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
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
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
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
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
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
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
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.