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GCSE Trigonometry: SOHCAHTOA Explained with Examples

Complete guide to GCSE trigonometry. Learn SOHCAHTOA, how to find missing sides and angles, and the exact values you need to memorise for every exam board.

What Is Trigonometry and Where It Appears on the Paper

Trigonometry is the study of the relationship between the angles and the side lengths of a triangle. At GCSE you work almost entirely with right-angled triangles, using three ratios: sine (sin), cosine (cos) and tangent (tan). Given one angle and one side, these ratios let you calculate any other side. Given two sides, they let you calculate any angle.

Every board tests it, on both tiers. On Foundation papers a SOHCAHTOA question is typically worth 2 to 4 marks and usually asks for a single missing side or angle. On Higher papers trigonometry gets combined with Pythagoras, bearings, area and 3D shapes, and the exact-value questions appear on the non-calculator paper. It sits in the same strand as Pythagoras' theorem, and the two are revised together on our Pythagoras and trigonometry topic page.

The good news is that every SOHCAHTOA question, however it is dressed up, comes down to the same three steps: label the sides, choose the ratio, solve the equation. This guide takes each step in turn.

Key points
  • SOHCAHTOA applies to right-angled triangles only
  • Foundation: find a side or an angle; Higher adds exact values, 3D problems and the sine and cosine rules
  • Every question reduces to: label the sides, choose the ratio, solve

Step One: Label the Sides Correctly

Most trigonometry mistakes happen before any calculation starts, at the labelling stage. The three sides of a right-angled triangle are named relative to the angle you are working with, so the same physical side can be "opposite" in one question and "adjacent" in another.

The hypotenuse (H) is the easiest: it is always the longest side, and it is always opposite the right angle. It never changes, whichever angle you use. The opposite (O) is the side directly across from the angle you are using, the one the angle "points at". The adjacent (A) is the side next to your angle that is not the hypotenuse.

Get into the habit of writing H, O and A on the diagram before doing anything else. Examiners cannot give you method marks for a correct ratio applied to the wrong sides, so ten seconds of labelling protects the whole question.

Key points
  • Hypotenuse: opposite the right angle, always the longest side
  • Opposite: across from the angle you are using
  • Adjacent: next to your angle, but never the hypotenuse
  • Opposite and adjacent swap if you switch to the other angle, so always label relative to the angle in the question

Step Two: Choose the Ratio with SOHCAHTOA

SOHCAHTOA is the mnemonic that stores all three ratios: SOH means sin⁡θ=OH\sin\theta = \frac{O}{H}, CAH means cos⁡θ=AH\cos\theta = \frac{A}{H}, and TOA means tan⁡θ=OA\tan\theta = \frac{O}{A}.

To choose the right one, look at which two sides the question involves: the side you know and the side you want (or, for an angle question, the two sides you know). Tick them off against the letters. Know the hypotenuse and want the opposite? That is O and H, so sin. Know the opposite and the adjacent? That is O and A, so tan. Exactly one ratio fits any pair of sides, so this step is mechanical once the labelling is right.

Many students like the formula-triangle version: write the three letters of the chosen ratio in a triangle with the top letter over the bottom two, cover the quantity you want, and what remains tells you whether to multiply or divide. Use whichever version you can do quickly under pressure; they are the same maths.

Key points
  • SOH: $\sin\theta = O \div H$
  • CAH: $\cos\theta = A \div H$
  • TOA: $\tan\theta = O \div A$
  • Identify the two sides involved in the question; only one ratio uses both

Finding a Missing Side (Unknown on Top)

When the side you want sits on top of the ratio (the O in sin or tan, or the A in cos), the rearrangement is a single multiplication. This is the friendlier of the two side-finding cases: write the ratio, substitute what you know, multiply.

Always write the ratio equation down before reaching for the calculator. On a 3-mark question, boards typically award a method mark for the correct ratio and substitution, so a slip on the calculator still leaves you with most of the marks if the working is on the page.

In a right-angled triangle, one angle is 35∘35^\circ and the hypotenuse is 10 cm. Find the side opposite the 35∘35^\circ angle, to 2 decimal places.

  1. 1

    Label: we know H = 10 and want O, so use SOH

  2. 2

    sin⁡35∘=O10\sin 35^\circ = \frac{O}{10}

  3. 3

    O=10×sin⁡35∘O = 10 \times \sin 35^\circ

  4. 4

    O=10×0.5736...=5.7357...O = 10 \times 0.5736... = 5.7357...

O ≈ 5.74 cm

Finding a Missing Side (Unknown on the Bottom)

When the unknown side is on the bottom of the ratio, you must divide instead of multiply, and this is where a large share of exam errors happen. If cos⁡40∘=12h\cos 40^\circ = \frac{12}{h}, then h=12cos⁡40∘h = \frac{12}{\cos 40^\circ}, not 12×cos⁡40∘12 \times \cos 40^\circ.

A quick sanity check catches the error every time: cos⁡40∘≈0.766\cos 40^\circ \approx 0.766, which is less than 1, so dividing by it must make the answer bigger than 12. If you are finding a hypotenuse, your answer must be the longest side; if it comes out shorter than a given side, you multiplied when you should have divided.

A right-angled triangle has an angle of 40∘40^\circ. The side adjacent to it is 12 cm. Find the hypotenuse, to 1 decimal place.

  1. 1

    Label: we know A = 12 and want H, so use CAH

  2. 2

    cos⁡40∘=12h\cos 40^\circ = \frac{12}{h}

  3. 3

    h=12cos⁡40∘=120.7660...h = \frac{12}{\cos 40^\circ} = \frac{12}{0.7660...}

  4. 4

    h=15.664...h = 15.664...

  5. 5

    Check: 15.7 > 12, correct for a hypotenuse

h ≈ 15.7 cm

Finding a Missing Angle with Inverse Functions

To find an angle from two known sides, you need the inverse trigonometric functions: sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1} and tan⁡−1\tan^{-1} (also written arcsin, arccos, arctan). These undo the ratio: if tan⁡θ=0.75\tan\theta = 0.75, then θ=tan⁡−1(0.75)\theta = \tan^{-1}(0.75). On most calculators they sit above the sin, cos and tan keys, accessed with SHIFT or 2nd.

Two practical warnings. First, check your calculator is in degrees mode (look for a small D on screen). In radians mode tan⁡−1(0.75)\tan^{-1}(0.75) gives 0.6435, which is not a GCSE answer, and the working looks identical, so the error is easy to miss. Second, keep the fraction exact inside the inverse function rather than rounding early: tan⁡−1(6÷8)\tan^{-1}(6 \div 8) is safer than rounding 0.75 first (harmless here, but with a value like 5÷75 \div 7 early rounding shifts the final angle).

A right-angled triangle has opposite side 6 cm and adjacent side 8 cm. Find the angle θ\theta between the adjacent side and the hypotenuse, to 1 decimal place.

  1. 1

    We know O and A, so use TOA

  2. 2

    tan⁡θ=68=0.75\tan\theta = \frac{6}{8} = 0.75

  3. 3

    θ=tan⁡−1(0.75)\theta = \tan^{-1}(0.75)

  4. 4

    θ=36.869...∘\theta = 36.869...^\circ

θ ≈ 36.9°

Exact Values for 30°, 45° and 60°

Both tiers now require certain trig values without a calculator: sin, cos and tan of 0∘0^\circ, 30∘30^\circ, 45∘45^\circ, 60∘60^\circ and 90∘90^\circ (tan 90∘90^\circ is undefined). These turn up on the non-calculator paper, usually as a short question in the form "write down the exact value" or embedded in a triangle problem.

If memorising a table feels brittle, learn the two triangles they come from instead. Cutting an equilateral triangle of side 2 in half gives a right-angled triangle with sides 1, 3\sqrt{3} and 2, which produces all the 30∘30^\circ and 60∘60^\circ values. A right-angled isosceles triangle with legs of 1 has hypotenuse 2\sqrt{2}, which produces the 45∘45^\circ values. Sketching one of these on the exam takes twenty seconds and is more reliable than a half-remembered list.

For example: a right-angled triangle has an angle of 60∘60^\circ and the adjacent side is 5 cm. The opposite side is 5×tan⁡60∘=535 \times \tan 60^\circ = 5\sqrt{3} cm, and on a non-calculator paper you leave the answer exactly in that surd form.

Key points
  • $\sin 30^\circ = \frac{1}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\tan 30^\circ = \frac{1}{\sqrt{3}}$
  • $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$, $\tan 45^\circ = 1$
  • $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 60^\circ = \frac{1}{2}$, $\tan 60^\circ = \sqrt{3}$
  • $\sin 0^\circ = 0$, $\cos 0^\circ = 1$, $\sin 90^\circ = 1$, $\cos 90^\circ = 0$
  • Derive them from the half-equilateral and isosceles right-angled triangles if you forget

When SOHCAHTOA Fails: Triangles Without a Right Angle

SOHCAHTOA only works when the triangle has a right angle, because the three ratios are defined using the hypotenuse, and only right-angled triangles have one. If a question gives you a triangle with angles like 50∘50^\circ, 60∘60^\circ and 70∘70^\circ and no right angle marked, SOHCAHTOA is the wrong tool, and forcing it will produce a confident-looking wrong answer.

Higher tier has the tools for this situation: the sine rule (asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B}) links each side to the angle opposite it, and the cosine rule (a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A) handles the cases the sine rule cannot, such as two sides with the angle between them. There is also the area formula 12absin⁡C\frac{1}{2}ab\sin C. These are separate techniques with their own practice questions, but the habit that matters here is the check itself: before writing SOH, CAH or TOA, confirm the right angle is actually marked on the diagram.

On Foundation you will not be asked about non-right-angled triangles in this way, so if you are on Foundation and a trig question appears, it is safe to assume SOHCAHTOA (or Pythagoras) is the intended method.

Key points
  • No right angle means no hypotenuse, so the three GCSE ratios do not apply
  • Higher tier uses the sine rule and cosine rule for general triangles
  • Check the diagram for the right-angle mark before starting
  • Foundation trigonometry questions always involve a right angle

Exam Technique and How to Practise

Trigonometry rewards routine. The questions vary their context (ladders, ramps, bearings, shadows) but not their structure, so the fastest way to improve is to drill the three-step routine until choosing the ratio is automatic. Pair it with Pythagoras practice, because papers love combining them in one diagram: use Pythagoras when you know two sides and want the third with no angle involved, and trigonometry the moment an angle other than the right angle enters the question. Our guide to Pythagoras' theorem covers that side of the strand.

On marks: show the ratio, the substitution and the rearrangement as separate lines. Round only at the final step, and give at least 3 significant figures unless the question specifies. If the question says "exact value", a rounded decimal scores nothing; leave the surd in.

Exam Ladder generates trigonometry questions at every difficulty with fully worked solutions, so you can practise the routine on fresh numbers each time, and mock exams are graded on real exam-board grade boundaries so you can see what your trigonometry marks are worth. Check how long you have left with the exam calendar, and start with a 7-day free trial.

Key points
  • Write ratio, substitution and rearrangement as separate lines for method marks
  • Round once, at the end, to at least 3 significant figures
  • Angle plus side: trigonometry. Two sides, no angle needed: Pythagoras
  • "Exact value" means leave the surd; a decimal scores nothing

Frequently asked questions

Do I need a calculator for GCSE trigonometry?

For most questions, yes — use the sin, cos and tan buttons, and make sure the calculator is in degrees mode. On non-calculator papers you will only be asked about the exact values (0°, 30°, 45°, 60°, 90°), which you must memorise or derive.

What's the difference between the sine rule and SOHCAHTOA?

SOHCAHTOA only works for right-angled triangles. The sine rule (a/sin A = b/sin B) and the cosine rule work for any triangle and are Higher tier techniques. If the triangle has no right angle, SOHCAHTOA cannot be used.

How do I know whether to multiply or divide when finding a side?

Look at where the unknown sits in the ratio. If it is on top (the opposite in sin, for example), multiply the known side by the trig value. If it is on the bottom (the hypotenuse in sin or cos), divide the known side by the trig value. Sanity check: a hypotenuse must come out as the longest side.

How do I remember SOHCAHTOA?

A popular sentence is "Some Old Horses Can Always Hear Their Owners Approaching". But the more reliable route is repetition: after enough practice questions, writing SOH CAH TOA at the top of your working becomes automatic.

Why is my calculator giving a weird answer for an angle?

Almost always because it is in radians mode. Look for a small D (degrees) on the display; if it shows R or RAD, switch modes and repeat the calculation. An answer like 0.64 for an angle that should be about 37° is the classic symptom.