Why does 90^ give an error?

As the angle approaches 90^ the adjacent side shrinks towards zero, so OA grows without limit — it is undefined.

G20, G21Geometry & MeasuresFoundation & HigherGrades 4–8

Trigonometry

Trigonometry links the angles of a right-angled triangle to the ratios of its sides. Pythagoras needs two sides to find a third; trigonometry lets you use one side and one angle — or two sides to find an angle. The three ratios, sine, cosine and tangent, are remembered as SOH CAH TOA.

50 min Video by Maths Genie AQA GCSE Maths

The big picture

In every right-angled triangle with a $35^\circ$ angle, the ratio $\frac{\text{opposite}}{\text{hypotenuse}}$ is the same number, whatever the size of the triangle — because all such triangles are similar. That fixed number is $\sin 35^\circ$ . Your calculator simply looks it up.

Every question follows the same routine: label the sides from the angle's point of view, choose the ratio that uses the two sides involved, write the equation, then solve — multiplying or dividing for a side, or using the inverse ( $\sin^{-1}$ ) for an angle.

Using SOHCAHTOA to Find a Missing AngleCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Label the opposite, adjacent and hypotenuse relative to an angle
Use sine, cosine and tangent to find missing sides
Use inverse trigonometric functions to find missing angles
Recall exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°

Labelling and choosing a ratio

The $\textbf{hypotenuse}$ is opposite the right angle. The $\textbf{opposite}$ is across from the angle you are using. The $\textbf{adjacent}$ is next to that angle (but not the hypotenuse).

$\sin\theta = \frac{O}{H}$ , $\cos\theta = \frac{A}{H}$ , $\tan\theta = \frac{O}{A}$ . Cross out the side you do not know and do not need; the remaining pair tells you the ratio.

Opposite is across from

\theta

; adjacent is next to it; the hypotenuse is opposite the right angle.

Finding a missing side

Write the ratio with the numbers in. If the unknown is on top, multiply: $x = 12 \sin 40^\circ$ . If the unknown is on the bottom, rearrange: $x = \frac{12}{\sin 40^\circ}$ .

Example 1In a right-angled triangle, the hypotenuse is 15 cm and angle

\theta = 32^\circ

. Find the side opposite

\theta

to 1 d.p.

1O and H: use sine.

\sin 32^\circ = \frac{x}{15}

x = 15 \sin 32^\circ = 7.948\ldots

Answer

7.9

Your turnSame method, new numbers — try it before moving on.

The side adjacent to a $50^\circ$ angle is 8 cm. Find the opposite side, to 1 decimal place.

Your answer

Show worked solution

Solution

1O and A: use tangent. $\tan 50^\circ = \frac{x}{8}$ .
2 $x = 8 \tan 50^\circ = 9.534\ldots$ , so $9.5$ cm.

Answer: $9.5$ cm

Example 2The side adjacent to a

28^\circ

angle is 20 cm. Find the hypotenuse to 1 d.p.

1A and H: use cosine.

\cos 28^\circ = \frac{20}{h}

h = \frac{20}{\cos 28^\circ} = 22.651\ldots

Answer

22.7

Your turnSame method, new numbers — try it before moving on.

The side opposite a $35^\circ$ angle is 6 cm. Find the hypotenuse, to 1 decimal place.

Your answer

Show hint

The unknown is on the bottom of the fraction, so divide.

Show worked solution

Solution

1 $\sin 35^\circ = \frac{6}{h}$ .
2 $h = \frac{6}{\sin 35^\circ} = 10.460\ldots$ , so $10.5$ cm.

Answer: $10.5$ cm

Tip — Make sure your calculator is in degrees (a small D or DEG on the screen). In radians, $\sin 30$ gives $-0.988$ instead of $0.5$ .

Finding a missing angle

Write the ratio as a fraction of the two known sides, then use the inverse function: $\theta = \sin^{-1}\left(\frac{O}{H}\right)$ , and so on. On most calculators, press SHIFT then sin.

Example 1A right-angled triangle has opposite 9 cm and adjacent 14 cm relative to

\theta

. Find

\theta

to 1 d.p.

1O and A:

\tan\theta = \frac{9}{14}

\theta = \tan^{-1}\left(\frac{9}{14}\right) = 32.735\ldots^\circ

Answer

32.7^\circ

Your turnSame method, new numbers — try it before moving on.

A ladder 6 m long reaches 5.2 m up a vertical wall. Find the angle the ladder makes with the ground, to 1 decimal place.

Your answer

Show worked solution

Solution

1The height is opposite the angle; the ladder is the hypotenuse: $\sin\theta = \frac{5.2}{6}$ .
2 $\theta = \sin^{-1}(0.8\dot{6}) = 60.07\ldots^\circ$ , so $60.1^\circ$ .

Answer: $60.1^\circ$

Elevation, depression and exact values

An $\textbf{angle of elevation}$ is measured up from the horizontal; an $\textbf{angle of depression}$ is measured down from the horizontal. They are equal when looking between the same two points (alternate angles).

Exact values must be learned for the non-calculator paper: $\sin 30^\circ = \cos 60^\circ = \frac{1}{2}$ ; $\sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}$ ; $\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}$ ; $\tan 45^\circ = 1$ ; $\tan 30^\circ = \frac{\sqrt{3}}{3}$ ; $\tan 60^\circ = \sqrt{3}$ ; $\sin 0^\circ = \cos 90^\circ = 0$ ; $\sin 90^\circ = \cos 0^\circ = 1$ ; $\tan 0^\circ = 0$ .

Example 1From the top of a 40 m cliff, the angle of depression of a boat is

18^\circ

. How far is the boat from the foot of the cliff?

1The angle at the boat (elevation) is also

18^\circ

. The cliff is opposite, the distance adjacent.

\tan 18^\circ = \frac{40}{d}

, so

d = \frac{40}{\tan 18^\circ} = 123.1\ldots

Answer

123

Example 2Without a calculator, find

x

if a right-angled triangle has hypotenuse 12 cm and

x

is opposite a

30^\circ

angle.

x = 12 \sin 30^\circ = 12 \times \frac{1}{2}

Answer

6

Your turnSame method, new numbers — try it before moving on.

Without a calculator, a right-angled triangle has adjacent side 5 cm next to a $60^\circ$ angle. Find the hypotenuse, in cm.

Your answer

Show worked solution

Solution

1 $\cos 60^\circ = \frac{5}{h}$ and $\cos 60^\circ = \frac{1}{2}$ .
2 $h = 5 \div \frac{1}{2} = 10$ cm.

Answer: $10$ cm

Where exact values come from

The exact values come from two triangles. Half an equilateral triangle of side 2 has sides $1$ , $\sqrt{3}$ and $2$ , giving the $30^\circ$ and $60^\circ$ values. A right-angled isosceles triangle with sides $1$ , $1$ , $\sqrt{2}$ gives the $45^\circ$ values. Draw them and you never need to memorise a table.

Think like an examiner

Write the ratio with numbers substituted, e.g. $\cos 28 = \frac{20}{h}$ — it is the method mark.
Give angles to 1 d.p. and lengths to 3 s.f. unless told otherwise; write the unrounded value first.
For elevation and depression questions, sketch the triangle and mark the horizontal.

SOH CAH TOA

\sin\theta = \frac{O}{H}

Opposite over hypotenuse.

\cos\theta = \frac{A}{H}

Adjacent over hypotenuse.

\tan\theta = \frac{O}{A}

Opposite over adjacent.

Watch out for these

Labelling the hypotenuse as the adjacent.

The hypotenuse is always opposite the right angle; the adjacent is the other side touching

\theta

\cos 28 = \frac{20}{h} \Rightarrow h = 20 \cos 28

The unknown is on the bottom:

h = \frac{20}{\cos 28}

Finding an angle with

\tan(9 \div 14)

Use the inverse:

\tan^{-1}(9 \div 14)

Calculator in radians giving strange answers.

Switch to degree mode.

Stretch yourself

A kite on a straight 50 m string makes an angle of $62^\circ$ with the ground. Nadia reels in some string. The kite stays at the same height, and the string now makes an angle of $75^\circ$ with the ground. How long is the string now, to 1 decimal place?

Hint — Find the height first using the $62^\circ$ triangle, then use it in the $75^\circ$ triangle.

Reveal solution

SolutionHeight

= 50 \sin 62^\circ = 44.147\ldots

m. New string length

= \frac{44.147\ldots}{\sin 75^\circ} = 45.7

m (to 1 d.p.). Reeling in about 4.3 m of string makes the angle steeper while the height stays the same.

\blacksquare

Questions students ask

How do I know whether to use trigonometry or Pythagoras?

If an angle is given or asked for, use trigonometry. If only sides are involved (three sides, two known), use Pythagoras.

Can trigonometry be used in triangles without a right angle?

SOH CAH TOA needs a right angle. For other triangles, Higher students use the sine rule and cosine rule.

Why does

\tan 90^\circ

give an error?

As the angle approaches

90^\circ

the adjacent side shrinks towards zero, so

\frac{O}{A}

grows without limit — it is undefined.

Key takeaways

Label O, A, H from the angle in question.
Choose sin, cos or tan from the two sides involved.
Unknown on top: multiply. Unknown on bottom: divide.
Angles: use $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ .
Learn exact values for 0°, 30°, 45°, 60° and 90°.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

In triangle $ABC$ , angle $B = 90^\circ$ , $AC = 18$ cm and angle $A = 41^\circ$ . Work out the length of $BC$ . Give your answer to 3 significant figures.[3]
Show mark scheme
- M1 $\sin 41 = \frac{BC}{18}$
- M1 $BC = 18 \sin 41$
- A1 $11.8$ cm (11.80…)
Answer: $11.8$ cm

Question 2Grade 63 marks

A ramp rises 1.2 m over a horizontal distance of 7.5 m. Building rules say a ramp must not be steeper than $10^\circ$ .

Does the ramp meet the rules? Show your working.[3]
Show mark scheme
- M1 $\tan\theta = \frac{1.2}{7.5}$
- A1 $\theta = 9.09^\circ$
- B1Yes, because $9.09^\circ < 10^\circ$

Question 3Grade 73 marks

Without a calculator, show that the exact area of a right-angled triangle with hypotenuse 8 cm and one angle of $30^\circ$ is $8\sqrt{3}$ cm².[3]
Show mark scheme
- M1Opposite $= 8 \sin 30 = 4$
- M1Adjacent $= 8 \cos 30 = 4\sqrt{3}$
- A1Area $= \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3}$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Which ratio uses opposite and hypotenuse?
- sine
- cosine
- tangent
- Pythagoras
Show answer and explanation
Answer: sine
SOH.
What is $\tan 45^\circ$ ?
- $1$
- $\frac{1}{2}$
- $\sqrt{2}$
- $0$
Show answer and explanation
Answer: $1$
Opposite and adjacent are equal.
What is $\sin 30^\circ$ ?
- $\frac{1}{2}$
- $\frac{\sqrt{3}}{2}$
- $1$
- $\frac{\sqrt{2}}{2}$
Show answer and explanation
Answer: $\frac{1}{2}$
An exact value to learn.
Hypotenuse 10, angle $40^\circ$ . Adjacent side (1 d.p.)?
- 7.7
- 6.4
- 8.4
- 13.1
Show answer and explanation
Answer: 7.7
$10 \cos 40$ .
Opposite 5, adjacent 5. The angle is…
- $45^\circ$
- $30^\circ$
- $60^\circ$
- $90^\circ$
Show answer and explanation
Answer: $45^\circ$
$\tan^{-1}(1)$ .
Opposite 4, hypotenuse 9. The angle (1 d.p.)?
- $26.4^\circ$
- $63.6^\circ$
- $24.0^\circ$
- $0.5^\circ$
Show answer and explanation
Answer: $26.4^\circ$
$\sin^{-1}(4 \div 9)$ .
Adjacent 12, angle $25^\circ$ . Hypotenuse (1 d.p.)?
- 13.2
- 10.9
- 5.6
- 28.4
Show answer and explanation
Answer: 13.2
$12 \div \cos 25$ .
What is $\cos 30^\circ$ ?
- $\frac{\sqrt{3}}{2}$
- $\frac{1}{2}$
- $\sqrt{3}$
- $\frac{\sqrt{3}}{3}$
Show answer and explanation
Answer: $\frac{\sqrt{3}}{2}$
From half an equilateral triangle.

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