# Lesson Video: Trigonometry in Three Dimensions Mathematics

In this video, we will learn how to solve trigonometry problems in three dimensions.

17:04

### Video Transcript

In this video, we’re going to review the trigonometric ratios of sine, cosine, and tangent. And then we’ll see how we can use these to find unknown sides or angles in three dimensions.

The first important thing to note is that our trig ratios apply in right triangles. And if we’re working with an angle of 𝜃, then the labeling of the sides of our right triangle is very important. The longest side on a right triangle is called the hypotenuse. And in some diagrams where it’s harder to see, then we could always find it by the fact that it’s opposite the right angle. The labeling of the other two sides depends on the position of the angle. The side that’s opposite the angle in question is the opposite side. The side that’s next to or adjacent to the angle is called the adjacent side.

We can define the trigonometric ratios for an angle 𝜃 in the following way. Sine, which is abbreviated so that it looks like “sin” but still pronounced sine, is sin of 𝜃 equals the opposite over the hypotenuse. cosine, abbreviated to cos, gives us cos of 𝜃 equals adjacent over hypotenuse. Finally, tangent, abbreviated to tan 𝜃, is calculated by the opposite over the adjacent sides.

It’s common to see SOH CAH TOA written as a way to help us remember these ratios. So when it comes to applying the trig ratios in 3D, we still follow the same principles, but we need to find right triangles. If, for example, we had this cuboid, then this triangle would be a right triangle in two dimensions, so would this triangle. But would this triangle be? No, it wouldn’t be. This triangle is not in two dimensions. It covers three dimensions. Very often when we’re solving trigonometry problems in three dimensions, we find that we need to use two or more different 2D triangles to help us find a missing length or angle.

We may also find that we need to use the Pythagorean theorem too. We recall that this tells us that the square of the hypotenuse is equal to the sum of the squares on the other two sides. We’ll now look at some trigonometry problems in three dimensions. And we remember that we’re looking for right triangles in two dimensions.

Using the trigonometric ratios, find tan of 𝜃.

In this question, we have this rectangular prism or cuboid, and we can see that 𝜃 is the angle between the line 𝐹𝐷 and 𝐹𝐺. In order to use the trigonometric ratios, we need to have a right triangle. We could create a right triangle with the triangle 𝐹𝐷𝐺 and the right angle here at vertex 𝐺. Note that this triangle, 𝐹𝐷𝐺, would be a two-dimensional triangle as it sits on the face of our cuboid.

It’s often very helpful to draw our triangles separately so that we can visualize the problem. We’d have vertex 𝐷 at the top and 𝐹 and 𝐺 at the base of this triangle. 𝐷𝐺 is given on the diagram as four centimeters, and 𝐹𝐺 is three centimeters. The angle 𝜃 is the angle here at 𝐷𝐹𝐺. When we’re using the trigonometric ratios, we often use the phrase SOH CAH TOA to help us remember them. The TOA part helps us to remember tan of the angle, so we’d have tan of 𝜃 equals the opposite over the adjacent sides.

The longest side or hypotenuse isn’t needed for the tan ratio. The side that’s opposite the angle 𝜃 is the length 𝐷𝐺. The side that’s adjacent to our angle 𝜃 is the length 𝐹𝐺. So we begin by saying that tan 𝜃 equals O over A. That’s opposite over adjacent. And we fill in the lengths that we’re given. The opposite side is four centimeters, and the adjacent side is three centimeters. Our answer then for tan 𝜃 is the fraction four-thirds.

Note that we weren’t actually asked to calculate the size of the angle 𝜃 but just to find tan 𝜃. If we did want to find the value of 𝜃, we’d need to use the inverse tan function on our calculator.

In the next question, we’ll see our first example where we need to use two different two-dimensional triangles to help us find a missing angle.

A cuboid is shown in the figure. Calculate the angle between 𝐵𝐻 and 𝐻𝐹. Give your answer to two decimal places.

We’re asked to consider the length 𝐵𝐻 here. That would be the space diagonal of the cuboid. The diagonal here would cross through three dimensions. The other length here is 𝐻𝐹. We’re asked to calculate the angle between 𝐵𝐻 and 𝐻𝐹. Let’s call this angle 𝜃. So how might we go about calculating this angle? Well, we can see that we have a triangle 𝐵𝐻𝐹. And we could also say that this will be a right triangle. Since we have a cuboid, we know that the length 𝐵𝐹 will meet 𝐻𝐹 at 90 degrees.

We might be familiar with two types of mathematics that we can apply in right triangles. We have the Pythagorean theorem, and we have trigonometry. As we have an unknown angle here, then we know that at some point, we’ll need to apply trigonometry here. So let’s take a closer look at this triangle 𝐵𝐻𝐹. We’re given that 𝐵𝐹 is 3.5 centimeters and the angle that we need to find out is this one at 𝐵𝐻𝐹, which we’ve called 𝜃.

We don’t quite have enough information in order to be able to use trigonometry. We’d need to know the length of the hypotenuse 𝐵𝐻 or the length of the other side 𝐻𝐹. So before we can attempt to find 𝜃, we’ll need to find one of these other two lengths. Let’s have a look at this length 𝐻𝐹. We can form another triangle 𝐻𝐺𝐹 on the base of this cuboid. We also know that this triangle too would be a right triangle.

So here’s our other triangle drawn. We can see that 𝐹𝐺 on the diagram is three centimeters and 𝐻𝐺 is given as four centimeters. Remember that it’s this side, 𝐻𝐹, that we wish to find out. So on our pink diagram, we have this length here. So let’s define this with the letter 𝑥. When we have a right triangle, two sides that we know, and one side that we wish to find out, we can use the Pythagorean theorem. This tells us that the square on the hypotenuse is equal to the sum of the squares on the other two sides.

The first step then to finding our unknown 𝐹𝐻 and then to find the angle is to apply the Pythagorean theorem. The longest side here, our hypotenuse, is 𝑥. And the other two sides are three and four. And it doesn’t matter which way round we write these. So we have 𝑥 squared equals three squared plus four squared. As three squared is nine and four squared is 16, we’ll have that 𝑥 squared is equal to 25. To find the value of 𝑥, we take the square root of both sides of our equation. So 𝑥 is the square root of 25. And that’s equal to five centimeters.

We have now found the value that 𝐻𝐹 is five centimeters. So we can then go ahead and find our angle 𝜃. In order to work out which of our trigonometric ratios we need to use out of sine, cosine, or tangent, we need to look carefully at the sides that we have and wish to find out. We have the side opposite our angle 𝜃. And we have the side that’s adjacent to it. Notice that for our hypotenuse, the longest side, we don’t have the value of it, and we don’t want to calculate it.

Using the phrase SOH CAH TOA, we can see that we have the O for opposite and the A for adjacent. So that means that we need our tan ratio. We can write this trig ratio as tan 𝜃 equals opposite over adjacent. We can then plug in the values that we have for the opposite and adjacent and solve for the angle. This gives us tan of 𝜃 equals 3.5 over five. In order to find 𝜃 then, we need to use the inverse tan. As we’re asked to give our answer to two decimal places, we can reasonably use a calculator here.

This inverse function of tan on our calculator is usually found above the tan button. Pressing shift or second function will allow us to type in this calculation. We can obtain the value that 𝜃 equals 34.99202 and so on. Rounding our answer to two decimal places means that we check our third decimal digit to see if it’s five or more. And so our answer is that the angle between 𝐵𝐻 and 𝐻𝐹 is 34.99 degrees to two decimal places.

Let’s look at another question.

The figure shows a right triangular prism. Find the angle between 𝐴𝐹 and 𝐴𝐶, giving your answer to two decimal places.

The first thing we can do in this question is identify our two lengths 𝐴𝐹 and 𝐴𝐶. The length 𝐴𝐹 will cut across the rectangular face here. The length 𝐴𝐶 will be the diagonal of the base of this right triangular prism. The angle between them will be the angle 𝐹𝐴𝐶 created here. And we can define this as the angle 𝜃. We can create a triangle 𝐴𝐹𝐶 in order to help us calculate our unknown angle 𝜃.

Let’s take a closer look at this triangle 𝐴𝐹𝐶. We know that the length 𝐹𝐶 is four centimeters and our angle here is 𝜃 at 𝐹𝐴𝐶. In order to apply trigonometry in this triangle, we need to be sure if we have a right triangle. We’re told in the question that this is a right triangular prism, which means that we have a right angle here at angle 𝐸𝐵𝐴 and at 𝐹𝐶𝐷. And since this plane on the bottom, 𝐵𝐴𝐷𝐶, meets our other plane 𝐸𝐵𝐶𝐹 at right angles, then we know that we have a right angle here at 𝐹𝐶𝐴.

We note, however, that as we look at our triangle 𝐴𝐹𝐶, we don’t quite have enough information. We’re going to need to find the length of one of these other two sides. If, for example, we look at 𝐴𝐶, we can see on our diagram that this length 𝐴𝐷, which is eight centimeters, is different to the length of 𝐴𝐶. We’ll need to create another right triangle in two dimensions to help us find the length of this side 𝐴𝐶.

We can, in fact, create this triangle 𝐴𝐶𝐷, which will have a right angle at angle 𝐴𝐷𝐶. We can draw out triangle 𝐴𝐶𝐷 on the bottom of our right triangular prism. We can see that 𝐶𝐷 is three centimeters and 𝐴𝐷 is eight centimeters. Don’t worry if your diagrams aren’t perfectly accurate; they don’t have to be to scale. They’re just there to help us visualize the problem. Remember why we’re doing this. We’re trying to find the length 𝐴𝐶, which is common to both triangles. We can define this as anything, but let’s call it the letter 𝑥.

When we find this length 𝑥 on our first triangle, we can fill in the information into our second triangle. As we have a right triangle and two known sides and one unknown side, we can apply the Pythagorean theorem, which tells us that the square of the hypotenuse is equal to the sum of the squares on the other two sides. So we take our Pythagorean theorem, often written as 𝑐 squared equals 𝑎 squared plus 𝑏 squared. The hypotenuse, 𝑐, is our length 𝑥. So we’ll have 𝑥 squared equals three squared plus eight squared. That’s the length of our two other sides. And it doesn’t matter which way round we write those.

We can evaluate three squared is nine, eight squared is 64, and adding those gives us 𝑥 squared equals 73. To find 𝑥, we take the square root of both sides of our equation. So we have 𝑥 equals the square root of 73. It’s very tempting at this point to pick up our calculator and find a decimal answer for the square root of 73. But as we haven’t finished with this value, we’re going to keep it in this square root form.

Now that we have found 𝑥, that means we’ve found our length of 𝐴𝐶. And so we can use this to find our angle 𝜃. As we’re interested in the angle here, that means we’re not going to use the Pythagorean theorem again, but we’ll need to use some trigonometry. In order to work out which of the sine, cosine, or tangent ratios we need, we’ll need to look at the sides that we have.

The length 𝐹𝐶 is opposite our angle 𝜃. 𝐴𝐶 is adjacent to the angle. And the hypotenuse is always the longest side. Now, we’re not given the hypotenuse, and we’re not interested in calculating it, so we can remove it from this problem. Using SOH CAH TOA can be useful to help us figure out which ratio we want. We have the opposite and the adjacent sides, so that means that we’re going to use the tan or tangent ratio. tan of 𝜃 is given by the opposite over the adjacent sides. We now fill in the values that we have.

The opposite length is four centimeters, and the adjacent length is given by root 73. So tan of 𝜃 is equal to four over root 73. In order to find 𝜃 by itself, we need the inverse operation to tan. And that’s finding the inverse tan. This function, written as tan with a superscript negative one, can usually be found on our calculator above the tan button. Using our calculator to evaluate this will give us 𝜃 equals 25.0873 and so on. And the units here will be degrees as, of course, this is an angle, not a length.

We’re asked to round our answer to two decimal places. So that means we check our third decimal digit to see if it’s five or more. And as it is, then our answer is given as 25.09 degrees. And so the angle between 𝐴𝐹 and 𝐴𝐶 is 25.09 degrees.

Before we finish with this question, let’s just review what we could have done instead. When we started this question, we had this large triangle 𝐴𝐹𝐶 which cut through our triangular prism. We were told that 𝐹𝐶 was four centimeters. And we worked out this length of 𝐴𝐶. But could we have done it by working out the length of 𝐴𝐹 instead?

If we look at our triangular prism, in order to work out the length of 𝐴𝐹, we’d need another triangle. We do have a right triangle here. And this length of 𝐸𝐹 will be eight centimeters. However, if we were trying to work out this length of 𝐴𝐹, we’d also need to work out this length of 𝐴𝐸. In order to find 𝐴𝐸, we’d need to create yet another right triangle. We would know that 𝐴𝐵 is four centimeters and 𝐴𝐵 is also three centimeters. So we would eventually get the correct answer for 𝐴𝐹 and, therefore, our angle 𝜃. It’s just that that second method would involve three triangles instead of the two triangles that we used.

Let’s now summarize what we’ve learned in this video. Firstly, we reviewed our trigonometric ratios. sin of 𝜃 is equal to the opposite over hypotenuse sides. cos of 𝜃 equals the adjacent over the hypotenuse. And the tan of 𝜃 equals the opposite over adjacent sides. All of these are valid for an angle 𝜃 in a right triangle. These ratios can be easily recalled by using the phrase SOH CAH TOA.

When using the trigonometric ratios in three dimensions, we need to be using two-dimensional right triangles to help us. And we often need to use more than one of these two-dimensional triangles. A handy tip is that if we are using more than one triangle, we should keep our answers to our calculation in the square root form until we get to our final answer where we can round the decimal value if required.

And finally, as we saw in a few questions, when we’re solving trigonometry in three dimensions, we might also require the Pythagorean theorem as well. In the same way as for our trigonometric ratios in 3D, the same applies for the Pythagorean theorem in 3D. That is, we need to be using it in two-dimensional right triangles only.