Lesson video

Hi, I'm Mr. Bond, and in this lesson, we're going to learn how to apply trigonometry within three-dimensional shapes.

We already know how to apply trigonometry within two-dimensional shapes.

Now we're going to apply trigonometry in three-dimensional shapes.

Take a look at the cuboid on the screen.

We can measure angles between two line segments in 3-D shapes.

In this example, we'd be finding the angle between AG and EG.

Now we're going to work through an example.

We want to calculate the size of the angle BHF within this cuboid.

Let's start by marking the angle on the diagram.

The angle BHF will be here.

So the angle BHF lies between the line segments BH and FH.

And you might have also noticed that this angle lies within the triangle BFH, like this.

What do we know about this triangle? Well, we know that the line segment BF is three centimetres.

So we want to know what are the other two line segments.

The one that I've chosen is FH.

We can use Pythagoras' Theorem to find the length of FH because we know the length of FE and EH.

So FH will be equal to the square root of seven squared and five squared.

This is equal to the square root of 74.

So now, looking at our triangle, given that we want the angle theta, we can say that the tangent of theta is equal to the opposite over the adjacent side.

So this is tan theta is equal to three over root 74.

So to find the value of the angle theta, we can do the inverse tangent of three over root 74, and this is 19.

2 degrees to three significant figures.

Here's a question for you to try.

Pause the video to have a go and resume the video once you finished.

Here's the solution to question number one.

This question is very similar to the example that we've just been through, but it's worth noting how the question was asked.

You'll see that, in part a, it asked us to find the length BH, and this is very often how a GCSE question is structured.

They use a preliminary question to help you with the final answer to the next part of the question.

So we have to apply Pythagoras' Theorem within a three-dimensional shape, within the cuboid, in order to find the length of BH.

Following this, we have to find the size of the angle BHF in a very similar way as I said to the previous example.

So far, we've looked at finding the angle between two line segments.

Have a look at this diagram.

What's this showing the angle between? This diagram shows that we can also measure angles between line segments and faces in 3-D shapes, and we're going to have a look at an example of this next.

In this example, we have a three-dimensional shape with four triangular faces.

Take a look at the diagram.

What do you notice? Well, if we have a look at the vertex O, we can see that there are three right angles at this vertex.

So we know that three of the faces are right angle triangles.

Let's see what else we're given.

We're told OB is eight centimetres, AB is equal to 11 centimetres, and AC is equal to six centimetres.

ABC is an isosceles triangle, so that's the non-right angle triangle facing us.

We want to calculate the angle between the line segment OB and the face ABC.

What would that look like on the diagram? Well, we can see the face ABC, That's the face that's facing us, and we can imagine the line segment OB is almost at the back of the 3-D shape.

So the angle that we're looking for is here.

Let's label the other information on the diagram.

We were told that AB is equal to 11 centimetres, AC is equal to six centimetres, and OB is equal to eight centimetres.

To find the angle between the face and the line segment OB means that we might need to consider the point directly between A and C.

I'm going to call this M.

So to find the angle MBO, I first need to find the length of the line segment BM.

We know that ABC is an isosceles triangle.

The line segment BM splits this triangle into two right angle triangles.

So I'm going to draw a separate diagram showing triangle ABM.

We can work out the length of BM using Pythagoras, just like we have done in previous examples.

So BM is equal to the square root of 11 squared subtract three squared 'cause we're looking for the length of a shorter side.

So this is equal to the square root of 112.

Now that we know the length of the line segment BM, we can consider another triangle, another right angle triangle, BOM, like this.

This contains the angle that we're looking for.

We know two of the side lengths of this triangle.

We know that line segment OB is eight centimetres long and we've just worked out that the length of the line segment BM is root 112.

So we know the adjacent and the hypotenuse.

So this means that we can use cosine.

Cosine theta is equal to eight over root 112.

So we can use the inverse cosine to find the angle.

Using your calculator, we can calculate that the angle is equal to 40.

9 degrees to one decimal place.

There's another way that we can measure angles in 3-D shapes.

We could also measure the angles between two faces in 3-D shapes.

It will be useful to look at the side elevation.

You can see this in the triangle that I've drawn on the right.

So if we could find the height of this triangle and some other information, we could calculate the size of the angle using trigonometry.

I won't go into any more detail than this for now, as you're about to try a question like this for yourselves.

Here's a question for you to try.

Pause the video to have a go and resume the video once you finished.

Here's the answer to question number two.

This one was quite a complex one.

You'll see that I've drawn an additional triangle.

The triangle that I've drawn shows the angle between the two faces, so that's the angle that we're looking for.

You'll see that I've used a point O.

I've called the point, O, the point that's directly beneath point E, which it says in the question is directly above the centre of the square base.

So we can think of O as the centre of the square base.

So the first thing that we needed to do was we need to consider the triangular face ABE, and we need to find the length of AE on the additional triangle that I've drawn there.

To do that, we'd simply use Pythagoras' Theorem because, of course, the face ABE is an isosceles triangle, which we can split into two right angle triangles just like we did in the previous example.

So, we then know that the distance AO is 2.

9 millimetres.

This is half of AB, so half of the length of one of the square's sides.

Finally, we just need to apply some trigonometry.

So we can see that we have the hypotenuse and the adjacent side.

So we're going to use cosine theta is equal to adjacent over hypotenuse, and then work out the angle between the two faces as 71.

3 degrees.

That's all for this lesson.

Thanks for watching.