### Video Transcript

In this video, we’re going to see
how to use the cosine ratio to calculate the length of either the adjacent or the
hypotenuse in a right-angled triangle. So first of all, a reminder of what
the cosine ratio is. I have on the screen here a diagram
of a right-angled triangle. And I’ve chosen one of the other
two angles to label as 𝜃. I’ve then labeled the three sides
of this triangle with their names in relation to this angle 𝜃. So we have the hypotenuse. We have the opposite. And we have the adjacent. The cosine ratio or cos, as it’s
often abbreviated to, is the ratio between the adjacent and the hypotenuse of this
triangle. So its definition is that cos of
the angle 𝜃 is equal to the adjacent divided by the hypotenuse. And if you’re using SOHCAHTOA to
help you with trigonometry, then this is the CAH part of that. We’re now going to see how we can
use this ratio in order to calculate the length of missing sides in right-angled
triangles.

So here is our first problem.

We’re given a diagram of a
right-angled triangle. We can see we’ve been given the
length of one side and one angle. And we’re asked to calculate 𝑥,
which is the length of another side. So for me, the first step for any
problem involving trigonometry is to label the three sides of the triangle with
their labels in relation to this angle of 28 degrees. So using the first out of each of
these words, you can see there I have the hypotenuse, the opposite, and the
adjacent. Now looking at those three sides,
you can see I’ve got the six centimeters, which is the hypotenuse. And I want to calculate 𝑥, which
is the adjacent. So I have the 𝐴 and 𝐻 involved in
this ratio. And if you think about SOHCAHTOA,
then 𝐴 and 𝐻 appear together in the cos ratio, which is how we know that it’s the
cos ratio that we’re going to need for this particular question.

So I recall the definition of the
cos ratio. And it’s this that it’s the
adjacent divided by the hypotenuse. What I’m going to do then is that
I’m going to write down this ratio. But I’m gonna fill in the
information I know. So I’m gonna replace 𝜃, which is
the angle, with 28. I’m gonna replace the adjacent with
𝑥 because that’s its letter in the diagram. And I’m gonna replace the
hypotenuse with six. So this gives me cos of 28 is equal
to 𝑥 over six. Now this is an equation that I’m
looking to solve in order to work out the value of 𝑥. So the first step to do in this,
well there’s a six in the denominator. So I need to multiply both sides of
this equation by six. When I do that, I get that 𝑥 is
equal to six cos 28. This just means six multiplied by
cos 28. But we don’t need the
multiplication sign there.

Now, I’m gonna evaluate this using
my calculator, making sure it’s in degree mode because the angle in the question was
specified in degrees. So evaluating this, I get that 𝑥
is equal to 5.29768. The question asked me to find this
value to the nearest tenth. So I need to round my answer. And therefore I have that 𝑥 is
equal to 5.3. So a quick summary, in this
question then, what we did was we identified that it was the cosine ratio we needed
by labeling the sides of the triangle. We then wrote this ratio out, but
using the information in this particular question, and then solved the resulting
equation in order to work out that missing value 𝑥.

Right, here is the second example
we’re going to look at.

It says, calculate the length of
the hypotenuse of triangle 𝐴𝐵𝐶, to two decimal places. And we can see that we’re given a
diagram of a right-angled triangle where we’ve got the length of one side and the
size of one of the other angles.

So as before, my first step is to
label the three sides with the labels in relation to this angle of 65 degrees. So we have their labels here. And again we can see that it’s the
cosine ratio we’re going to need because I’ve been given the length of the adjacent
side. It’s 5.1 millimeters. And it’s the hypotenuse I want to
calculate. So 𝐴 and H, remember, is the
cosine ratio. So as before, what I’m going to do
is I’m gonna write out the cosine ratio. But I’m gonna fill in the
information I know. So 𝜃 is gonna become 65. The adjacent is gonna become
5.1. And the hypotenuse I will refer to
as 𝐵𝐶 as that’s the notation for this particular triangle. So I have cos of 65 is equal to 5.1
over 𝐵𝐶. Now again, this is an equation that
I’d like to solve in order to work out the value of 𝐵𝐶. It’s slightly more complex because,
this time, the unknown length is in the denominator of a fraction.

So my first step in solving this
equation is going to be to multiply both sides of the equation by 𝐵𝐶. So doing this gives me 𝐵𝐶
multiplied by cos 65 is equal to 5.1. Now, I want to work out 𝐵𝐶. So the next step is to divide both
sides of the equation by cos of 65. Cos of 65 is just a number. So it’s perfectly okay for me to do
that. This gives me then that 𝐵𝐶 is
equal to 5.1 over cos of 65. This stage here is when I’m gonna
use my calculator to evaluate this. And this tells me that 𝐵𝐶 is
equal to 12.067 and so on. Now the question asked me to give
my answer to two decimal places.

So finally, I need to round this
answer. And I need to include the units
which are going to be millimeters. So my final answer then is that the
length of the hypotenuse 𝐵𝐶 is 12.07 millimeters.

So the process for this question
was very similar to the previous one. We labeled the three sides,
identified the need for the cosine ratio, we wrote it down using the information in
the question, and then solved the resulting equation. The equation was just slightly more
complex this time because the length we were looking to find was in the denominator
of a fraction. So it involved two steps in order
to solve it rather than just one.

Right, our next example looks a
little bit different.

We’re given a circle. And we’re told that 𝐴𝐵 is the
diameter of this circle. We’re then asked to calculate the
radius of this circle to the nearest hundredth.

Now, we can see that, within this
circle, we have a triangle. And we’ve got one of the angles of
the triangle which is 40 degrees and one side length which is four centimeters. The question is, well is this a
right-angled triangle? Because if it is, then we can apply
trigonometry to this problem. Now in order to see that this is in
fact a right-angled triangle, you need to make a connection with some work in
another area of mathematics. And that is the area of circle
theorems. One of the key circle theorems
tells us that if you have a scenario like this, then the angle that is at the
circumference of the circle is half the angle that’s at the center. Now here the angle at the center is
a straight line. So it’s 180 degrees. And therefore the angle at the
circumference, the angle 𝐴𝐶𝐵, is half of that. So it’s 90 degrees, which means
that this angle here is in fact a right angle.

If you’ve perhaps studied topics in
a different order and you haven’t met circle theorems yet, then you’d expect that
right angle to be marked on the diagram for you. But ultimately once you’ve met
circle theorems, you would be able to deduce that for yourself. So what this tells us is that we do
have a right-angled triangle. And therefore, we are able to use
trigonometry in order to answer this problem. Therefore, my first step is gonna
be to label the three sides of this triangle with their names, the opposite, the
adjacent, and the hypotenuse, in relation to the angle of 40 degrees. And so what we can see is we know
the length of the adjacent. If we want to calculate the radius
of this circle, then we need to calculate the length of 𝐴𝐵, which is the diameter,
and then halve it. So again we need to know the
hypotenuse. So A and 𝐻 are involved, which
tells us it’s the cosine ratio that we need.

So as in the previous questions,
we’re going to write the cosine ratio right again but filling in the information in
this particular question. So we have that cos of the angle
which is 40 is equal to four, the adjacent of 𝐴𝐵, the hypotenuse. Now, what we’re looking to do is
solve this equation in order to work out the value of 𝐴𝐵. So this is very similar to what we
saw in the previous example. 𝐴𝐵 is in the denominator of this
fraction. So first of all we need to multiply
both sides of the equation by 𝐴𝐵. And doing so gives me 𝐴𝐵
multiplied by cos 40 is equal to four. The next step then is I need to
divide both sides of this equation by cos of 40. So I have 𝐴𝐵 is equal to four
over cos 40. Now, I’m not going to evaluate this
quite yet because the question asked me to calculate the radius of the circle. And 𝐴𝐵 is the diameter. So I need to halve this in order to
answer the question.

So 𝑟, which I’m using to represent
the radius of the circle, is four over cos 40 multiplied by a half. And I’ll use my calculator now to
evaluate this, which tells me that 𝑟 is equal to 2.6108 and so on. Now the question asked me to find
this radius to the nearest hundredth. So I’ll round my answer. So we have then that the radius of
the circle is equal to 2.61 centimeters. So in this question, the actual
process is very similar to what we’ve already seen previously. The only difference was that, this
time, we were combining our knowledge of trigonometry with our knowledge of circle
theorems as well and using circle theorems in order to identify that the angle
𝐴𝐶𝐵 was 90 degrees. And therefore, we could apply
trigonometry to the problem.

Right, the final question we’re
going to look at is a noncalculator question.

So we have a diagram of a
right-angled triangle. And we’re told to use the
information in the table in order to calculate 𝐵𝐶. And in the table, we can see that
we’ve been given the values of the sine, cosine, and tangent ratios for this angle
of 35 degrees.

So when doing trigonometry, you
almost always have access to a calculator because you need your calculator in order
to tell you the values of these sine, cosine, and tangent ratios. There are some particular angles,
so 30 degrees, 45 degrees, and 60 degrees, for which these values are quite simple
ratios in terms of surds. And so you can do trigonometry
without a calculator. But the majority of the time we do
need it. However, this question has got
round the need for a calculator by giving us the values of sin, cos, and tan within
the question itself. And so we’ll see how to use those
in a minute. First step then as it’s probably
involving trigonometry is to label the three sides in relation to this angle of 35
degrees. So we have the three labels
here. And again, as with all the examples
in this video, we can see that it’s the cosine ratio we’re going to need because we
have the length of the hypotenuse 𝐻. And we’re looking to calculate 𝐵𝐶
which is the adjacent. So 𝐴 and 𝐻 tells us it’s the
cosine ratio we need.

So as in the previous questions, we
are going to write out the cosine ratio but filling in the information for this
question, so replacing the angle with 35, replacing the adjacent with 𝐵𝐶, and
replacing the hypotenuse with 10. We have then that cos of 35 is
equal to 𝐵𝐶 over 10. Now, I need to solve this equation
for 𝐵𝐶. So I’m going to multiply both sides
of the equation by 10. And I have that 𝐵𝐶 is equal to 10
cos 35. Now, this is where the information
in the question comes in because remember we haven’t got a calculator. But by looking at the table, we can
see that we have the value of cos 35 here. It’s 0.819. So I can use that value at this
stage in the question. So this tells me that 𝐵𝐶 is equal
to 10 multiplied by 0.819. And that’s a straightforward
multiplication to do without a calculator. So we have that the length of 𝐵𝐶
is 8.19 centimeters.

Now that’s not its exact value
because cos of 35 isn’t exactly 0.819; it’s a longer decimal. But we’ve only been given it to
three decimal places. So this value for 𝐵𝐶, this is
exact to two decimal places or to the nearest hundredth. So this question then just gives
you one example of how you might use trigonometry when you don’t have access to a
calculator.

In summary then, we’ve recalled the
definition of the cosine ratio as the adjacent divided by the hypotenuse. And we’ve seen how to apply this
ratio to four different questions where we’re looking to calculate either the
adjacent or the hypotenuse in a right-angled triangle.