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How do sine, cosine and tangent connect the angles and sides of a right-angled triangle?

Use the sine, cosine and tangent ratios to find unknown sides and angles in right-angled triangles

A focused answer to the N(A)-Level Mathematics outcome on trigonometry. Labelling opposite, adjacent and hypotenuse, the SOH-CAH-TOA ratios, finding a side, and finding an angle.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to use the three trigonometric ratios - sine, cosine and tangent - to find unknown sides and angles in right-angled triangles. The first skill is labelling the sides correctly relative to the angle; after that, SOH-CAH-TOA tells you which ratio to use.

The answer

Labelling the sides

For a chosen angle (not the right angle), the three sides are named relative to it:

The hypotenuse is the longest side, opposite the right angle.
The opposite is the side directly across from the chosen angle.
The adjacent is the side next to the chosen angle (between it and the right angle).

The opposite and adjacent swap if you choose the other non-right angle, so always label with respect to the angle in the question.

The three ratios: SOH-CAH-TOA

The ratios connect the angle to two of the sides:

\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan \theta = \frac{\text{opposite}}{\text{adjacent}}

The memory aid SOH-CAH-TOA records these: Sine = Opposite over Hypotenuse, Cosine = Adjacent over Hypotenuse, Tangent = Opposite over Adjacent.

Finding a side

To find a missing side: label the sides, pick the ratio that uses the two sides you care about, substitute, and rearrange. For a $40^\circ$ angle with hypotenuse $12$ , finding the opposite uses sine: $\text{opposite} = 12 \sin 40^\circ$ .

Finding an angle

To find a missing angle: form the ratio from the two known sides, then use the inverse function ( $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ ) on the calculator. If $\tan \theta = \dfrac{3}{4}$ , then $\theta = \tan^{-1}\!\left(\dfrac{3}{4}\right) = 36.9^\circ$ .

Calculator and rounding

Make sure the calculator is set to degrees. Round angles and lengths to the accuracy the question asks for, usually $1$ decimal place for angles or $3$ significant figures for lengths.

Worked example

A ramp rises at an angle of $15^\circ$ to the horizontal. The sloping length of the ramp (the hypotenuse) is $8\ \text{m}$ . Find the height the ramp rises, correct to $3$ significant figures.

Step 1: Label the sides

The height is opposite the $15^\circ$ angle, and the sloping ramp is the hypotenuse.

Step 2: Choose the ratio

Opposite and hypotenuse means sine (SOH):

\sin 15^\circ = \frac{\text{height}}{8}.

Step 3: Rearrange to make height the subject

\text{height} = 8 \sin 15^\circ.

Step 4: Calculate

8 \times 0.2588 = 2.07\ \text{m (to 3 significant figures)}.

So the ramp rises about $2.07\ \text{m}$ .

Examples in context

Example 1. Height of a tree. Standing $20\ \text{m}$ from a tree, the angle up to the top is $35^\circ$ . The height is opposite and the $20\ \text{m}$ is adjacent, so tangent applies: height $= 20 \tan 35^\circ \approx 14.0\ \text{m}$ . Trigonometry lets you find a height you cannot measure directly.

Example 2. Angle of a slope. A road climbs $50\ \text{m}$ vertically over a horizontal distance of $400\ \text{m}$ . The angle of the slope uses tangent: $\theta = \tan^{-1}\!\left(\dfrac{50}{400}\right) \approx 7.1^\circ$ . Real gradients and ramps are described by exactly this kind of angle calculation.

Try this

Cue. A right-angled triangle has a $50^\circ$ angle and adjacent side $6\ \text{cm}$ . The opposite is $6 \tan 50^\circ \approx 7.15\ \text{cm}$ .
Cue. Find $\theta$ if $\sin \theta = 0.5$ . Use the inverse: $\theta = \sin^{-1}(0.5) = 30^\circ$ .
Cue. With hypotenuse $10\ \text{cm}$ and a $60^\circ$ angle, the adjacent side is $10 \cos 60^\circ = 10 \times 0.5 = 5\ \text{cm}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksIn a right-angled triangle, the angle is

30^\circ

and the hypotenuse is

10\ \text{cm}

. Find the length of the side opposite the

30^\circ

angle.

Show worked answer →

The opposite and hypotenuse are involved, so use sine (SOH): $\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ .

$\sin 30^\circ = \dfrac{\text{opposite}}{10}$ .

Since $\sin 30^\circ = 0.5$ : opposite $= 10 \times 0.5 = 5\ \text{cm}$ .

What markers reward: choosing sine because the opposite and hypotenuse are the sides involved, rearranging to make the opposite the subject, and the correct length. Picking the wrong ratio is the main error.

Original4 marksA right-angled triangle has the side opposite an angle equal to

7\ \text{cm}

and the adjacent side equal to

24\ \text{cm}

. Find the size of the angle, correct to

1

decimal place.

Show worked answer →

Opposite and adjacent are involved, so use tangent (TOA): $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$ .

$\tan \theta = \dfrac{7}{24}$ .

$\theta = \tan^{-1}\!\left(\dfrac{7}{24}\right) = 16.3^\circ$ (to $1$ decimal place).

What markers reward: choosing tangent for opposite over adjacent, using the inverse tangent to find the angle, and rounding as asked. Forgetting to use the inverse function (and instead reading $\tan$ of the fraction) is the usual slip.