In this video, we’re going to see how to use the sine ratio to calculate the length of
either the opposite or the hypotenuse in a right-angled triangle. First of all, a quick
reminder of the sine ratio. So I have here a right-angled triangle. And I’ve labelled one
of the other angles as 𝜃. I’ve then labelled the three sides of the triangle with their
names — the opposite, the adjacent, and the hypotenuse — in relation to this angle 𝜃.
The sine ratio remember is the ratio of the opposite and the hypotenuse. So it’s defined
as sine of 𝜃 is equal to opposite divided by hypotenuse. And if you’re familiar with
SOHCAHTOA to help with trigonometry, then this is of course the SOH part of that.
So let’s see how to apply this to our first question. We have here a right-angled
triangle, where we’re given the length of one of the sides and the size of an angle. And
we’re asked to find the value of 𝑥 to the nearest tenth. And from the diagram, we can
see that 𝑥 represents a length. So the first step for me for any problem involving
trigonometry is to label the three sides of the triangle with their names. So in
relation to this 42 degrees, I’m gonna label the hypotenuse of the triangle which is
this side here, the opposite which is this side here, and the adjacent. So what I can
see is that I know I’ve got one angle. I’ve been given the length of the hypotenuse. And
I want to work out the length of 𝑥, which is the opposite. So it’s O and H that I’m
interested in. And if you recall SOHCAHTOA, O and H appear together in the sine ratio,
which is how I know it’s sine that I’m going to be using. Now this video is of course
specifically about using the sine ratio. But in a general context, that’s how you would
identify whether it’s sine, cosine, or tangent that you’re interested in.
So let’s just write down the definition of the sine ratio. And remember it’s the sine of
𝜃, the angle, is equal to the opposite divided by the hypotenuse. So what I’m going to do
is I’m gonna write this ratio out again. But I’m gonna fill in the pieces of information
I know. So I’m going to replace 𝜃 with 42. I’m gonna replace the opposite with 𝑥 cause
that’s its letter in this question. And I’m gonna replace the hypotenuse with 10. So I
have sin of 42 is equal to 𝑥 over 10. Now this is an equation that I can solve in
order to work out the value of 𝑥. So as there is a 10 in the denominator, I’m going to
multiply both sides of this equation by 10. Now I’ve swapped the two sides of the
equation round because I prefer 𝑥 to be the subject on the left-hand side. And what it
gives me is that 𝑥 is equal to 10 sin 42. That’s how you write 10 multiplied by sin 42.
There’s no need for the multiplication sign.
Now this is something that I can evaluate on my calculator. So I can type 10 sin 42 into
my calculator, making sure it’s in degree mode. And when I do that, I get an answer of
6.691. So this is my answer then for the value of 𝑥. The question asked for it to the
nearest tenth. So I need to round my answer. And therefore I have that 𝑥 is equal to
6.7. So in this question, we labelled the sides first of all. We identified that it was
the sine ratio that we needed. We recalled the definition of the sine ratio, wrote it
down using the information in the question, and then solved the resulting equation in
order to work out this value 𝑥.
Right, our second question, we’re given a right-angled triangle. And we’re asked to find
the value of 𝑦, this time to two decimal places. So as before, first step for problem
involving trigonometry is to label the three sides of this right-angled triangle. So I
have the opposite, the adjacent, and the hypotenuse. Now I can see that I know the
opposite this time. It’s 22 centimetres. And it’s the hypotenuse that I’m looking to
work out. So again, O and H are involved in this ratio. So we know it’s the sine ratio.
Let’s recall its definition. And remember it was that sine of 𝜃 is equal to the
opposite divided by the hypotenuse. So as before, I’m gonna write this ratio out again,
but filling in the information I know. I’m gonna replace 𝜃 with the angle 31 degrees.
I’m gonna replace the opposite with 22. And I’m gonna replace the hypotenuse with 𝑦. So
I have sin of 31 is equal to 22 over 𝑦. And this is the equation that I’m looking to
solve in order to work out the value of 𝑦.
Now this is a slightly more complicated equation than the one we had previously because,
this time, 𝑦 appears in the denominator of a fraction. So it’s gonna involve two steps.
The first step is to multiply both sides of the equation by 𝑦 cause that will bring
it out of the denominator. So when I do that, it will give me 𝑦 multiplied by sin 31,
which remember I can just write as 𝑦 sin 31, is equal to 22.
Now the next step to solving this equation is- well, I want to get 𝑦 on its own. So I
need to divide both sides of the equation by sin 31. That was just a number. So it’s
perfectly okay for me to do that. So when I do that, I get 𝑦 is equal to 22 over sin
31. And at this point, I’m gonna use my calculator to evaluate that. And I have that 𝑦
is equal to 42.71528. Now the question asked me to find this value to two decimal
places. So I’m gonna round my answer. And this gives me that 𝑦 is equal to 42.72. So we
followed a very similar process to the previous question. Because we’re looking to find
the hypotenuse though and that was in the denominator of a fraction, we just had a more
complicated equation to solve once we got to that point. But the initial stages of
setting up the equation were almost identical.
Okay, now we’re gonna look at a couple of worded problems. So the first problem says, a
ladder six metres long leans against a vertical wall, making an angle of 15 degrees with
the wall. How far is the foot of the ladder from the wall? So often with worded
problems, you’re given a description of the situation. But you’re not given a diagram.
And it’s always a good idea to start off by drawing your own diagram. So we’re gonna
have a diagram of a ladder, a wall, and the floor. Now we are assuming here that the
floor is horizontal. And therefore we have a right angle between the wall and the floor.
Let’s put the information on. The ladder is six metres long. So this length here is six
metres. And the angle between the ladder and the wall is 15 degrees. That’s this angle.
We’re asked how far is the foot of the ladder from the wall. So we’re interested in this
distance, which I’ll call 𝑥 metres.
So by drawing that diagram, we’ve now made this problem look very similar to the
previous questions that we’ve already looked at. So the first step, we’re gonna label
the three sides: the adjacent, the opposite, and the hypotenuse. And as before and
throughout this video, we can see that it’s the opposite and the hypotenuse that have
been involved in the ratio we want here, the 𝑥 and the six. So because it’s the
opposite and the hypotenuse, we’re gonna be using that sine ratio. And remember its
definition is that sine of the angle is equal to the opposite divided by the hypotenuse.
So writing it out using the information in this question, well we’ve then got sin of 15
is equal to 𝑥 over six. So we want to solve this equation to work out the value of 𝑥.
There is a six in the denominator. So we want to multiply both sides of the equation by
And if we do that, we have 𝑥 is equal to six sin 15. Now I’m gonna evaluate that using
my calculator. And this gives me a value of 1.55291. Now I haven’t been given a specific
level of accuracy for rounding this. So I’ll round it to the nearest hundredth, which
because 𝑥 is measured in metres will also be to the nearest centimetre. And therefore I
have the distance between the foot of the ladder and the wall is 1.55 metres. So once we
had the sine ratio written down, this problem was no more complicated than the first
problem that we looked at. The extra step was just making sure we’ve read the
information in the question carefully and drew the appropriate diagram to help.
Okay, our final problem is another worded question. It says, Joe is flying a kite. The
string makes an angle of 61 degrees with the ground. And the kite is 48 metres
vertically above the ground. We are asked to calculate the length of the kite’s string.
So again, this question isn’t accompanied by a diagram, which means we need to draw our
own of the ground and then the kite flying above it.
So here we have that diagram of Joe and the kite. And let’s fill in the relevant
information. The string makes an angle of 61 degrees with the ground. So this is 61
degrees. And the kite is 48 metres vertically above the ground. We’re looking to work
out the length of the kite’s string. So that’s this length here. And I’ll label it as 𝑦
metres. So we have a right-angled triangle. We have one known length and one known
angle. And we’re looking to calculate another length, which means we can apply
trigonometry to this problem. As before first step, label those three sides: the
opposite, the adjacent, and the hypotenuse in relation to the angle of 61 degrees. So as
before, we can see that we have the opposite, 48 metres, and the hypotenuse, 𝑦, involved in
this ratio. So again, it’s gonna be the sine ratio that we need to use for this
So we have our definition of the sine ratio. And we’re going to write it out using the
values in this question. We have then that sine of 61 is equal to 48 over 𝑦. And now
this problem is similar to the second problem that we looked at, where the unknown value
𝑦 is in the denominator of this fraction. So if you recall the first step to solving
this equation was to multiply both sides by 𝑦. And when we do that, we have 𝑦 sin
61 is equal to 48. Now to work out the value of 𝑦, we need to divide both sides of
the equation by sin 61. And this gives us 𝑦 is equal to 48 over sin 61. This I’m gonna
evaluate using my calculator. And I have 𝑦 is equal to 54.88099. Now I haven’t been
asked for a specific degree of rounding. So again, the nearest hundredth would be
equivalent to the nearest centimetre here. So that’s how I round my answer. So I have
54.88 metres for the length of the kite’s string.
Now, it is always worth just doing a quick sanity check on your answer to make sure you
haven’t used the ratio upside down, for example. We were looking to calculate 𝑦 which
was the hypotenuse of the triangle. And the hypotenuse remember is the longest side. So
this measurement that we get should certainly be bigger than the other sides in the
triangle. And it is bigger than that 48. So that gives us a little bit of confidence
that we’ve used the sine ratio the right way up at least.
So to summarise then, for each of these questions involving the sine ratio, recall the
definition of the sine ratio first of all. Write it down using the information in the
particular question you’re answering. And then solve the resulting equation. Sometimes
that will be straightforward. Sometimes it will involve two steps of working out. If
you’re given a worded problem, read it through very carefully and draw your own diagram
first to help you visualise the situation.