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Published August 9, 2026

Trigonometry in Right-Angled Triangles (SOH CAH TOA)

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Anne Wood
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Trigonometry: SOH CAH TOA

Trigonometry helps us understand the relationship between angles and side lengths in right-angled triangles. It’s a key skill used in architecture, navigation, and engineering, and this guide will show you how to master the three core trigonometric ratios: Sine, Cosine, and Tangent.

Labelling the Triangle

The first and most important step is to correctly label the sides of a right-angled triangle in relation to a specific angle ($\theta$).

The Three Sides

  • The Hypotenuse is always the longest side, opposite the right angle.
  • The Opposite side is directly across from the angle $\theta$.
  • The Adjacent side is next to the angle $\theta$ (but is not the hypotenuse).
Right-angled triangle with labelled sides A right-angled triangle with the right angle at the bottom left, angle theta at the bottom right, the vertical side labelled Opposite, the horizontal side labelled Adjacent, and the diagonal side labelled Hypotenuse. θ Opposite Adjacent Hypotenuse

The Trigonometric Ratios: SOH CAH TOA

This handy mnemonic helps you remember the three essential trigonometric ratios.

SOH

Sine = Opposite / Hypotenuse
$\sin(\theta) = \frac{\text{Opp}}{\text{Hyp}}$

CAH

Cosine = Adjacent / Hypotenuse
$\cos(\theta) = \frac{\text{Adj}}{\text{Hyp}}$

TOA

Tangent = Opposite / Adjacent
$\tan(\theta) = \frac{\text{Opp}}{\text{Adj}}$

Worked Examples

Example 1: Finding a Missing Side

Find the length of side $x$ to 1 decimal place.
Right-angled triangle with a 30 degree angle A right-angled triangle with the right angle at the bottom left, a 30 degree angle at the bottom right, the hypotenuse labelled 12 centimetres, and the vertical opposite side labelled x. 30° x 12 cm
  1. Label the sides: $x$ is the Opposite, 12cm is the Hypotenuse.
  2. Choose the ratio: We have O and H, so we use SOH (Sine).
  3. Substitute and solve: $\sin(30^\circ) = \frac{x}{12}$ $x = 12 \times \sin(30^\circ) = 6$
Answer: 6.0 cm

Example 2: Finding a Missing Angle

Find the size of angle $\theta$ to 1 decimal place.
Right-angled triangle with an unknown angle theta A right-angled triangle with the right angle at the bottom left, angle theta at the bottom right, the adjacent side labelled 7 centimetres, and the hypotenuse labelled 10 centimetres. θ 7 cm 10 cm
  1. Label the sides: 7cm is the Adjacent, 10cm is the Hypotenuse.
  2. Choose the ratio: We have A and H, so we use CAH (Cosine).
  3. Substitute: $\cos(\theta) = \frac{7}{10} = 0.7$
  4. Use the inverse function to find the angle: $\theta = \cos^{-1}(0.7) \approx 45.6^\circ$
Answer: 45.6°

Tutor Insights

🤔 Common Misunderstandings

  • Mixing up Opposite and Adjacent. These sides depend entirely on which angle you’re looking from. The hypotenuse is the only one that never changes.
  • Using the wrong ratio. Forgetting SOH CAH TOA or picking the wrong one for the given sides.

📝 Common Exam Mistakes

  • Calculator in the wrong mode. It MUST be in Degree (DEG) mode. Radian (RAD) mode will give you the wrong answers.
  • Not using the inverse function ($\sin^{-1}, \cos^{-1}, \tan^{-1}$) when finding an angle.
  • Rounding too early or to the wrong number of decimal places.

Practice Questions

  1. Find the length of side $a$ to 1 decimal place.
    Right-angled triangle with a 35 degree angle A right-angled triangle with the right angle at the bottom left, a 35 degree angle at the bottom right, the hypotenuse labelled 15 centimetres, and the vertical opposite side labelled a. 35° a 15 cm
  2. Find the size of angle $\alpha$ to 1 decimal place.
    Right-angled triangle with an unknown angle alpha A right-angled triangle with the right angle at the bottom left, angle alpha at the bottom right, the adjacent side labelled 12 kilometres, and the vertical opposite side labelled 18 kilometres. α 12 km 18 km
Show Answers
  1. Working: Sides are Opposite (a) and Hypotenuse (15). Use SOH. $\sin(35^\circ) = \frac{a}{15} \implies a = 15 \times \sin(35^\circ) \approx 8.6$ Answer: 8.6 cm.
  2. Working: Sides are Opposite (18) and Adjacent (12). Use TOA. $\tan(\alpha) = \frac{18}{12} = 1.5 \implies \alpha = \tan^{-1}(1.5) \approx 56.3$ Answer: 56.3°.

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