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

Sine rule Cosine ruleTangent of a circle

Pythagoras’ Theorem

Plans and elevations

Exact trigonometric values

Surds

Decimal placesMidpoints

This 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.

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

- Volume of 3D shapes
- Surface area of 3D shapes

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