📐 Trigonometry I: sine, cosine and tangent
How tall is a tree, a cell tower or a cliff that you cannot climb? With one angle, one distance and a calculator, you can work it out from the ground. Trigonometry has measured mountains, guided ships and now helps phones, games and drones know exactly where things are.
- I can name the hypotenuse, opposite and adjacent sides for a given angle.
- I can explain why sin, cos and tan are fixed for each angle, using similar triangles.
- I can choose the right ratio using SOH CAH TOA.
- I can find a missing side, whether it is on the top or the bottom of the fraction.
- I can use my calculator in degrees mode and round sensibly.
- hypotenuse
- the longest side of a right-angled triangle, opposite the right angle
- opposite side
- the side facing the angle θ you are using
- adjacent side
- the side next to θ that is not the hypotenuse
- sine, cosine, tangent
- the three fixed ratios of sides for an angle: sin, cos and tan
- similar triangles
- triangles with the same angles; one is an enlargement of the other
- θ (theta)
- a Greek letter often used to name an angle
1 Learn it
Naming the sides of a right-angled triangle
Trigonometry is the maths of right-angled triangles. It links the angles to the lengths, so you can measure things you cannot reach.
The hypotenuse is the longest side. It is always opposite the right angle.
Now pick one of the other angles and call it θ (the Greek letter theta). The opposite side faces θ. The adjacent side is next to θ (and is not the hypotenuse).
Why the ratios never change: similar triangles
O = opposite, A = adjacent, H = hypotenuse. These three triangles have the same angles. Each one is an enlargement of the small one.
Divide the opposite side by the hypotenuse in each triangle. You get the same answer every time, because enlarging multiplies both sides by the same scale factor.
So the ratio depends only on the angle, not on the size of the triangle. That is the big idea of trigonometry.
| Sides O, A, H | O ÷ H | A ÷ H | O ÷ A |
|---|---|---|---|
| 3, 4, 5 | \(\dfrac{3}{5}\) | \(\dfrac{4}{5}\) | \(\dfrac{3}{4}\) |
| 6, 8, 10 | \(\dfrac{6}{10}\) | \(\dfrac{8}{10}\) | \(\dfrac{6}{8}\) |
| 9, 12, 15 | \(\dfrac{9}{15}\) | \(\dfrac{12}{15}\) | \(\dfrac{9}{12}\) |
| as a decimal | 0.6 | 0.8 | 0.75 |
Sine, cosine and tangent: SOH CAH TOA
The three fixed ratios have names. We write them as sin, cos and tan.
Remember them with the word SOH CAH TOA (say “so-ka-toe-a”).
- \(\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\)
- \(\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\)
- \(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}\)
Using your calculator correctly
Your calculator must be in degrees mode. Look for a small D (or DEG) at the top of the screen.
Quick test: type sin 30. If you get 0.5, you are ready. If you get −0.988…, the calculator is in radians (R). If you get 0.454…, it is in gradians (G). Use the SETUP or MODE menu to choose degrees.
Keep the full calculator value while you work. Round only the final answer: lengths to 3 significant figures, angles to 1 decimal place.
| Angle | sin | cos | tan |
|---|---|---|---|
| 30° | 0.5 | 0.866… | 0.577… |
| 45° | 0.707… | 0.707… | 1 |
| 60° | 0.866… | 0.5 | 1.732… |
Finding a missing side: x on the top
Step 1: label the sides from the angle you are given.
Step 2: tick the side you know and the side you want. Choose the ratio that uses exactly those two sides.
Step 3: write the equation and solve it. When x is on the top of the fraction, multiply.
- \(\sin 35^\circ = \dfrac{x}{12}\)
- \(x = 12 \times \sin 35^\circ\)
- \(x = 6.8829\ldots\)
- \(x = 6.88\) cm (3 s.f.)
Finding a missing side: x on the bottom
Sometimes the side you want ends up on the bottom of the fraction.
Multiply both sides by x, then divide by the trig value. Now the answer is the known side divided by sin, cos or tan.
- \(\tan 34^\circ = \dfrac{7}{x}\)
- \(x \times \tan 34^\circ = 7\)
- \(x = \dfrac{7}{\tan 34^\circ}\)
- \(x = 10.3779\ldots\)
- \(x = 10.4\) cm (3 s.f.)
2 Watch it work
Naming the sides
- The right angle is at Q, so the hypotenuse is the side facing Q: PR
- The angle at R faces side PQ, so PQ is the opposite
- The other side touching R is QR, so QR is the adjacent
How high is the kite?
- We know the hypotenuse (40 m) and want the opposite (h): SOH, use sin
- \(\sin 52^\circ = \dfrac{h}{40}\)
- \(h = 40 \times \sin 52^\circ\)
- \(h = 31.5204…\)
- \(h = 31.5\) m (3 s.f.)
Height of a tree
- We know the adjacent (15 m) and want the opposite (h): TOA, use tan
- \(\tan 28^\circ = \dfrac{h}{15}\)
- \(h = 15 \times \tan 28^\circ\)
- \(h = 7.9756…\)
- \(h = 7.98\) m (3 s.f.)
x on the bottom
- We know the adjacent (9 cm) and want the hypotenuse (x): CAH, use cos
- \(\cos 50^\circ = \dfrac{9}{x}\)
- \(x \times \cos 50^\circ = 9\)
- \(x = \dfrac{9}{\cos 50^\circ}\)
- \(x = 14.0015…\)
- \(x = 14.0\) cm (3 s.f.)
A ladder against a wall
- The ladder is the hypotenuse (6 m). The height is opposite the 72° angle: SOH, use sin
- \(\sin 72^\circ = \dfrac{h}{6}\)
- \(h = 6 \times \sin 72^\circ\)
- \(h = 5.7063…\)
- \(h = 5.71\) m (3 s.f.)
3 Your turn
Try each question in your book first. Then tap the button to check every step.
👀 Show the answer
- The right angle is at L, so the hypotenuse is KM
- θ is at K and faces side LM, so LM is the opposite
- The other side touching K is KL, the adjacent
👀 Show the answer
- Hypotenuse 13, opposite 5, adjacent 12
- \(\sin \theta = \dfrac{5}{13}\)
- \(\cos \theta = \dfrac{12}{13}\)
- \(\tan \theta = \dfrac{5}{12}\)
👀 Show the answer
- \(\tan 40^\circ = 0.8391…\)
- \(\cos 72^\circ = 0.3090…\)
- Round each to 3 significant figures
👀 Show the answer
- Opposite and hypotenuse are O and H
- SOH contains O and H
👀 Show the answer
- We know the hypotenuse and want the opposite: SOH, use sin
- \(\sin 42^\circ = \dfrac{x}{8}\)
- \(x = 8 \times \sin 42^\circ\)
- \(x = 5.3530…\)
- \(x = 5.35\) cm (3 s.f.)
👀 Show the answer
- We know the hypotenuse and want the adjacent: CAH, use cos
- \(\cos 25^\circ = \dfrac{x}{14}\)
- \(x = 14 \times \cos 25^\circ\)
- \(x = 12.6883…\)
- \(x = 12.7\) cm (3 s.f.)
👀 Show the answer
- We know the adjacent and want the opposite: TOA, use tan
- \(\tan 58^\circ = \dfrac{x}{6}\)
- \(x = 6 \times \tan 58^\circ\)
- \(x = 9.6020…\)
- \(x = 9.60\) cm (3 s.f.)
👀 Show the answer
- We know the opposite and want the hypotenuse: SOH, use sin
- \(\sin 40^\circ = \dfrac{11}{x}\)
- \(x = \dfrac{11}{\sin 40^\circ}\)
- \(x = 17.1130…\)
- \(x = 17.1\) cm (3 s.f.)
👀 Show the answer
- The slope is the hypotenuse (5 m). The height is opposite the 6° angle: use sin
- \(h = 5 \times \sin 6^\circ\)
- \(h = 0.5226…\) m
- \(h = 0.5226… \times 100\) cm
- \(h = 52.3\) cm (3 s.f.)
👀 Show the answer
- We know the adjacent (1.5 m) and want the hypotenuse (L): CAH, use cos
- \(\cos 75^\circ = \dfrac{1.5}{L}\)
- \(L = \dfrac{1.5}{\cos 75^\circ}\)
- \(L = 5.7956…\)
- \(L = 5.80\) m (3 s.f.)
👀 Show the answer
- sin 30° should be exactly 0.5
- −0.988 is the sine of 30 radians, a different unit for angles
- Change the calculator to degrees mode (D or DEG on the screen)
👀 Show the answer
- The height splits the roof into two right-angled triangles with hypotenuse 4 m
- \(h = 4 \times \sin 30^\circ\)
- \(h = 2\) m
- \(\text{half width} = 4 \times \cos 30^\circ\)
- \(\text{half width} = 3.4641…\) m
- \(\text{width} = 2 \times 3.4641…\)
- \(\text{width} = 6.93\) m (3 s.f.)
👀 Show the answer
- Left triangle: the mast is opposite 60°, the distance is adjacent: use tan
- \(d_1 = \dfrac{25}{\tan 60^\circ}\)
- \(d_1 = 14.4338…\) m
- Right triangle: the mast is opposite 45°
- \(d_2 = \dfrac{25}{\tan 45^\circ}\)
- \(d_2 = 25\) m
- \(AB = 14.4338… + 25\)
- \(AB = 39.4\) m (3 s.f.)
👀 Show the answer
- \(\sin 60^\circ = 0.8660…\), not 1
- Doubling the angle does not double the ratio: the opposite side grows more and more slowly
- sin = opposite ÷ hypotenuse, and the hypotenuse is the longest side
- So the top is always smaller than the bottom, and the answer is less than 1
👀 Show the answer
- Find the two shorter sides first: they are the base and height
- \(\text{opposite} = 10 \times \sin 35^\circ\)
- \(\text{opposite} = 5.7358…\) cm
- \(\text{adjacent} = 10 \times \cos 35^\circ\)
- \(\text{adjacent} = 8.1915…\) cm
- Area of a triangle = half of base × height
- \(A = 0.5 \times 5.736 \times 8.192\)
- \(A = 23.5\) cm² (3 s.f.)
4 Watch out!
Opposite and adjacent depend on which angle you are using. Put your finger on θ: the opposite is the side across from your finger.
When x is on the bottom you must divide. Sense check: x is the hypotenuse, so it must be longer than 9 cm. 5.79 cm cannot be right.
Rounding tan 28° too early loses accuracy. Keep the full calculator value and round only the final answer.
The calculator was in radians. Switch it to degrees (D on the screen) and test with sin 30 = 0.5.
5 In your exams
In JCE Paper 1 (no calculator) you are often asked to name the hypotenuse, opposite or adjacent side, or to write sin, cos or tan of an angle as a fraction from a triangle with whole-number sides. In Paper 2 (calculator) there is usually a structured question where you find a missing side of a right-angled triangle, often in a real-life setting such as a ladder, ramp, roof or kite. Show the ratio you used and the substitution before the answer.
Right-angled triangle trigonometry is in both IGCSE 0580 Core and Extended. Core questions ask for missing sides and angles, often in context. Extended builds on it with the sine and cosine rules, the area formula \(\dfrac{1}{2}ab\sin C\), three-dimensional problems and the graphs of sin, cos and tan. Lengths are normally given to 3 significant figures.
👀 Reveal the solution
- \(\text{opposite} = 1 \times \sin \theta\)
- \(\text{adjacent} = 1 \times \cos \theta\)
- Pythagoras: opposite² + adjacent² = hypotenuse²
- \((\sin \theta)^2 + (\cos \theta)^2 = 1\), because the hypotenuse is 1
- \((\sin 40^\circ)^2 = 0.4132…\)
- \((\cos 40^\circ)^2 = 0.5868…\)
- \(0.4132 + 0.5868 = 1\)
- The hypotenuse is opposite the right angle. The opposite faces θ. The adjacent is next to θ.
- Similar triangles have the same ratios, so sin, cos and tan depend only on the angle.
- SOH CAH TOA: sin = O ÷ H, cos = A ÷ H, tan = O ÷ A.
- x on top → multiply. x on the bottom → divide.
- Calculator in degrees (check sin 30 = 0.5). Round only at the end: 3 s.f.