## What you need to know

Before stepping into the third dimension, it’s important to be comfortable with the following two rules of trigonometry (specifically, the ones for right-angled triangles) in 2-dimensions.

Firstly, Pythagoras’ Theorem.

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

where c is the longest side of a right-angled triangle (the hypotenuse), and a, b are the other two sides.

Secondly, the rules of \sin, \cos, and \tan, as outlined in our handy acronym: SOHCAHTOA. These can be used to make the handy equation trianlges you see below.

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

Where the letters represent **O**pposite, **A**djacent, and **H**ypotenuse.

Now that’s out of the way, we can get onto the fun stuff. If you feel you might be a little rusty on these facts though, it’s worth having another look at them here (trigonometry revision).

Fortunately for you, there’s nothing new to learn for 3-dimensional trigonometry. The 2D rules are all we’ll be using, we just have to figure out how to apply them. Typically, it will be useful to break our 3D picture up into smaller 2D ones in to help us figure it out. Let’s have a look.

**Example: **Shape ABCDEFG is a cuboid. Find the length of side FC, marked in red, to 3sf.

Firstly, the shape is a cuboid, which means it is full of right-angles. Every corner is a right-angle, in fact, which means our rules mentioned above will almost certainly come in useful.

So, we need to find FC. Let’s first we consider the triangle FCH on its own.

Because we’re in a cuboid, this is a right-angled triangle. We know that side CH is 5cm, but otherwise we have no information about this triangle. We need an extra side-length or angle for us to be able to find the length of FC. If we find the length of FH, for example, then we can use Pythagoras to find FC. It is very useful then to notice that FH is also a hypotenuse of a different triangle…

To find side-length FH, we’re going to now consider the triangle FEH. Fortunately, we do know a little more about this triangle.

As before, this is also a right-angled triangle, but the difference is we know one of the side-lengths as well as an angle. This is enough information for us to use SOHCAHTOA to find the length of FH.

FE is adjacent to the angle, and we’re looking for the hypotenuse FH, so we will use \cos.

\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:

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Firstly, note that the length of FH is expressed as 10.013… cm because when you’re putting it into your calculator, you should use the exact answer (by utilising the ANS button) and not a rounded-off decimal.

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

1) ABCDE is a square-based pyramid. The apex of the pyramid, E, is directly over the centre of the base. Calculate the perpendicular height of the pyramid. Leave your answer in surd form.

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 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}.

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2) ABCDEFGH is a cuboid. FE = 2.3mm, CH = 1.9mm, FC = 3.4mm. Calculate the size of angle EFH to 1 decimal place.

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.)}

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### Worksheets and Exam Questions

### Videos

#### 3D Pythagoras and Trigonometry Q1

GCSE MATHS#### 3D Pythagoras and Trigonometry Q2

GCSE MATHS### Other worksheets

## 3D Pythagoras and Trigonometry Revision and Worksheets

## 3D Pythagoras and Trigonometry Teaching Resources

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#### Trigonometry Questions, Revision and Worksheets

#### Areas of Shapes Worksheets, Questions and Revision

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