1MA1 · G20–G21Geometry & MeasuresFoundation & HigherGrades 4–7

Right-Angled Trigonometry

Pythagoras links the three sides of a right-angled triangle but says nothing about its angles. $\textbf{Trigonometry}$ does: sine, cosine and tangent connect an angle to the ratio of two sides. With them you can find the height of a building from the angle you look up at it, or the angle a ramp makes with the ground.

50 min Pythagoras & Trigonometry

The big picture

In any right-angled triangle with the same angle, the ratios of the sides are fixed, however large the triangle is — that is why they can be tabulated and built into calculators. Label the sides relative to the angle (opposite, adjacent, hypotenuse), choose the ratio that uses the two sides involved, and solve.

Edexcel expects exact values of sin, cos and tan for $0^{\circ}$ , $30^{\circ}$ , $45^{\circ}$ , $60^{\circ}$ and $90^{\circ}$ on the non-calculator paper, and applies trigonometry to elevation, depression, bearings and, at Higher tier, 3D shapes.

What you'll learn

Label opposite, adjacent and hypotenuse relative to an angle
Use sine, cosine and tangent to find missing sides and angles
Recall exact values of trigonometric ratios for key angles
Solve problems involving elevation, depression and 3D shapes

Labelling and choosing a ratio

Relative to the angle $\theta$ : the $\textbf{hypotenuse}$ is opposite the right angle; the $\textbf{opposite}$ is across from $\theta$ ; the $\textbf{adjacent}$ is next to $\theta$ (not the hypotenuse).

SOH CAH TOA: $\sin\theta = \dfrac{O}{H}$ , $\cos\theta = \dfrac{A}{H}$ , $\tan\theta = \dfrac{O}{A}$ .

Pick the ratio containing the side you know and the side you want.

\sin\theta = \tfrac{O}{H},\ \cos\theta = \tfrac{A}{H},\ \tan\theta = \tfrac{O}{A}

SOH CAH TOA.

Tip — Label all three sides before choosing a ratio. The opposite and adjacent swap if you use the other acute angle.

Finding sides and angles

To find a side, substitute into the ratio and rearrange: if $\cos 38^{\circ} = \dfrac{A}{15}$ , then $A = 15\cos 38^{\circ}$ .

If the unknown is on the bottom, rearrange carefully: $\tan 50^{\circ} = \dfrac{8}{A}$ gives $A = \dfrac{8}{\tan 50^{\circ}}$ .

To find an angle, use the inverse function: $\theta = \sin^{-1}\left(\dfrac{O}{H}\right)$ .

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

38^{\circ}

. Find the side adjacent to that angle.

\cos 38^{\circ} = \dfrac{A}{15}

A = 15\cos 38^{\circ}

Answer11.8 cm

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

The hypotenuse of a right-angled triangle is 20 cm and one angle is $41^\circ$ . Find the side adjacent to that angle, to 1 decimal place.

Your answer

To 1 decimal place.

Show worked solution

Solution

1 $\cos 41^\circ = \frac{a}{20}$ .
2 $a = 20\cos 41^\circ = 15.09\ldots$

Answer: $15.1$ cm

Example 2A right-angled triangle has opposite side 7 cm and hypotenuse 12 cm. Find the angle.

\sin\theta = \dfrac{7}{12}

\theta = \sin^{-1}\left(\dfrac{7}{12}\right)

Answer

35.7^{\circ}

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

A right-angled triangle has opposite side 5 cm and hypotenuse 9 cm. Find the angle, to 1 decimal place.

Your answer

To 1 decimal place.

Show worked solution

Solution

1 $\sin\theta = \frac{5}{9}$ .
2 $\theta = \sin^{-1}(0.5\dot{5}) = 33.7^\circ$ .

Answer: $33.7^\circ$

Exact values

$\sin 30^{\circ} = \tfrac12$ , $\cos 30^{\circ} = \tfrac{\sqrt3}{2}$ , $\tan 30^{\circ} = \tfrac{1}{\sqrt3}$ .

$\sin 45^{\circ} = \cos 45^{\circ} = \tfrac{\sqrt2}{2}$ , $\tan 45^{\circ} = 1$ .

$\sin 60^{\circ} = \tfrac{\sqrt3}{2}$ , $\cos 60^{\circ} = \tfrac12$ , $\tan 60^{\circ} = \sqrt3$ . Also $\sin 0^{\circ} = 0$ , $\cos 0^{\circ} = 1$ , $\sin 90^{\circ} = 1$ , $\cos 90^{\circ} = 0$ .

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

Write down the exact value of $\tan 60^\circ$ .

Your answer

Type √ as sqrt, e.g. sqrt3.

Show worked solution

Solution

1From half an equilateral triangle with sides 1, $\sqrt{3}$ , 2.
2 $\tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3}$ .

Answer: $\sqrt{3}$

Why this works

The values come from two triangles you can draw: half an equilateral triangle of side 2 (giving $30^{\circ}$ and $60^{\circ}$ ) and a right-angled isosceles triangle with shorter sides 1 (giving $45^{\circ}$ ). Redraw them rather than memorise blindly.

Elevation, depression and 3D

The $\textbf{angle of elevation}$ is measured up from the horizontal; the $\textbf{angle of depression}$ is measured down from the horizontal. They are equal for two points looking at each other (alternate angles).

In 3D, find a right-angled triangle containing the angle you want, often using Pythagoras first to find a missing side of that triangle.

Example 1A cuboid measures 6 cm by 4 cm by 3 cm. Find the angle between the space diagonal and the base.

1Base diagonal:

\sqrt{36 + 16} = \sqrt{52}

2In the vertical triangle:

\tan\theta = \dfrac{3}{\sqrt{52}}

Answer

22.6^{\circ}

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

From 50 m away on level ground, the angle of elevation of the top of a tower is $32^\circ$ . Find the height of the tower, to 1 decimal place.

Your answer

To 1 decimal place.

Show worked solution

Solution

1 $\tan 32^\circ = \frac{h}{50}$ .
2 $h = 50\tan 32^\circ = 31.24\ldots$ m.

Answer: $31.2$ m

Tip — Check your calculator is in degree mode before every trigonometry question.

Think like an examiner

Label sides and write the ratio used.
Show the rearrangement before calculating.
Give angles to 1 d.p. and lengths to 3 s.f. unless told otherwise.
Use exact values on the non-calculator paper.

Trigonometry

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

Sine.

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

Cosine.

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

Tangent.

Watch out for these

Labelling the hypotenuse as the adjacent.

The hypotenuse is always opposite the right angle.

Writing

\tan 50^{\circ} = \tfrac{8}{A}

A = 8\tan 50^{\circ}

A = \dfrac{8}{\tan 50^{\circ}}

Using

\sin

instead of

\sin^{-1}

to find an angle.

Inverse functions give angles.

Calculator in radian mode.

Use degrees for GCSE trigonometry.

Stretch yourself

From the top of an 80 m cliff, the angles of depression to two boats directly out to sea, one behind the other, are $38^{\circ}$ and $24^{\circ}$ . How far apart are the boats?

Hint — The angle of depression equals the angle of elevation from each boat. Find each boat’s distance from the foot of the cliff with tangent.

Reveal solution

SolutionFor each boat,

\tan(\text{angle}) = \dfrac{80}{d}

, so

d = \dfrac{80}{\tan(\text{angle})}

. Nearer boat:

\dfrac{80}{\tan 38^{\circ}} = 102.4

m. Further boat:

\dfrac{80}{\tan 24^{\circ}} = 179.7

m. The boats are

179.7 - 102.4 = 77.3

m apart.

\blacksquare

Questions students ask

How do I choose between sin, cos and tan?

Label the sides, identify which two are involved (known and unknown), and use the ratio that contains those two.

Why is the angle of elevation equal to the angle of depression?

They are alternate angles between two horizontal lines.

Do I need exact values?

Yes for Edexcel: they appear on the non-calculator paper.

Key takeaways

Label opposite, adjacent and hypotenuse relative to the angle.
SOH CAH TOA: choose the ratio with the two sides involved.
Rearrange for sides; use inverse functions for angles.
Know exact values for 0°, 30°, 45°, 60° and 90°.
Elevation and depression are measured from the horizontal.

Exam-style questions

Practise it the way it is examined

3 original questions · 6 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 42 marks

In a right-angled triangle, the side adjacent to a $28^\circ$ angle is 15 cm. Work out the opposite side, to 3 significant figures.[2]
Show mark scheme
- M1 $\tan 28 = \frac{x}{15}$
- A1 $7.98$ cm
Answer: $7.98$ cm

Question 2Grade 52 marks

From the top of a 40 m cliff, a boat is 120 m from the foot of the cliff. Work out the angle of depression, to 1 decimal place.[2]
Show mark scheme
- M1 $\tan\theta = \frac{40}{120}$
- A1 $18.4^\circ$
Answer: $18.4^\circ$

Question 3Grade 72 marks

Without a calculator, find the exact length of the side opposite a $60^\circ$ angle in a right-angled triangle with hypotenuse 10 cm.[2]
Show mark scheme
- M1 $10 \sin 60^\circ$
- A1 $5\sqrt{3}$ cm
Answer: $5\sqrt{3}$ cm

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.

Next step

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