G20, G21Right-angled trigonometry

Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Geometry and measures

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Revision notes

Sine, cosine and tangent link the angles and sides of a right-angled triangle. You use SOH CAH TOA to find a missing side or angle, learn exact values for 0°, 30°, 45°, 60° and 90° for the non-calculator paper and, on Higher, solve problems in 3D.

Grade by grade

What you need to be able to do, from the first marks up to the top grade.

4
Label the hypotenuse, opposite and adjacent sidesThe opposite is across from the angle, the hypotenuse is opposite the right angle, and the adjacent is the other side next to the angle.
5
Find a missing side using SOH CAH TOAChoose the ratio from the side you know and the side you want, then rearrange, e.g. x = 12 × sin 40°.
5
Find a missing angle using inverse trigUse sin⁻¹, cos⁻¹ or tan⁻¹ on your calculator, e.g. θ = tan⁻¹(5 ÷ 8).
5
Recall exact trig valuesKnow sin and cos of 0°, 30°, 45°, 60° and 90°, and tan of 0°, 30°, 45° and 60°, without a calculator.
5
Solve angle of elevation and depression problemsDraw the right-angled triangle with the angle measured from the horizontal.

Notes

Labelling the sides

The hypotenuse (H) is the longest side, opposite the right angle.
The opposite (O) is the side opposite the angle you are using.
The adjacent (A) is the side next to that angle that is not the hypotenuse.
The opposite and adjacent swap if you use the other angle, so always label from the angle in the question.

SOH CAH TOA

\(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\), \(\tan\theta = \dfrac{O}{A}\).
Pick the ratio that uses the side you know and the side you want.
Unknown on the top: multiply. \(\sin 40^\circ = \dfrac{x}{12}\) gives \(x = 12\sin 40^\circ = 7.71\).
Unknown on the bottom: divide. \(\cos 25^\circ = \dfrac{9}{x}\) gives \(x = \dfrac{9}{\cos 25^\circ} = 9.93\).
Missing angle: use the inverse. \(\tan\theta = \dfrac{5}{8}\) gives \(\theta = \tan^{-1}(0.625) = 32.0^\circ\).
Check your calculator is in degrees: sin 30 should give 0.5.

Exact values (non-calculator)

sin 30° = cos 60° = \(\frac{1}{2}\); sin 60° = cos 30° = \(\frac{\sqrt{3}}{2}\); sin 45° = cos 45° = \(\frac{\sqrt{2}}{2}\).
tan 30° = \(\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\); tan 45° = 1; tan 60° = \(\sqrt{3}\).
sin 0° = 0, sin 90° = 1, cos 0° = 1, cos 90° = 0, tan 0° = 0.

Elevation, depression and 3D

An angle of elevation is measured up from the horizontal. An angle of depression is measured down from the horizontal.

Cheatsheet

SOH CAH TOA: sin θ = O ÷ H, cos θ = A ÷ H, tan θ = O ÷ A
Unknown side on the top of the fraction: multiply; on the bottom: divide
Missing angle: θ = sin⁻¹(O ÷ H), cos⁻¹(A ÷ H) or tan⁻¹(O ÷ A)
sin 0°, 30°, 45°, 60°, 90° = 0, \(\frac{1}{2}\), \(\frac{\sqrt{2}}{2}\), \(\frac{\sqrt{3}}{2}\), 1
cos 0°, 30°, 45°, 60°, 90° = 1, \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{2}}{2}\), \(\frac{1}{2}\), 0
tan 0°, 30°, 45°, 60° = 0, \(\frac{\sqrt{3}}{3}\), 1, \(\sqrt{3}\)
Elevation: up from the horizontal; depression: down from the horizontal

How to answer each type of question

Calculate the length of a side

3 marks5

Label O, A and H from the given angle.
Choose the ratio that uses the known side and the unknown side, and write the equation.
Rearrange (multiply or divide) and calculate, then round.

Example. Triangle PQR has a right angle at Q. PR = 14 cm and angle QPR = 38°.
Work out the length of QR. Give your answer correct to 3 significant figures.

Show the model answer

QR is opposite the 38° angle and PR is the hypotenuse, so sin 38° = QR ÷ 14 (M1)
QR = 14 × sin 38° (M1)
= 8.62 cm (A1)

Calculate the size of an angle

3 marks5

Label the two sides you know as O, A or H from the angle you want.
Write the ratio as a fraction, e.g. tan θ = O ÷ A.
Use the inverse function and round, usually to 1 decimal place.

Example. A ramp rises 0.8 m vertically over a horizontal distance of 5.5 m.
Work out the angle the ramp makes with the horizontal. Give your answer correct to 1 decimal place.

Show the model answer

tan θ = 0.8 ÷ 5.5 (M1)
θ = tan⁻¹(0.8 ÷ 5.5) (M1)
= 8.3° (A1)

Non-calculator: use exact values

2 to 4 marks5

Write the trig equation as normal.
Replace the trig value with its exact value (e.g. sin 30° = \(\frac{1}{2}\)).
Simplify, or match a ratio to an exact value to find the angle.

Example. Do not use a calculator.
(a) ABC is a triangle with a right angle at B. Angle BAC = 30° and AC = 18 cm. Work out the length of BC. (2 marks)
(b) DEF is a triangle with a right angle at E. DE = 7 cm and EF = 7 cm. Show that angle EDF = 45°. (2 marks)

Show the model answer

(a) sin 30° = BC ÷ 18, so BC = 18 × \(\frac{1}{2}\) (M1) = 9 cm (A1)
(b) EF is opposite angle D and DE is adjacent to it, so tan D = 7 ÷ 7 = 1 (M1)
tan 45° = 1, so angle EDF = 45° (A1)

Shortcuts and memory tricks

Formula triangles: O on top with S and H below (SOH), A on top with C and H below (CAH), O on top with T and A below (TOA). Cover the one you want to find.
Exact sin values follow a pattern: \(\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}\) for 0°, 30°, 45°, 60°, 90°. Cos is the same list backwards.
Sin and cos are never bigger than 1. If sin⁻¹ or cos⁻¹ gives a maths error, you have divided the wrong way round.
Test your calculator mode at the start: sin 30 = 0.5 means degrees.

Where marks are lost

Calculator in radians or gradians mode.
Labelling the opposite and adjacent from the wrong angle.
Writing θ = tan(0.625) instead of θ = tan⁻¹(0.625).
Multiplying when the unknown is on the bottom: if cos 25° = 9 ÷ x, then x = 9 ÷ cos 25°.
Rounding the ratio before using the inverse: tan⁻¹(0.15) = 8.5°, but tan⁻¹(0.8 ÷ 5.5) = 8.3°.

Exam technique

Write the trig equation first (e.g. sin 38° = QR ÷ 14); it is usually the first method mark.
Unless told otherwise, give angles to 1 decimal place and lengths to 3 significant figures.
On the non-calculator paper, any trig with 30°, 45° or 60° needs the exact values, so learn them as a table.
Use a previous answer at full accuracy (the ANS key or calculator memory), not a rounded one.

Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy1 mark

Write down the exact value of \(\cos 0^\circ\).[1]

Show the answer and mark scheme

Answer: \(1\)

B1 for \(1\)

Worked solution: \(\cos 0^\circ = 1\).

Question 2Medium3 marks

A ladder 5 m long leans against a vertical wall. The ground is horizontal and the foot of the ladder is 1.2 m from the wall.
The ladder is safe to use when the angle between the ladder and the ground is between 70° and 80°.

Is the ladder safe to use?
You must show your working.[3]

Show the answer and mark scheme

Answer: Yes: \(\cos\theta = \frac{1.2}{5}\), so \(\theta = 76.1^\circ\), which is between 70° and 80°.

M1 for \(\cos\theta = \frac{1.2}{5}\) (or finding the height \(\sqrt{5^2 - 1.2^2}\) and using sin or tan)
A1 for 76.1... (76.11°)
C1 for 'yes' with 76.1° compared with 70° and 80°

Worked solution: \(\cos\theta = \frac{1.2}{5} = 0.24\), so \(\theta = \cos^{-1}(0.24) = 76.11\ldots^\circ\).
76.1° is between 70° and 80°, so the ladder is safe.

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