A relative to the given angle. Use the ratio containing the known and unknown sides, substitute the values, then rearrange the equation.

Find a missing angle

What the problem asks: Calculate one of the acute angles, usually to a stated accuracy.
Information given: A right-angled triangle gives two side lengths.
How to solve it: Label the two known sides, form the matching trig ratio, then use inverse sine, inverse cosine or inverse tangent.

Solve a contextual problem

What the problem asks: Find the requested real-world length or angle, sometimes with an extra step after the trig calculation.
Information given: A height, ramp, cable, angle of elevation or angle of depression creates a right-angled triangle.
How to solve it: Sketch and label the triangle, solve the missing side or angle with SOHCAHTOA, then return to the context and check what the question finally asks for.

Solve a 3D problem

Higher only

What the problem asks: Find a length or angle that may not lie on one visible face.
Information given: A cuboid, prism or other solid contains the relevant sides in a right-angled cross-section.
How to solve it: Draw the correct 2D cross-section, use Pythagoras first if it is missing a supporting length, then solve the triangle with SOHCAHTOA.

Applicability check: if the triangle is not right-angled, SOHCAHTOA does not apply directly. A Higher-tier question may instead need the sine rule, cosine rule or trigonometric area formula.

For right-angled triangle problems

The five-step SOHCAHTOA method

Use this sequence when a right-angled triangle question gives enough side and angle information to use sine, cosine or tangent. It makes the ratio choice and the algebra visible.

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.

Non-calculator exact values: after the right-triangle method is secure, learn the exact sine and cosine values for

0^\circ,30^\circ,45^\circ,60^\circ,90^\circ

and the exact tangent values for

0^\circ,30^\circ,45^\circ,60^\circ

. These are assessed at both Foundation and Higher tier.

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.

Recognise: A right-angled triangle gives an acute angle and the adjacent side, while the opposite side is unknown.
Find: A missing side, so use an ordinary trig ratio rather than an inverse function.
Method: Opposite and adjacent appear in TOA, so use tangent and rearrange for x.

\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 · Non-calculator

Use an exact trigonometric value

2 marks

Question

A right-angled triangle has hypotenuse $12$ cm and an angle of $30^\circ$ . The side opposite the $30^\circ$ angle is $x$ . Find $x$ exactly.

Recognise: A non-calculator question uses a standard angle and asks for the side opposite it.
Find: An exact side length, so keep the exact trig value rather than using a decimal.
Method: Opposite and hypotenuse appear in SOH. Use sin 30° = 1/2, then multiply by 12.

\sin30^\circ=\frac{x}{12}

\frac12=\frac{x}{12}

x=12\times\frac12

\boxed{x=6\text{ cm}}

Tier check: exact trigonometric values are Foundation and Higher content, not Higher-only.

Example 3

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.

Recognise: A right-angled triangle gives an acute angle and the adjacent side, while the hypotenuse is unknown.
Find: A missing side: the hypotenuse, which must be longer than the given adjacent side.
Method: Adjacent and hypotenuse appear in CAH, so use cosine. The unknown is in the denominator, so 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 4

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.

Recognise: A right-angled triangle gives the opposite and adjacent side lengths and asks for an acute angle.
Find: An angle, so the final calculator operation must be an inverse trigonometric function.
Method: Opposite and adjacent appear in TOA. Form their ratio, then use inverse tangent.

\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 5 · 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.

Recognise: The level ground, vertical tree and line of sight form a right-angled triangle, but the given horizontal line starts at eye level.
Find: The total tree height, not only the vertical side above the learner's eyes.
Method: Use tangent to find the height above eye level, then add the learner's 1.6 m 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 6 · 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.

Recognise: The required angle is between a space diagonal and its projection onto the rectangular base, so it lies in a vertical right-angled cross-section.
Find: An angle in 3D, with the base diagonal needed as an intermediate adjacent side.
Method: Use Pythagoras to find the base diagonal, then use inverse tangent in the exposed cross-section.

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.

13 original questions · total 34 marks

Trigonometry GCSE exam questions

The set moves from exact values and side labels to multi-step applications.

Before you startAllow about 45 minutes · show each exact value, ratio and rearrangement · answers start collapsed

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}

Use exact trigonometric values

2 marks

Work out $4\sin 30^\circ+6\cos 60^\circ$ without a calculator.

Show worked answer

Use the exact values $\sin 30^\circ=\frac12$ and $\cos 60^\circ=\frac12$ .

$4\left(\frac12\right)+6\left(\frac12\right)=2+3$

$\boxed{5}$

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}

Angle of depression

3 marks

A viewing platform is $32$ m above level ground. The angle of depression from the platform to a cyclist is $24^\circ$ . Calculate the horizontal distance, $d$ , from the point directly below the platform to the cyclist. Give your answer to 3 significant figures.

Show worked answer

The horizontal through the platform is parallel to the ground, so the angle of elevation from the cyclist is also $24^\circ$ .

The platform height is opposite this angle and the horizontal distance is adjacent.

\tan 24^\circ=\frac{32}{d}

d\tan 24^\circ=32

d=\frac{32}{\tan 24^\circ}

d=71.873176\ldots

\boxed{d=71.9\text{ m}}

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 $5$ cm by $12$ cm and a height of $9$ 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=5^2+12^2

d^2=169

d=13\text{ cm}

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

\tan\theta=\frac{9}{13}

\theta=\tan^{-1}\left(\frac{9}{13}\right)

\theta=34.695154\ldots^\circ

\boxed{\theta=34.7^\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.

Hypotenuse is not the longest side

The hypotenuse is opposite the right angle and must be the triangle’s longest side. If your labels or answer disagree, check the diagram again.

Inverse input greater than 1

If a sine or cosine calculation gives $\sin^{-1}(k)$ or $\cos^{-1}(k)$ with $k>1$ , the side ratio is impossible. Recheck the numerator and denominator.

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 trigonometry topics

Strengthen powers and roots

Indices

Connect square roots with powers and simplify exact expressions before using trigonometric ratios.

Revise indices

Exact values and Higher extensions

Extend beyond right triangles

Exact trigonometric values are assessed at Foundation and Higher. The sine rule, cosine rule and trigonometric area formula are Higher-only topics and need their own methods.

Revise exact trig values Revise the cosine rule

Content standards

Curriculum and rights review

Curriculum references checked 4 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

Practise GCSE trigonometry for free with an AI tutor

Trigonometry GCSE basics video

What you need to know about GCSE trigonometry

Right angle and hypotenuse

Opposite and adjacent

Labels can change

Calculator setup

What you need to be able to do in GCSE trigonometry

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 trigonometry topics

Indices

Extend beyond right triangles

Continue with Higher exam preparation

Curriculum and rights review