3D Pythagoras And Trigonometry Worksheets | Questions and Revision

# 3D Pythagoras and Trigonometry Worksheets, Questions and Revision

Level 6 Level 7

## 3D Pythagoras and Trigonometry

$3D$ Pythagoras and Trigonometry is just adding a $3$rd dimension.

First lets recap the basics.

Pythagoras’ Theorem.

$a^2 + b^2 = c^2$,

Secondly, trigonometry $SOHCAHTOA$

$\sin(x) = \dfrac{O}{H}, \cos(x) = \dfrac{A}{H}, \tan(x) = \dfrac{O}{A}$

Make sure you are happy with these topics before continuing

## 3D Pythagoras

There $3D$ Pythagoras there is a new equation we can use, which just uses Pythagoras’ theorem twice.

In the diagram shown, find the length of $\textcolor{red}{d}$

Equation 1:

$\textcolor{limegreen}{a}^2 + \textcolor{orange}{b}^2 = e^2$

and we can see that $edc$ also forms a right angled triangle so we know,

Equation 2:

$e^2 + \textcolor{blue}{c}^2 =\textcolor{red}{d}^2$

So this means we can combine equation 1 and 2, to give our 3D Pythagoras equation.

$\textcolor{limegreen}{a}^2 + \textcolor{orange}{b}^2 +\textcolor{blue}{c}^2 =\textcolor{red}{d}^2$

## 3D Trigonometry

With $3D$ Trigonometry, there is no trick, you need to solve each section in steps which makes it a harder topic.

Example: Shape $ABCDEFG$ is a cuboid.

Find the length of side $FC$, marked in red, to $3$sf.

Firstly, the shape is a cuboid, which means every corner is a right-angle.

First what we need to do is find $FH$, this will give us the base of the right-angled triangle $FHC$, which will let us find $FC$

To find side-length $FH$, we need to use trigonometry

Adjacent $FE$

Hypotenuses $=x$

This means we will be using $CAH$

$\cos(26) = \dfrac{\text{Adjacent}}{\text{Hypotenuse}} = \dfrac{9}{FH}$

$FH \times \cos(26) = 9$

$FH = 9 \div \cos(26) = 10.013… \text{cm}$

Now we know $FH$, our first triangle, $FCH$, looks like this:

Now, we know two side-lengths of this triangle, we can use Pythagoras’ theorem to find the third, FC, which is the answer to the whole question.

$(FC)^2 = 5^2 + (10.013…)^2$

$$FC = \sqrt{5^{2} + (10.013…)^{2}} = 11.2 \text{cm (3s.f.)}$$

### Example Questions

If we draw a line from the apex at E down to the centre of the base, then that line represents the perpendicular height, since we know the apex is directly above the centre. Consider the triangle formed by this line, the line which goes from the centre to C, and the line EC.

We know the hypotenuse, but we need further information. Here, we observe the distance from the centre to C is half the distance from A to C. Given that we know the width of the square-based triangle, we can find the length of AC, halve it, and then use the result as a part of Pythagoras’ theorem to find the perpendicular height.

For finding AC, consider the triangle ABC.

Therefore, the distance from the centre of the base to C is $5\sqrt{2}$. Finally, we consider again the first triangle, which we now know has base of $5\sqrt{2}$cm, and calculate the perpendicular height.

$12^2 = (\text{HEIGHT})^2 + (5\sqrt{2})^2$

$(\text{HEIGHT})^2 = 12^2 - (5\sqrt{2})^2 = 144 - 50 = 94$

$\text{HEIGHT } = \sqrt{94}\text{cm}$.

We are looking for angle EFH. Consider the triangle EFH.

We know the length of one side of this triangle (right) thus we need more information before we can calculate the angle. We must use the other 2 lengths we’re given to find that extra bit of information about triangle EFH. Given that we know FC and CH, we should consider the triangle CHF, as it includes both the sides we know, and it shares a side (FH) with the triangle on the left.

With this triangle, CFH (left), we can find FH given the information we have. Using Pythagoras, we get:

$3.4^2 = 1.9^2 + (FH)^2$

$(FH)^2 = 3.4^2 - 1.9^2 = 7.95$

$FH = \sqrt{7.95}$

Now we have FH, we can look back at our original triangle and use SOHCAHTOA to find angle EFH. The two sides we know, FE and FH, are the adjacent and hypotenuse respectively, so we should use $\cos$.

$cos(EFH) = \dfrac{A}{H} = \dfrac{2.3}{\sqrt{7.95}}$

$\text{Angle } EFH = \cos^{-1}\left(\dfrac{2.3}{\sqrt{7.95}}\right) = 35.3\degree \text{ (1d.p.)}$

Here we use 3D Pythagoras to find that XY is

$XY^2 =9^2 + 6^2+6^2$

$XY =\sqrt{81+36+36}=\sqrt{153}=3\sqrt{17} \text{ cm}$

Here again, we use 3D Pythagoras to find that XY is

$XY^2 =9^2 + 6^2+12^2$

$XY =\sqrt{81+36+144}=\sqrt{261}=3\sqrt{29} \text{ cm}$

First of all, we can work out the length of DB by Pythagoras or by recognizing that the diagonal of a square is $\sqrt{2} \times \text{ side length}$, thus:

$DB=14\sqrt{2}$

Hence the length from D to the centre of the square, $O$, is half this value

$DO=7\sqrt{2}$

Now we have enough information to find the required angle,

$\tan(EDB) = \dfrac{O}{A} = \dfrac{11}{7\sqrt{2}}=$

$angle EDB = tan^{-1} \dfrac{11}{7\sqrt{2}}=48.0\degree$

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