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In order to access this I need to be confident with:

Sine rule Cosine rule Tangent of a circle Plans and elevationsExact trigonometric values

Surds Decimal places How to find the midpointThis topic is relevant for:

Here we will learn about **3D trigonometry** including how to combine your knowledge of Pythagoras’ Theorem, Trigonometric Ratios, The Sine Rule and The Cosine Rule and apply it to find missing angles and sides of triangles in 3-dimensional shapes.

There are also 3D trigonometry worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

** 3D trigonometry **is an 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:

You may need to carry out this process several times in a question to fully answer what is being asked. You must be able to justify which rule or formulae you need to use.

*Top Tip: Look out for common angles or common sides.*

In order to find a missing angle or side within a

**Calculate the necessary missing angle or side of a triangle.****Sketch and label the second triangle using information from Step**1 .**Calculate the missing angle or side of the final triangle.**

Get your free 3D trigonometry worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free 3D trigonometry worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEThe diagram shows a cuboid.

Calculate the length of the line AH . Write your answer in the form a\sqrt{b}\mathrm{cm} where a and b are integers.

**Calculate the necessary missing angle or side of a triangle.**

The line

Here we need to find the value of

\[\begin{aligned}
H&=\frac{O}{\sin(\theta)}\\\\
H&=\frac{18}{\sin(30)}\\\\
H&=36\mathrm{cm}\\\\
FH&=36\mathrm{cm}
\end{aligned}
\]

2**Sketch and label the second triangle using information from step 1.**

3**Calculate the missing angle or side of the final triangle**.

\[\begin{aligned}
c^{2}&=a^{2}+b^{2}\\\\
x^{2}&=24^{2}+36^{2}\\\\
x^{2}&=1872\\\\
x&=12\sqrt{13}\mathrm{cm}
\end{aligned}
\]

^{o}^{o}

**Calculate the necessary missing angle or side of a triangle.**

The triangle

We need to find the value of

\[\begin{aligned}
&H=\frac{A}{\cos(\theta)}\\\\
&x=\frac{3.5}{ \cos(75)}\\\\
&x=13.52296157\mathrm{cm}
\end{aligned}
\]

**Sketch and label the second triangle using information from step 1.**

You must remember not to round your solution too early. Here we will continue to use the full decimal given for

**Calculate the missing angle or side of the final triangle**.

Here we have a right angle triangle so we can use another trigonometric ratio to find the length

\[\begin{aligned}
H&=\frac{O}{\sin(\theta)}\\\\
CE&=\frac{13.52296157}{\sin(40)}\\\\
CE&=21.03799352…\\\\
CD&=21.34\mathrm{cm}\qquad(3.d.p)
\end{aligned}
\]

- The front face of the cuboid is a square.
- The midpoint
M lies half-way betweenE andF . - The line
AC is at30 from the line^{o}AB . BCM is a triangle.AC = 52cm - CE=26\sqrt{3}

Using the diagram below, calculate the size of angle

**Calculate the necessary missing angle or side of a triangle.**

The line

Here we need to find the value of

\[\begin{aligned}
O&=H\sin(\theta)\\\\
O&=52\sin(30)\\\\
O&=26\mathrm{cm}\\\\
BC&=26\mathrm{cm}
\end{aligned}
\]

**Sketch and label the second triangle using information from step 1.**

As

**Calculate the missing angle or side of the final triangle**.

As we can split an isosceles triangle into two right-angle triangles, and we can use trigonometric ratios to find the angle \frac{\theta}{2} then multiply by 2 to find \theta :

\[\begin{aligned}
\frac{\theta}{2}&=\tan^{-1}(\frac{O}{A})\\\\
\frac{\theta}{2}&=\tan^{-1}(\frac{13}{26\sqrt(3)})\\\\
\frac{\theta}{2}&=16.10211375…\\\\
\theta&=32.20\mathrm{cm}\qquad(2dp)
\end{aligned}
\]

**Calculate the necessary missing angle or side of a triangle.**

Each side of the triangle

Here we need to find the value of

\[\begin{aligned}
x^{2}&=5^{2}+12^{2}\\\\
x&=\sqrt{25+144}\\\\
x&=13\mathrm{cm}\\\\
BF&=13\mathrm{cm}\\\\
\end{aligned}
\]

Here we need to find the value of

\[\begin{aligned}
y^{2}&=9^{2}+5^{2}\\\\
y&=\sqrt{81+25}\\\\
y&=\sqrt{106}\mathrm{cm}\\\\
BH&=\sqrt{106}\mathrm{cm}
\end{aligned}\]

Here we need to find the value of

\[\begin{aligned}
z^{2}&=9^{2}+12^{2}\\\\
z&=\sqrt{81+144}\\\\
z&=\sqrt{225}\mathrm{cm}\\\\
FH&=15\mathrm{cm}
\end{aligned}
\]

**Sketch and label the second triangle using information from step 1.**

Keeping BH=\sqrt{106} , we now have the triangle

**Calculate the missing angle or side of the final triangle**.

As we know the three side lengths of the triangle, we can use the cosine rule to find the missing angle \theta.

\[\begin{aligned}
\cos(A)&=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\\\
\cos(A)&=\frac{288}{390}\\\\
A&=cos^{-1}(\frac{288}{390})\\\\
A&=42.39947336…\\\\
A&=42.4^{\circ} (1.d.p)
\end{aligned}
\]

The diagram shows a triangular prism.

- Angle
DEC = 50º - Angle
EBC = 55º AE = AD

The angle *ϴ**ϴ*

**Calculate the necessary missing angle or side of a triangle.**

The line

Here we need to find the value of

\begin{aligned}
\\H&=\frac{O}{\sin(\theta)}\\
\\H&=\frac{25}{\sin(50)}\\
\\CE&=32.63518223…\mathrm{cm}\\
\end{aligned}

**Sketch and label the second triangle using information from step 1.**

**Calculate the missing angle or side of the final triangle**.

We know two sides and one of the two opposite angles so we need to use the sine rule to find the value for *ϴ*

\begin{aligned}
\\\sin(\theta)&=\frac{\sin(55)}{32.63518223…}\times{23}\\
\\\theta&=\sin^{-1}(0.5773063219)\\
\\\theta&=35.26130584…\\
\\\theta&=35.26^{\circ}\quad(2dp)\\
\end{aligned}

The diagram shows a cylinder:

A, B andD are points on the circumference of the circles.C is the centre of the circle.ABC is a triangleAD is the diameter of the cylinder.

By calculating the size of angle *ϴ*

**Calculate the necessary missing angle or side of a triangle.**

As we know two sides and one of the two opposite angles, we can use the sine rule to find the missing angle *ϴ*

Here we need to find the value of *ϴ*

\begin{aligned}
\\\sin(\theta)&=\frac{\sin(32)}{7.8}\times{12.4}\\
\\\theta&=\sin^{-1}(0.8424357534)\\
\\\theta&=57.39823007…^{\circ}\\
\end{aligned}

**Sketch and label the second triangle using information from step 1.**

As

As the angle

We now have enough information on the second triangle to calculate the height of the cylinder. Remember not to round too early.

**Calculate the missing angle or side of the final triangle**

To find the value of

\begin{aligned}
\\A&=\frac{O}{\tan(\theta)}\\
\\x&=\frac{15.6}{\tan(32.60176993…)}\\
\\x&=24.39138008…\\
\\x&=24.39cm\quad(2dp)\\
\end{aligned}

**Using Pythagoras’ Theorem instead of trigonometry**

Using two sides of a non right-angle triangle to find the third side instead of using the cosine rule.

**Incorrect trigonometric ratio used**

Incorrect labelling of any triangle can lead to the wrong trig function being used.

**Confusing the Sine Rule with the Cosine Rule**

Misunderstanding when to use the sine rule or cosine rule to find a missing side or angle.

**Using the inverse trig function instead and 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.

3D trigonometry is part of our series of lessons to support revision on trigonometry. You may find it helpful to start with the main trigonometry lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

1. ABCDEFGH is a cuboid. Calculate the length of DF to 2 decimal places.

9.04cm

10.33cm

7.53cm

15.82cm

Let’s first look at triangle FGH:

We want to know the length of FH which is the opposite.

Now that we know the length of FH , we can consider the triangle DFH:

Since we know two sides and we want to calculate the third side, we can use Pythagoras Theorem:

```
\begin{aligned}
c^{2}&=a^{2}+b^{2}\\\\
c^{2}&=5^{2}+9.039060338^{2}\\\\
c^{2}&=106.7046118\\\\
c&=\sqrt{106.7046118}\\\\
c&=10.33 \mathrm{cm}
\end{aligned}
```

2. ABCDEF is a triangular prism. Calculate the angle DAE . Give your answer to 1 dp.

43.7^{\circ}

21.0^{\circ}

35.8^{\circ}

39.3^{\circ}

First we need to look at the triangle DEF:

We need to find the length of DE , which is the opposite.

\begin{aligned} O&=H \sin(\theta)\\\\ O&=9sin(62)\\\\ O&=7.946528336 \mathrm{cm} \end{aligned}

Now that we know the length of DE, we can consider the triangle ADE :

We want to find the angle DAE. We know O and A .

\begin{aligned} \tan(\theta)&=\frac{O}{A}\\\\ \tan(\theta)&=\frac{7.946528336}{11}\\\\ \theta&=\tan^{-1}(\frac{7.946528336}{11})\\\\ \theta&=35.8^{\circ} \end{aligned}

3. ABCDE is a square based pyramid. By finding the value of x , calculate the perimeter of the base of the pyramid, correct to 2 decimal places.

282.84m

400m

565.69m

70.71m

First we need to work out the value of x:

\begin{aligned} O&= H \sin(\theta)\\\\ O&=50 \sqrt{2} \sin(45)\\\\ O &=50 \mathrm{m} \end{aligned}

Now we can look at the base:

The length of the sides can be found using Pythagoras Theorem:

\begin{aligned} &c^2=a^2+b^2\\\\ &c^2=50^2+50^2\\\\ &c^2=5000\\\\ &c=70.71m \end{aligned}

Each side is 70.71m therefore the perimeter is:

4 \times 70.71=282.84 \mathrm{m}4. Three satellites leave Earth on three different trajectories. 2 hours after launch, satellite A is 800km from Earth, satellite B is 500km from Earth and satellite C is 750km from Earth.

Use the cosine rule to calculate the size of angle CAB at this point in time.

76.60^{\circ}

38.22^{\circ}

32.01^{\circ}

51.78^{\circ}

AB, AC and BC can all be worked out using Pythagoras Theorem:

\begin{aligned} &AB^2=500^2+800^2\\ &AB^2=890000\\ &AB=943.3981132\\\\ &AC^2=750^2+800^2\\ &AC^2=1202500\\ &AC=1096.58561\\\\ &BC^2=500^2+750^2\\ &BC^2=812500\\ &BC=901.388189\\ \end{aligned}

We can then apply the cosine rule:

The length of the sides can be found using Pythagoras Theorem:

\begin{aligned} \cos(A)&=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\\\ \cos(A)&=\frac{1096.58561^{2}+943.3981132^{2}-901.388189^{2}}{2 \times 1096.58562 \times 943.3981132}\\\\ \cos(A)&=0.6186459892\\\\ A&=\cos^{-1}(0.6186459892)\\\\ A&=51.87^{\circ} \end{aligned}

5. Given that GH=10cm, work out the size of the angle GEH . Give your answer to 1 dp.

38.7^{\circ}

38.5^{\circ}

51.3^{\circ}

36.4^{\circ}

First we need to calculate the length of EH. We can do this using Pythagoras Theorem:

\begin{aligned} a^{2}&=c^{2}-b^{2}\\\\ a^{2}&=20^{2}-12^{2}\\\\ a^{2}&=256\\\\ a&=16 \mathrm{cm} \end{aligned}

We can now look at the triangle EGH:

We can calculate angle GEH using the sine rule:

\begin{aligned} \frac{\sin(A)}{a} &= \frac{sin(B)}{b}\\\\ \frac{\sin(\theta)}{10} &= \frac{sin(85)}{16}\\\\ \sin(\theta)&= \frac{sin(85)}{16} \times 10\\\\ \sin(\theta)&= 0.6226216863\\\\ \theta&= \sin^{-1}(0.6226216863)\\\\ \theta&=38.5^{\circ} \end{aligned}

6. ABCDEF is a triangular prism. X, Y, and Z are midpoints on each edge of the prism and triangle XYZ is isosceles. Using this information and the diagram to help you, calculate the size of angle XYZ.

36.5^{\circ}

51.3^{\circ}

9.2^{\circ}

18.0^{\circ}

We need to calculate the length XZ using the triangle CXZ:

\begin{aligned} O&=A \tan(\theta)\\\\ O&=5 \tan(76)\\\\ O&=20.05390467 \mathrm{cm} \end{aligned}

We can now look at triangle XYZ:

Using the cosine rule:

\begin{aligned} \cos(A) &= \frac{b^{2}+c^{2}-a^{2}}{2bc}\\\\ \cos(\theta)&= \frac{32^{2}+32^{2}-20.05390467^{2}}{2 \times 32 \times 32}\\\\ \cos(\theta)&=0.8036332556\\\\ \theta&= \cos^{-1}(0.8036332556)\\\\ \theta&=36.5^{\circ} \end{aligned}

1. ABCDEFGH is a cuboid.

Calculate the angle between the diagonal DF and the base AEHD .

Give your answer to 3 sf.

**(4 marks)**

Show answer

Triangle ADE:

DE^{2}=8^{2}+12^{2}

**(1)**

\begin{aligned} DE&=\sqrt{208}\\\\ DE&=14.4222\mathrm{cm} \end{aligned}

**(1)**

Triangle DEF:

\tan(\theta)=\frac{8}{14.4222}

**(1)**

\begin{aligned} \theta&=\tan^{-1}(\frac{8}{14.4222})\\\\ \theta&=29.0^{\circ} \end{aligned}

**(1)**

2. ABCDEF is a triangular prism.

The cross-section of the prism is an isosceles triangle.

M is the midpoint of AC .

Calculate the length of EM .

**(4 marks)**

Show answer

\tan(70)=\frac{x}{4.5}

**(1)**

\begin{aligned} 4.5\tan(70)&=x\\\\ 12.364&=x \end{aligned}

**(1)**

\begin{aligned} EM^{2}&=16^{2}+12.364^{2}\\\\ EM^{2}&=408.868 \end{aligned}

**(1)**

\begin{aligned} EM&=\sqrt{408.868}\\\\ EM&=20.2\mathrm{cm} \end{aligned}

**(1)**

3. Find the size of the angle AFH .

**(6 marks)**

Show answer

\begin{aligned}
AF^{2}&=8^{2}+5^{2}\\\\
AF&=9.434
\end{aligned}

**(1)**

\begin{aligned} FH^{2}&=17^{2}+5^{2}\\\\ FH&=17.720 \end{aligned}

**(1)**

\begin{aligned} AH^{2}&=17^{2}+8^{2}\\\\ AH&=18.788 \end{aligned}

**(1)**

\cos(x)=\frac{9.434^{2}+17.720^{2}-18.788^{2}}{2\times9.434\times17.720}

**(1)**

\cos(x)=0.1096

**(1)**

\begin{aligned} x&=\cos^{-1}(0.1496)\\\\ x&=81.4^{\circ} \end{aligned}

**(1)**

You have now learned how to:

- Apply Pythagoras’ Theorem and trigonometric ratios to find angles and lengths in right-angled triangles (and, where possible, general triangles) in 2 (and 3) dimensional figures

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