Pick any angle in a right triangle and the three sides lock into fixed proportions — proportions so reliable that knowing one angle and one side lets you recover every length you cannot reach with a ruler. That is the whole engine of trigonometry.
The three sides, named by your angle
A right triangle has one \( 90^\circ \) corner, marked with a small square. The longest side, lying directly across from that right angle, is the hypotenuse. It never moves — it is the hypotenuse no matter which other corner you are looking at.
The other two sides get their names from a choice you make: pick one of the two acute angles to focus on, and call it \( \theta \) (the Greek letter "theta"). Now the remaining sides split into roles relative to \( \theta \):
- The opposite side is the one across the triangle from \( \theta \) — it does not touch the angle at all.
- The adjacent side is the one that, together with the hypotenuse, forms the angle \( \theta \). ("Adjacent" means "next to".)
This is the part beginners most often trip on, so say it slowly: opposite and adjacent are not fixed labels. Switch your attention to the other acute angle and the two sides swap roles. Only the hypotenuse stays put.
In words Find the right angle, and the side facing it is the hypotenuse. Then look at your chosen angle \( \theta \): the side reaching away from it is the opposite, and the side it sits on is the adjacent. Choose the other angle and those two trade places.
Three ratios: SOH-CAH-TOA
A trig ratio is nothing more exotic than a fraction built from two of those side lengths. For a chosen angle \( \theta \), there are three that matter:
\[ \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \]The three names are sine, cosine and tangent, abbreviated \( \sin \), \( \cos \) and \( \tan \). The classic mnemonic packs all three into one chant:
\[ \textbf{SOH} \;\;\textbf{CAH}\;\; \textbf{TOA} \]Read it as Sine = Opposite over Hypotenuse, Cosine = Adjacent over Hypotenuse, Tangent = Opposite over Adjacent. Notice that \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \) too, since dividing opposite-over-hypotenuse by adjacent-over-hypotenuse cancels the hypotenuse.
Here is the quiet miracle that makes all of this work. If you scale a right triangle up or down — same angles, bigger or smaller — every side grows by the same factor, so the ratios do not budge. A \( 37^\circ \) angle gives the very same \( \sin 37^\circ \) whether the triangle fits on your thumbnail or spans a football field. The ratio depends only on the angle, never on the size. (This is the idea of similar triangles doing all the heavy lifting.)
Step the angle below and watch the three ratios update as live fractions of the side lengths. Notice how \( \sin\theta \) and \( \cos\theta \) always land between \( 0 \) and \( 1 \), while \( \tan\theta \) climbs without limit as \( \theta \) approaches \( 90^\circ \).
Two special triangles are worth memorising, because their ratios come out exact. The 45-45-90 triangle has two equal legs; if each leg is \( 1 \), the hypotenuse is \( \sqrt{2} \), so
\[ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \qquad \tan 45^\circ = 1 \]The 30-60-90 triangle has sides in the ratio \( 1 : \sqrt{3} : 2 \) (short leg, long leg, hypotenuse). From it,
\[ \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2} \]Finding a missing side
Once you know one angle and one side, a single trig ratio unlocks any other side. The recipe is always the same: pick the ratio whose two sides are exactly "the one you know" and "the one you want", then solve.
Suppose a right triangle has an angle of \( 37^\circ \) and the hypotenuse measures \( 10 \). To find the side opposite that angle, the pairing "opposite and hypotenuse" points straight at sine:
\[ \sin 37^\circ = \frac{\text{opposite}}{10} \quad\Longrightarrow\quad \text{opposite} = 10 \times \sin 37^\circ \]A calculator gives \( \sin 37^\circ \approx 0.602 \), so the opposite side is about \( 10 \times 0.602 = 6.02 \). The move that makes this painless is choosing the ratio that contains both the side you know and the side you want — then it is one multiplication away.
Tip Before reaching for the calculator, label the sides relative to your angle: opposite, adjacent, hypotenuse. Then ask which two sides the problem involves — the known one and the unknown one — and SOH-CAH-TOA tells you exactly which ratio to use. Make sure the calculator is in degree mode when your angle is in degrees.
- The two sides in play are the adjacent (wanted) and the hypotenuse (known). That pairing is cosine: \( \cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} \).
- Substitute the numbers: \( \cos 40^\circ = \dfrac{\text{adjacent}}{8} \).
- Multiply both sides by \( 8 \): \( \text{adjacent} = 8 \times \cos 40^\circ \).
- Evaluate with \( \cos 40^\circ \approx 0.766 \): \( 8 \times 0.766 \approx 6.13 \).
The adjacent side is \( \mathbf{\approx 6.13} \).
Finding a missing angle
What if you know two sides and want the angle? You run the ratios backwards using the inverse trig functions, written \( \sin^{-1} \), \( \cos^{-1} \), \( \tan^{-1} \) (on a calculator these are usually labelled asin, acos, atan). They answer the question "which angle has this ratio?".
If a right triangle has an opposite side of \( 3 \) and a hypotenuse of \( 5 \), then \( \sin\theta = \dfrac{3}{5} = 0.6 \). To recover \( \theta \), apply the inverse sine:
\[ \theta = \sin^{-1}\!\left(\frac{3}{5}\right) = \sin^{-1}(0.6) \approx 36.87^\circ \]The logic mirrors the side problem exactly. Identify which two sides you have, let SOH-CAH-TOA name the ratio, compute the fraction, then hit the inverse key to turn the ratio back into an angle.
In words A trig function takes an angle and returns a ratio; its inverse takes a ratio and returns the angle. So if you know the sides, build the ratio first, then use \( \sin^{-1} \), \( \cos^{-1} \) or \( \tan^{-1} \) to read off the angle.
- We know the opposite and the adjacent. That pairing is tangent: \( \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} \).
- Substitute: \( \tan\theta = \dfrac{4}{3} \approx 1.333 \).
- Apply the inverse tangent to undo it: \( \theta = \tan^{-1}\!\left(\dfrac{4}{3}\right) \).
- Evaluate: \( \theta \approx 53.13^\circ \).
The angle is \( \theta \mathbf{\approx 53.13^\circ} \). (This is the famous \( 3 \)–\( 4 \)–\( 5 \) right triangle.)
A real application: the angle of elevation
Trigonometry earns its keep the moment a length is out of reach. Stand \( 30 \) metres from the base of a tall tree and tilt your gaze up to its top. The angle your line of sight makes above the horizontal is the angle of elevation; say a clinometer reads \( 50^\circ \).
Sketch the situation as a right triangle: the horizontal distance to the tree (\( 30 \) m) is adjacent to your \( 50^\circ \) angle, and the tree's height is opposite. Opposite and adjacent together call for tangent:
\[ \tan 50^\circ = \frac{\text{height}}{30} \quad\Longrightarrow\quad \text{height} = 30 \times \tan 50^\circ \]With \( \tan 50^\circ \approx 1.192 \), the height is about \( 30 \times 1.192 \approx 35.8 \) metres — measured without ever leaving the ground. The same trick sizes a ladder against a wall, the slope of a wheelchair ramp, or the altitude of a kite, all from one angle and one length you can reach.
Practice
Try each one yourself, then reveal the full solution. Round angles and lengths to two decimal places.
1. A right triangle has a \( 30^\circ \) angle and a hypotenuse of \( 12 \). Find the side opposite the \( 30^\circ \) angle.
The opposite side and the hypotenuse are involved, so use sine: \( \sin 30^\circ = \dfrac{\text{opposite}}{12} \).
Rearrange: \( \text{opposite} = 12 \times \sin 30^\circ \).
The special value \( \sin 30^\circ = \dfrac{1}{2} \), so \( \text{opposite} = 12 \times \dfrac{1}{2} = \mathbf{6} \).
2. A right triangle has the side adjacent to \( \theta \) equal to \( 5 \) and the hypotenuse equal to \( 10 \). Find \( \theta \).
You know the adjacent and the hypotenuse, so use cosine: \( \cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{5}{10} = \dfrac{1}{2} \).
Undo the cosine with its inverse: \( \theta = \cos^{-1}\!\left(\dfrac{1}{2}\right) \).
Since \( \cos 60^\circ = \dfrac{1}{2} \), we get \( \theta = \mathbf{60^\circ} \).
3. A \( 6 \)-metre ladder leans against a wall, making a \( 65^\circ \) angle with the ground. How high up the wall does it reach?
The ladder is the hypotenuse, and the height up the wall is the side opposite the \( 65^\circ \) angle. Opposite and hypotenuse call for sine: \( \sin 65^\circ = \dfrac{\text{height}}{6} \).
Rearrange: \( \text{height} = 6 \times \sin 65^\circ \).
With \( \sin 65^\circ \approx 0.906 \), the height is \( 6 \times 0.906 \approx \mathbf{5.44} \) metres.