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Pythagoras’ Theorem

This topic is relevant for:

Here we will learn about **trigonometry **including how to use SOHCAHTOA, inverse trigonometric functions, exact trigonometric values and the hypotenuse. We’ll also learn about the sine rule, the cosine rule, how to find the area of a triangle using

There are also trigonometry worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

**Trigonometry **is the relationship between angles and side lengths within triangles; it is derived from the greek words “trigōnon” meaning triangle and “metron” meaning measure.

Trigonometry was originally used by the Babylonians, over

The higher GCSE curriculum expands the use of trigonometric functions for non right-angle triangles, developing from the fundamental knowledge of the three trigonometric ratios (expressed as the mnemonic SOHCAHTOA) and exact trigonometric values in right angle triangles.

Get your free trigonometry worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOONGet your free trigonometry worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOON**SOHCAHTOA **is the abbreviation used to describe the three trigonometric ratios for the **sine**, **cosine **and **tangent **functions.

To determine which trigonometric function you need to use to answer a question, it depends on the location of the angle and the sides of the triangle that will be used.

The trigonometric functions apply to right-angle triangles.

- If you know the
**hypotenuse**and the**opposite**side of the angle, you would use the**sine**function. - If you know the
**hypotenuse**and the**adjacent**side (next to) the angle, you would use the**cosine**function. - If you know the
**opposite**and**adjacent**sides to the angle, you use the**tangent**function.

We can use SOHCAHTOA to calculate lengths and angles in 2D and 3D shapes by recognising right-angle triangles.

E.g.

We can find the length

**Step-by-step guide:** SOHCAHTOA

The **hypotenuse **is the longest side of a right angle triangle. It is the side opposite the right angle.

The hypotenuse does not occur for other types of triangles unless we know more information (such as an isosceles triangle can be made from 2 identical right angle triangles, back-to-back).

Once we know which angle we are using, we can label the sides opposite

We know the **hypotenuse** is opposite the right angle.

The **opposite **side is opposite the angle we are using.

The **adjacent **side is next to the angle we are using.

The triangle below is labelled based on using the angle

**Step-by-step guide:** __Hypotenuse__ (coming soon)

Calculate the value of

Labelling the sides

\[ \begin{aligned}
cos(\theta)&=\frac{A}{H}\\\\
cos(60)&=\frac{x}{16} \\\\
16\times cos(60&)=x\\\\
x&=8cm.
\end{aligned} \]

Inverse trigonometric functions allow us to calculate the size of angle

The inverse trigonometric functions look like this:

\[ \begin{aligned}
&\quad \;\; \text{sine} \quad \quad \quad \quad \quad \quad \quad \text{cosine} \quad \quad \quad \quad \quad \;\; \quad \text{tangent} \\\\
\text{trigonometric} \quad \quad \quad &\ sin(\theta) = \frac{O}{H} \quad \quad \quad \quad \;\; \cos(\theta)=\frac{A}{H}
\quad \quad \quad \quad \;\; \tan(\theta)=\frac{O}{A} \\
\text{function} \quad \quad \quad \quad &\\\\\\
\text{other} \quad \quad \quad \quad \quad &H=\frac{O}{\sin(\theta)} \quad \quad \quad \quad \;\; H=\frac{A}{\cos(\theta)} \quad \quad \quad \quad \;\; A=\frac{O}{\tan(\theta)} \\
\text{forms} \quad \quad \quad \quad \quad &\\
\quad \quad \quad \quad &O=H\times\sin(\theta) \quad \quad \quad A=H\times\cos(\theta) \quad \quad \quad O=A\times\tan(\theta) \\\\\\
\text{inverse} \;\; \quad \quad \quad \quad & \\
\text{trigonometric} \quad \quad \quad &\theta=\sin^{-1}(\frac{O}{H}) \quad \quad \quad \;\; \theta=\cos^{-1}(\frac{A}{H}) \quad \quad \quad \;\;\; \theta=\tan^{-1}(\frac{O}{A})\\
\text{function} \quad \quad \quad \quad &
\end{aligned}\]

**Step-by-step guide:** Trigonometric functions

Calculate the size of angle

The two sides that can be used to calculate the value of

\[ sin(\theta)=\frac{8}{10}. \]

In order to calculate

We therefore have:

\[ \begin{aligned}
\sin^{-1}(\sin(\theta))&=\sin^{-1}(\frac{8}{10})\\
\\\theta&=\sin^{-1}(\frac{8}{10})\\
\\\theta&=53.13^{\circ}\quad(2dp).\\
\end{aligned} \]

**Exact trigonometric values **are found when the relationship between the sides and the angles in a triangle have a specific relationship.

Summarising these values, we obtain the exact trigonometric values for sine, cosine and tangent for

\[ \begin{aligned}
& 0^{\circ} \quad \quad \quad \quad 30^{\circ} \quad \quad \quad \quad 45^{\circ} \quad \quad \quad \quad 60^{\circ} \quad \quad \quad \quad 90^{\circ} \\\\
\text{sin}(x) \quad\quad \quad \quad &0 \; \quad \quad \quad \quad \frac{1}{2} \quad \quad \quad \quad \frac{1}{\sqrt{2}}\quad \quad \quad \quad \frac{\sqrt{3}}{2} \quad \quad \quad \quad 1 \\\\
\text{cos}(x) \quad\quad \quad \quad &1 \; \quad \quad \quad \;\; \frac{\sqrt{3}}{2} \; \; \quad \quad \quad \frac{1}{\sqrt{2}}\;\; \quad \quad \quad \quad \frac{1}{2} \quad \quad \quad \quad \; 0 \\\\
\text{tan}(x) \quad\quad \quad \quad &0 \; \quad \quad \quad \;\; \frac{1}{\sqrt{3}} \; \quad \quad \quad \quad 1 \;\;\; \quad \quad \quad \quad \sqrt{3} \quad \quad \quad \text{Undefined}
\end{aligned}\]

We therefore have the triangle:

\[ \begin{aligned}
\sin(\theta)&=\frac{O}{H}\\
\\\sin(\theta)&=\frac{3}{6}\\
\\\sin(\theta)&=\frac{1}{2}\\
\end{aligned} \]

Using the table above,

\[ \begin{aligned}
\text{if } \sin(\theta) = \frac{1}{2}, \text{ then } \theta=30^{\circ}\\
\end{aligned} \]

We need to be able to interpret problems and recognise whether we need to use Pythagoras’ Theorem in

This flow chart describes the information you need to know about a shape in order to solve the problem.

It is important to recognise that with most of these problems, you may need to use the Pythagorean Theorem, or trigonometry, or both within the same question so you must be confident with these topics individually to access this topic fully.

Below is a summary of methods that can be used for right angled triangles:

**Name**

Pythagoras’ Theorem in 2D

Sine function

Cosine function

Tangent function

Inverse sine function

Inverse cosine function

Inverse tangent function

Pythagoras’ Theorem in 3D

**Used to find…**

The hypotenuse (c)

A shorter side (a)

If the triangle contains a right angle or not

Missing side:

hypotenuse (H)

or opposite (O)

Missing side:

hypotenuse (H)

or adjacent (A)

Missing side:

adjacent (A)

or opposite (O)

Missing angle

(\theta)

Missing angle

(\theta)

Missing angle

(\theta)

The diagonal (D)

**Rule or formula**

\quad\quad c=\sqrt{a^{2}+b^{2}}

\quad\quad a=\sqrt{c^{2}-b^{2}}

\quad\quad a^{2} + b^{2} = c^{2}

\quad\quad H=\frac{O}{\sin(\theta)}

\quad\quad O=H\times\sin(\theta)

\quad\quad H=\frac{A}{\cos(\theta)}

\quad\quad A=H\times\cos(\theta)

\quad\quad A=\frac{O}{\tan(\theta)}

\quad\quad O=A\times\tan(\theta)

\quad\quad \theta=\sin^{-1}(\frac{O}{H})

\quad\quad \theta=\cos^{-1}(\frac{A}{H})

\quad\quad \theta=\tan^{-1}(\frac{O}{A})

\quad\quad D=\sqrt{x^{2}+y^{2}+z^{2}}

Calculate the length of the hypotenuse of a right triangle,

The two important sides in this question are the opposite side (

\[ \begin{aligned}
H&=\frac{O}{\sin(\theta)}\\
\\H&=\frac{56}{\sin(63)}\\
\\H&=62.9m\quad(1dp)\\
\end{aligned} \]

The **sine rule** (or the law of sines) is a relationship between the size of an angle in a triangle and the opposite side. There are three relationships in a triangle as there are

Pythagoras’ Theorem cannot be used to find the third side of a non-right angled triangle. Instead we can use the sine rule or the cosine rule, depending on the information we know about the triangle.

To find a missing angle:

\frac{\sin (A)}{a}=\frac{\sin (B)}{b}

To find a missing side:

\frac{a}{\sin (A)}=\frac{b}{\sin (B)}

**Step-by-step guide: **Sine rule

Calculate the length

Label each angle A, B and C and each side a, b and c:

Here we know side

\begin{array}{l}
\frac{a}{\sin (A)}=\frac{c}{\sin (C)}\\
\\\frac{6}{\sin (55)}=\frac{c}{\sin (73)}\\
\\c=\frac{6}{\sin (55)}\times\sin(73)\\
\\c=\frac{6\sin(73)}{\sin (55)}\\
\\c=7.00\quad(2dp)
\end{array}

Here, the length

The **cosine rule** (or the law of cosines) is a formula which can be used to calculate the missing sides of a triangle or to find a missing angle. To do this we need to know the two arrangements of the formula and what each variable represents.

To find a missing side:

a^{2}=b^{2}+c^{2}-2bc\cos(A)

To find a missing angle:

A=\cos^{-1}(\frac{b^2+c^2-a^2}{2bc})

**Step-by-step guide:** Cosine rule

Find the length of

The vertices are already labelled with

Here, we need to find the missing side ^{2}

\begin{array}{l}
a^{2}=b^{2}+c^{2}-2bc\cos(A)\\
\\x^{2}=7.1^{2}+6.5^{2}-2\times7.1\times6.5\times\cos(32)\\
\\x^{2}=50.41+42.25-92.3\times\cos(32)\\
\\x^{2}=92.66-78.27483928…\\
\\x^{2}=14.38516072…\\
\\x=\sqrt{14.38516072…}\\
\\x=3.79cm\quad(2dp)\\
\end{array}

\frac{1}{2}abSin(C) is a formula to calculate the area of any triangle.

\[ Area =\frac{1}{2}ab\sin(C). \]

**Step-by-step guide:** Area of a triangle trig

Calculate the area of the triangle

Here, we label each side

\begin{array}{l}
A=\frac{1}{2}ab\sin(C)\\
\\A=\frac{1}{2}(12\times7\times\sin(77))\\
\\A=\frac{81.84708544}{2}\\
\\A=40.92cm^{2}\quad(2dp)\\
\end{array}

** 3D trigonometry **is the application of the trigonometric skills developed for

To find missing sides or angles in

The flowchart below can help determine which function you need to use:

Once you can justify which rule or formulae you need to use, you may need to carry out this process again for another triangle in the question.

Top tip: Look out for common angles or common sides.

**Step-by-step guide:** 3D trigonometry

Calculate the size of the angle in the triangular prism

We can see that triangle

This triangle contains no information about the angles so we need to use Pythagoras Theorem.

\begin{aligned}
a^{2}+b^{2}&=c^{2}\\
4^{2}+8^{2}&=c^{2}\\
80&=c^{2}\\
c&=\sqrt{80}\\
c&=8.94427191
\end{aligned}

This is a right angled triangle involving angles so we need to use SOHCAHTOA.

Since we know

\begin{aligned}
\tan(\theta)&=\frac{O}{A}\\
\tan(\theta) &= \frac{5}{8.94427191}\\
\theta &= \tan^{-1}(\frac{5}{8.94427191})\\
\theta &= 29.21^{\circ}
\end{aligned}

**Trigonometric graphs** are a visual representation of the sine, cosine and tangent functions. The horizontal axis represents the angle, usually written as

See below for all three trigonometric graphs for all angles of

**The graph of y = sin(θ)**

**Step-by-step guide:** Sin graph

**The graph of y = cos(θ)**

**Step-by-step guide:** Cos graph

**The graph of y = tan(θ)**

**Step-by-step guide:** Tan graph

Use the graph of

Here we draw the vertical line at

As the scale for each mark on the

**Labelling a triangle incorrectly**

E.g.

This triangle has been incorrectly labelled with the side next to the angle.

This will have an impact on the formula for the sine rule, the cosine rule, and the area of the triangle.

**Using the incorrect trigonometric function**

If the triangle is incorrectly labelled it can lead to the use of the incorrect standard or inverse trigonometric function.

**Rounding the decimal too early**

This can lose accuracy marks. Always use as many decimal places as possible throughout the calculation, then round the solution.

**Pythagoras’ Theorem or trigonometry?**

Use the flowchart to help you to recognise when to use Pythagoras’ Theorem and when to use trigonometry. Remember, you may need to use both.

**Using the sine rule instead of cosine rule**

In order to use the sine rule we need to have pairs of opposite angles and sides.

**Not using the included angle**

For the cosine rule and the area of a triangle using

**Using**A = b × h ÷ 2

If the vertical height of a triangle is not available then we cannot calculate the area by halving the base times the height.

**Using the inverse trig function instead, inducing a mathematical error**

If the inverse trig function is used instead of the standard trig function, the calculator may return a maths error as the solution does not exist.

**Sine and cosine graphs switched**

The sine and cosine graphs are very similar and can easily be confused with one another. A tip to remember is that you “sine up” from

**Asymptotes are drawn incorrectly for the graph of the tangent function**

The tangent function has an asymptote at

**The graphs are sketched using a ruler**

Each trigonometric graph is a curve and therefore the only time you are required to use a ruler is to draw a set of axes. Practice sketching each curve freehand and label important values on each axis.

1. Calculate the length of the side BC:

17.87cm

13.96cm

0.06cm

0.21cm

This is a right angled triangle involving angles so use SOHCAHTOA.

First we need to label the sides O, A and H.

We know A and we want to find H so we need to use cos.

\begin{aligned} H&=\frac{A}{\cos(\theta)}\\\\ H&=\frac{11}{\cos(52)}\\\\ H&=17.87\mathrm{cm} \end{aligned}

2. Using your knowledge of exact trigonometric values, work out the size of the angle marked . \theta

30^{\circ}

45^{\circ}

60^{\circ}

90^{\circ}

First we need to label the sides O, A and H.

Since we know O and A , we need to use tan.

\begin{aligned} \tan(\theta) &= \frac{O}{A}\\\\ \tan(\theta) &= \frac{\sqrt{3}}{1}\\\\ \tan(\theta) &=\sqrt{3} \end{aligned}

Using the exact trigonometric values,

\tan(\theta)=\sqrt{3} \text{ when } \theta=60^{\circ}

3. Work out the size of angle \theta .

60.67^{\circ}

54.53^{\circ}

46.69^{\circ}

60.07^{\circ}

This is not a right angled triangle and we know an angle and its opposite side so we need to use the sine rule.

\begin{aligned} \frac{\sin(A)}{a}&=\frac{\sin(B)}{b}\\\\ \frac{\sin(\theta)}{13}&=\frac{sin(70)}{15} \\\\ \sin(\theta)&= \frac{sin(70)}{15} \times 13\\\\ \sin(\theta)&=0.8144002713\\\\ \theta &=\sin^{-1}(0.8144002713)\\\\ \theta&=54.53^{\circ} \end{aligned}

4. Calculate the area of the following triangle:

168cm^{2}

91.50cm^{2}

23.71cm^{2}

140.90cm^{2}

We do not know the base or the height so we need to use:

A=\frac{1}{2}ab \sin(C)

\begin{aligned} A&=\frac{1}{2} \times 12 \times 28 \times \sin(57)\\\\ A&= 140.90cm^{2} \end{aligned}

5. Calculate the length of AE.

9.11cm

7.81cm

10.39 cm

10.91cm

The triangles AGH and AEH share the line AH .

Using the triangle AGH we can calculate the length of the line AH.

\begin{aligned} A &= \frac{O}{\tan(\theta)}\\\\ A &= \frac{6}{\tan(30)}\\\\ A &= 6\sqrt{3} \mathrm{cm} \end{aligned}

Using the triangle AEH we can calculate the length of AE .

\begin{aligned} a^{2}&+b^{2}=c^{2}\\\\ (AE)^{2}+5^{2}&=(6\sqrt{3})^{2}\\\\ (AE)^{2}&=(6\sqrt{3})^{2}-5^{2}\\\\ (AE)^{2}&=83\\\\ AE&=\sqrt{3}\\\\ AE&=9.11 \mathrm{cm} \end{aligned}

6. Write down the coordinates of a minimum point on the graph of y=cos(\theta) for 0^{\circ} \leqslant \theta \leqslant 360^{\circ}

(90, -1)

(180, 0)

(180, -1)

(90, 0)

The minimum point occurs at (180, -1)

1. Below is a sketch of a football pitch ABCD .

(a) Player F is standing exactly 60m perpendicular to Player E on the goal line and 75m from the corner where Player A is standing.

Player A kicks the football directly to Player F . Calculate the bearing of F from A . Write your answer correct to 2 decimal places.

(b) Player F then passes the ball to Player G (the goal keeper). Player G is standing at the midpoint of BC at a bearing of 060^{\circ} from F .

How far is Player G from Player F ? Write your answer to 2 decimal places.

**(7 Marks)**

Show answer

(a)

Bearing of F from A = 180 – angle EAF

**(1)**

Angle AEF = \sin^{-1}(\frac{60}{75})

**(1)**

Angle AEF = 53.13^{\circ}

**(1)**

Bearing of F from A = 180 – 53.13 =126.87^{\circ}

**(1)**

(b)

**(1)**

FG = \frac{45}{sin(60)}

**(1)**

FG = 51.96m

**(1)**

2. Triangle ABE and ACD are similar with AB:BC = 1:3. Using the information on the diagram, calculate the area of the shaded region BCDE .

State the units in your answer.

**(8 Marks)**

Show answer

Area (ABE) = ½ × a × b × sin(C) = ½ × 4 × 6 × sin(30)

**(1)**

Sin(30) = ½

**(1)**

Area (ABE) = 6cm^2

**(1)**

As AB:BC=1:3, BC= 4 × 3 = 12cm and DE = 6 × 3 = 18cm

**(1)**

AC = 12+4 = 16cm and AD = 6+18 = 24cm

**(1)**

Area (ACD) = ½ × a × b × sin(C) = ½ × 16 × 24 × sin(30)

**(1)**

Area (ACD) = 96cm^2

**(1)**

Area BCDE = 96 – 6 = 90cm^2

**(1)**

3. (a) The cuboid ABCDEFGH is shown below. ADEF is a square face with the side length of 2m , and the length AB = 8cm.

Calculate the length of the line AE. Write your answer as a surd in its simplest form.

(b) Given that the point X lies on the line EH so that XH = 3EX and angle XAB = 54.7^{\circ}, calculate the length of the line BX .

**(10 Marks)**

Show answer

(a)

AE^2 = AF^2 + EF^2

**(1)**

AE^2 = 2^2 + 2^2 = 8

**(1)**

AE = \sqrt{8} = 2\sqrt{2}

**(1)**

(b)

EX = 8 / 4 = 2m

**(1)**

AX = \sqrt{2^{2}+2\sqrt{2}^{2}}

**(1)**

AX = 2\sqrt{3}

**(1)**

Cosine rule stated to find BX :

BX^{2}=AX^{2}+AB^{2}-2\times{AX}\times{AB}\times\cos(A)

**(1)**

BX^{2}=(2\sqrt{3})^{2}+8^{2}-2\times{2\sqrt{3}}\times{8}\times\cos(54.7)

**(1)**

BX^{2}=43.97m

**(1)**

BX=6.63m

**(1)**

You have now learned how to:

- use trigonometric ratios in similar triangles to solve problems involving right-angled triangles
- recognise, sketch and interpret graphs of trigonometric functions (with arguments in degrees) y = sin x, y = cos x and y = tan x for angles of any size
- apply trigonometric ratios to find angles and lengths in right-angled triangles and, where possible, general triangles in 2 and 3 dimensional figures
- know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60° and 90°; know the exact value of tan θ for θ = 0°, 30°, 45°, 60°
- know and apply the sine rule, and cosine rule, to find unknown lengths and angles
- know and apply area = 1/2absinC to calculate the area, sides or angles of any triangle

- SOHCAHTOA
- Inverse trigonometric functions
- Exact trigonometric values
- The hypotenuse
- Sine rule
- Cosine rule
- The area of a triangle using 1/2absinC
- 3D trigonometry
- Sine, cosine and tangent graph

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