### 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, the 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 twenty-eight 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 twenty-eight, 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 twenty-eight 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 twenty-eight. This just means six
multiplied by cos twenty-eight, 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 five point two nine seven six eight.
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 five point three.

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 sixty-five 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 five point
one 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 sixty-five, the adjacent is
gonna become five point one, and the hypotenuse I will refer to as 𝐵𝐶 as that’s the notation for
this particular triangle.

So I have cos of sixty-five is equal to five point one 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 sixty-five is equal to five point
one. Now I want to work out 𝐵𝐶, so the next step is to divide both sides of the equation by cos
of sixty-five. Cos of sixty-five is just a number, so it’s perfectly okay for me to do that.

This gives me then that 𝐵𝐶 is equal to five point one over cos of sixty-five.
This stage here is when I’m gonna use my calculator to evaluate this. And this tells me that 𝐵𝐶 is equal to twelve point zero six seven 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 twelve point zero
seven 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 forty 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 a hundred and eighty
degrees, and therefore the angle at the circumference, the angle ACB, is half of that, so it’s
ninety 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 would 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 forty 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 forty 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 forty is equal to four. The next step
then is I need to divide both sides of this equation by cos of forty. So I have 𝐴𝐵 is equal to four over cos forty. 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
forty multiplied by a half, and I’ll use my calculator now to evaluate this, which tells me that 𝑟 is equal to two point six one zero eight 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 two point six one
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 without our knowledge of circle theorems as well and using circle theorems in order
to identify that the angle ACB was ninety degrees and therefore we could apply trigonometry to
the problem.

Right the final question we’re going to look at is a non-calculator
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 thirty-five 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 thirty degrees, forty-five degrees, and sixty 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 thirty-five 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 thirty-five, replacing
the adjacent with 𝐵𝐶, and replacing the hypotenuse with ten.

We have then that cos of thirty-five is equal to 𝐵𝐶 over ten. Now I need to solve
this equation for 𝐵𝐶, so I’m going to multiply both sides of the equation by ten. And I have that 𝐵𝐶 is equal to ten cos thirty-five. 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 thirty-five here. It’s nought point
eight one nine. So I can use that value at this stage in the question.

So this tells me that 𝐵𝐶 is equal to ten multiplied by zero point eight one nine,
and that’s a straightforward multiplication to do without a calculator. So we have that the length of 𝐵𝐶 is eight point one nine centimeters. Now that’s
not its exact value because cos of thirty-five isn’t exactly nought point eight one nine; 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.