Module 3 of 3Lesson 1 of 12 · Unit 22 · Trigonometry I: sine, cosine and tangent
Mathematics Form 3 · Term 3 · Unit 22

📐 Trigonometry I: sine, cosine and tangent

GeometryIGCSE path ✓JC ✓
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Why this matters

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.

🎯 By the end of this unit

  • 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.
📖 Key words

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).

θoppositeadjacenthypotenuseθoppositeadjacenthypotenuseθ at the bottomθ at the top
Same triangle, two different angles: the opposite and adjacent sides swap, but the hypotenuse stays the same.
⭐ Key ideaAlways find the hypotenuse first (opposite the right angle). Then look at your angle θ: the side it points at is the opposite.

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, HO ÷ HA ÷ HO ÷ 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 decimal0.60.80.75
354610891512
Sides 3, 4, 5 then 6, 8, 10 then 9, 12, 15. The marked angle is 36.9° in all three.
⭐ Key ideaSame angle → same ratios, whatever the size. Your calculator has these ratios stored for every angle.

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”).

  1. \(\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\)
  2. \(\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\)
  3. \(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}\)
OSHSOHACHCAHOTATOA
Formula triangles: cover the side you want. Side by side means multiply; one above the other means divide.
⭐ Key ideaSOH: Sin = Opposite ÷ Hypotenuse. CAH: Cos = Adjacent ÷ Hypotenuse. TOA: Tan = Opposite ÷ 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.

Anglesincostan
30°0.50.866…0.577…
45°0.707…0.707…1
60°0.866…0.51.732…
Dsin(30)0.5sincostanD meansdegreessin 30°is 0.5
Always check for the D before an exam starts.
⭐ Key ideaTest with sin 30 = 0.5 every time you pick up a calculator.

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.

  1. \(\sin 35^\circ = \dfrac{x}{12}\)
  2. \(x = 12 \times \sin 35^\circ\)
  3. \(x = 6.8829\ldots\)
  4. \(x = 6.88\) cm (3 s.f.)
35°x12 cm
We know the hypotenuse and want the opposite: that is SOH, so use sin.
⭐ Key ideax on top → multiply: x = known side × (sin, cos or tan).

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.

  1. \(\tan 34^\circ = \dfrac{7}{x}\)
  2. \(x \times \tan 34^\circ = 7\)
  3. \(x = \dfrac{7}{\tan 34^\circ}\)
  4. \(x = 10.3779\ldots\)
  5. \(x = 10.4\) cm (3 s.f.)
34°7 cmx
We know the opposite and want the adjacent: that is TOA, so use tan.
⭐ Key ideax on the bottom → divide: x = known side ÷ (sin, cos or tan). Sense check: the hypotenuse must be the longest side.

2 Watch it work

EXAMPLE 1

Naming the sides

In triangle PQR, the right angle is at Q and the marked angle is at R. Name the hypotenuse, the opposite side and the adjacent side.
40°RQP
  1. The right angle is at Q, so the hypotenuse is the side facing Q: PR
  2. The angle at R faces side PQ, so PQ is the opposite
  3. The other side touching R is QR, so QR is the adjacent
Answer Hypotenuse PR, opposite PQ, adjacent QR
EXAMPLE 2

How high is the kite?

Lesego flies a kite on a string 40 m long. The string makes an angle of 52° with the ground. How high is the kite above her hand? (Treat the string as straight.)
52°h40 m
  1. We know the hypotenuse (40 m) and want the opposite (h): SOH, use sin
  2. \(\sin 52^\circ = \dfrac{h}{40}\)
  3. \(h = 40 \times \sin 52^\circ\)
  4. \(h = 31.5204…\)
  5. \(h = 31.5\) m (3 s.f.)
Answer 31.5 m
EXAMPLE 3

Height of a tree

From a point 15 m from the foot of a tree, the angle up to the top of the tree is 28°. Find the height of the tree.
28°h15 m
  1. We know the adjacent (15 m) and want the opposite (h): TOA, use tan
  2. \(\tan 28^\circ = \dfrac{h}{15}\)
  3. \(h = 15 \times \tan 28^\circ\)
  4. \(h = 7.9756…\)
  5. \(h = 7.98\) m (3 s.f.)
Answer 7.98 m
EXAMPLE 4

x on the bottom

Find the length of the hypotenuse x.
50°9 cmx
  1. We know the adjacent (9 cm) and want the hypotenuse (x): CAH, use cos
  2. \(\cos 50^\circ = \dfrac{9}{x}\)
  3. \(x \times \cos 50^\circ = 9\)
  4. \(x = \dfrac{9}{\cos 50^\circ}\)
  5. \(x = 14.0015…\)
  6. \(x = 14.0\) cm (3 s.f.)
Answer 14.0 cm
EXAMPLE 5

A ladder against a wall

A ladder 6 m long leans against a vertical wall. It makes an angle of 72° with the level ground. How far up the wall does it reach?
72°h6 m
  1. The ladder is the hypotenuse (6 m). The height is opposite the 72° angle: SOH, use sin
  2. \(\sin 72^\circ = \dfrac{h}{6}\)
  3. \(h = 6 \times \sin 72^\circ\)
  4. \(h = 5.7063…\)
  5. \(h = 5.71\) m (3 s.f.)
Answer 5.71 m

3 Your turn

Try each question in your book first. Then tap the button to check every step.

1🟢 Warm-up
For the angle θ, which side is the hypotenuse, which is opposite and which is adjacent?
θKLM
👀 Show the answer
Step by step
  1. The right angle is at L, so the hypotenuse is KM
  2. θ is at K and faces side LM, so LM is the opposite
  3. The other side touching K is KL, the adjacent
Answer Hypotenuse KM, opposite LM, adjacent KL
2🟢 Warm-up
Write \(\sin \theta\), \(\cos \theta\) and \(\tan \theta\) as fractions.
θ51213
👀 Show the answer
Step by step
  1. Hypotenuse 13, opposite 5, adjacent 12
  2. \(\sin \theta = \dfrac{5}{13}\)
  3. \(\cos \theta = \dfrac{12}{13}\)
  4. \(\tan \theta = \dfrac{5}{12}\)
Answer \(\sin \theta = \dfrac{5}{13}\), \(\cos \theta = \dfrac{12}{13}\), \(\tan \theta = \dfrac{5}{12}\)
3🟢 Warm-up
Use your calculator to find, correct to 3 significant figures: (a) tan 40° (b) cos 72°
👀 Show the answer
Step by step
  1. \(\tan 40^\circ = 0.8391…\)
  2. \(\cos 72^\circ = 0.3090…\)
  3. Round each to 3 significant figures
Answer (a) 0.839 (b) 0.309
4🟢 Warm-up
You know the hypotenuse and you want the side opposite the angle. Which ratio do you use: sin, cos or tan?
👀 Show the answer
Step by step
  1. Opposite and hypotenuse are O and H
  2. SOH contains O and H
Answer sin (SOH)
5🔵 Core
Find the length of the side marked x.
42°x8 cm
👀 Show the answer
Step by step
  1. We know the hypotenuse and want the opposite: SOH, use sin
  2. \(\sin 42^\circ = \dfrac{x}{8}\)
  3. \(x = 8 \times \sin 42^\circ\)
  4. \(x = 5.3530…\)
  5. \(x = 5.35\) cm (3 s.f.)
Answer 5.35 cm
6🔵 Core
Find the length of the side marked x.
25°x14 cm
👀 Show the answer
Step by step
  1. We know the hypotenuse and want the adjacent: CAH, use cos
  2. \(\cos 25^\circ = \dfrac{x}{14}\)
  3. \(x = 14 \times \cos 25^\circ\)
  4. \(x = 12.6883…\)
  5. \(x = 12.7\) cm (3 s.f.)
Answer 12.7 cm
7🔵 Core
Find the length of the side marked x.
58°x6 cm
👀 Show the answer
Step by step
  1. We know the adjacent and want the opposite: TOA, use tan
  2. \(\tan 58^\circ = \dfrac{x}{6}\)
  3. \(x = 6 \times \tan 58^\circ\)
  4. \(x = 9.6020…\)
  5. \(x = 9.60\) cm (3 s.f.)
Answer 9.60 cm
8🔵 Core
Find the length of the side marked x.
40°11 cmx
👀 Show the answer
Step by step
  1. We know the opposite and want the hypotenuse: SOH, use sin
  2. \(\sin 40^\circ = \dfrac{11}{x}\)
  3. \(x = \dfrac{11}{\sin 40^\circ}\)
  4. \(x = 17.1130…\)
  5. \(x = 17.1\) cm (3 s.f.)
Answer 17.1 cm
9🔵 Core
The sloping surface of a wheelchair ramp at the entrance of a shop is 5 m long. It rises at an angle of 6° to the ground. How high is the top of the ramp? Give your answer in centimetres.
6°h5 m
👀 Show the answer
Step by step
  1. The slope is the hypotenuse (5 m). The height is opposite the 6° angle: use sin
  2. \(h = 5 \times \sin 6^\circ\)
  3. \(h = 0.5226…\) m
  4. \(h = 0.5226… \times 100\) cm
  5. \(h = 52.3\) cm (3 s.f.)
Answer 52.3 cm
10🔵 Core
Kabelo leans a ladder against a wall of his family’s house in Serowe. The foot of the ladder is 1.5 m from the wall and the ladder makes an angle of 75° with the ground. How long is the ladder?
75°1.5 mL
👀 Show the answer
Step by step
  1. We know the adjacent (1.5 m) and want the hypotenuse (L): CAH, use cos
  2. \(\cos 75^\circ = \dfrac{1.5}{L}\)
  3. \(L = \dfrac{1.5}{\cos 75^\circ}\)
  4. \(L = 5.7956…\)
  5. \(L = 5.80\) m (3 s.f.)
Answer 5.80 m
11🔵 Core
Neo types sin 30 into his calculator and gets −0.988. What has gone wrong, and what should the answer be?
👀 Show the answer
Step by step
  1. sin 30° should be exactly 0.5
  2. −0.988 is the sine of 30 radians, a different unit for angles
  3. Change the calculator to degrees mode (D or DEG on the screen)
Answer The calculator is in radians mode. In degrees mode, sin 30° = 0.5
12🟠 Stretch
The end of a roof is an isosceles triangle. Each sloping side is 4 m long and makes an angle of 30° with the horizontal. Find (a) the height of the roof (b) the total width of the roof.
30°30°4 m4 mhwidth
👀 Show the answer
Step by step
  1. The height splits the roof into two right-angled triangles with hypotenuse 4 m
  2. \(h = 4 \times \sin 30^\circ\)
  3. \(h = 2\) m
  4. \(\text{half width} = 4 \times \cos 30^\circ\)
  5. \(\text{half width} = 3.4641…\) m
  6. \(\text{width} = 2 \times 3.4641…\)
  7. \(\text{width} = 6.93\) m (3 s.f.)
Answer (a) 2 m (b) 6.93 m
13🟠 Stretch
A mast is 25 m tall. Two straight wires go from the top of the mast to points A and B on level ground, on opposite sides of the mast. One wire makes 60° with the ground and the other makes 45°. How far apart are A and B?
60°45°25 mAB
👀 Show the answer
Step by step
  1. Left triangle: the mast is opposite 60°, the distance is adjacent: use tan
  2. \(d_1 = \dfrac{25}{\tan 60^\circ}\)
  3. \(d_1 = 14.4338…\) m
  4. Right triangle: the mast is opposite 45°
  5. \(d_2 = \dfrac{25}{\tan 45^\circ}\)
  6. \(d_2 = 25\) m
  7. \(AB = 14.4338… + 25\)
  8. \(AB = 39.4\) m (3 s.f.)
Answer 39.4 m
14🟠 Stretch
Mpho says: “sin 30° = 0.5, so sin 60° must be 1, because 60 is double 30.” Is she right? Also explain why the sine of an angle in a right-angled triangle can never be more than 1.
👀 Show the answer
Step by step
  1. \(\sin 60^\circ = 0.8660…\), not 1
  2. Doubling the angle does not double the ratio: the opposite side grows more and more slowly
  3. sin = opposite ÷ hypotenuse, and the hypotenuse is the longest side
  4. So the top is always smaller than the bottom, and the answer is less than 1
Answer No. sin 60° ≈ 0.866. Sine is opposite ÷ hypotenuse, and the hypotenuse is the longest side, so sine is always less than 1.
15🟠 Stretch
A right-angled triangle has a hypotenuse of 10 cm and one angle of 35°. Find its area.
35°10 cm
👀 Show the answer
Step by step
  1. Find the two shorter sides first: they are the base and height
  2. \(\text{opposite} = 10 \times \sin 35^\circ\)
  3. \(\text{opposite} = 5.7358…\) cm
  4. \(\text{adjacent} = 10 \times \cos 35^\circ\)
  5. \(\text{adjacent} = 8.1915…\) cm
  6. Area of a triangle = half of base × height
  7. \(A = 0.5 \times 5.736 \times 8.192\)
  8. \(A = 23.5\) cm² (3 s.f.)
Answer 23.5 cm²

4 Watch out!

✗ WrongIn a right-angled triangle, the angle θ is at A. Kagiso labels the side next to θ (not the hypotenuse) as the opposite.
✓ RightThe side next to θ is the adjacent. The opposite is the side that does not touch θ at all.

Opposite and adjacent depend on which angle you are using. Put your finger on θ: the opposite is the side across from your finger.

✗ Wrong\(\cos 50^\circ = \dfrac{9}{x}\), so \(x = 9 \times \cos 50^\circ = 5.79\) cm
✓ Right\(x = \dfrac{9}{\cos 50^\circ} = 14.0\) cm

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.

✗ Wrong\(\tan 28^\circ \approx 0.53\), so \(h = 15 \times 0.53 = 7.95\) m
✓ Right\(h = 15 \times \tan 28^\circ = 7.98\) m (3 s.f.)

Rounding tan 28° too early loses accuracy. Keep the full calculator value and round only the final answer.

✗ Wrongsin 30 = −0.988
✓ Rightsin 30° = 0.5

The calculator was in radians. Switch it to degrees (D on the screen) and test with sin 30 = 0.5.

5 In your exams

🇧🇼 Junior Certificate (JCE)

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.

🎓 Later in IGCSE

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.

🏆 Challenge
A right-angled triangle has a hypotenuse of length 1. One of its angles is θ. Explain why its other two sides are \(\sin \theta\) and \(\cos \theta\). Then use Pythagoras to check that \((\sin 40^\circ)^2 + (\cos 40^\circ)^2 = 1\).
θsin θcos θ1
👀 Reveal the solution
Step by step
  1. \(\text{opposite} = 1 \times \sin \theta\)
  2. \(\text{adjacent} = 1 \times \cos \theta\)
  3. Pythagoras: opposite² + adjacent² = hypotenuse²
  4. \((\sin \theta)^2 + (\cos \theta)^2 = 1\), because the hypotenuse is 1
  5. \((\sin 40^\circ)^2 = 0.4132…\)
  6. \((\cos 40^\circ)^2 = 0.5868…\)
  7. \(0.4132 + 0.5868 = 1\)
Answer With hypotenuse 1, the opposite is sin θ and the adjacent is cos θ. By Pythagoras, (sin θ)² + (cos θ)² = 1 for every angle, and 0.4132 + 0.5868 = 1 confirms it for 40°.
📝 Remember

  • 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.

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