### Video Transcript

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
the 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 forty-two 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 is that 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 forty-two, I’m gonna
replace the opposite with 𝑥 cause that’s its letter in this question, and I’m gonna replace
the hypotenuse with ten. So I have sine of forty-two is equal to 𝑥 over ten. Now this is an equation that
I can solve in order to work out the value of 𝑥. So as there is a ten in the denominator, I’m
going to multiply both sides of this equation by ten. Now I’ve swapped the two sides of the equation around because I prefer 𝑥 to be
the subject on the left-hand side. And what it gives me is that 𝑥 is equal to ten sin
forty-two. That’s how you write ten multiplied by sin forty-two; there’s no need for the
multiplication sign.

Now this is something that I can evaluate on my calculator. So I can type ten
sin forty-two into my calculator, making sure it’s in degree mode. And when I do that, I get an
answer of six point six nine one. 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 the 𝑥 is equal to six point seven. 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 twenty-two 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 thirty-one degrees, I’m gonna replace the opposite
with twenty-two, and I’m gonna replace the hypotenuse with 𝑦. So I have sine of thirty-one is equal to twenty-two 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 because that will bring it out
of the denominator. So when I do that, it will give me 𝑦 multiplied by sine thirty-one, which remember
I can just write as 𝑦 sin thirty-one is equal to twenty-two.

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 thirty-one. 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 twenty-two over sin thirty-one. And at
this point, I’m gonna use my calculator to evaluate that. And I have that 𝑦 is equal to forty-two point seven one five two eight. 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 forty-two point seven two. 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 meters long leans against a vertical wall, making an angle of fifteen 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 floor. Let’s put the information on. The ladder is six meters
long, so this length here is six meters. And the angle between the ladder and the wall is
fifteen 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 call 𝑥 meters.

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 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 sine of fifteen 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 six.

And if we do that, we have 𝑥 is equal to six sin fifteen. Now I’m gonna
evaluate that using my calculator. And this gives me a value of one point five five two nine one. 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 meters will also be to the nearest centimetre. And therefore I have the distance between the foot of the ladder and the wall is
one point five five 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 sixty-one degrees with the ground. And the kite is forty-eight 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 sixty-one degrees with the ground, so this is
sixty-one degrees. And the kite is forty-eight 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 sixty-one degrees. So as before, we can see that we have the opposite forty-eight meters and the
hypotenuse 𝑦 involved in this ratio. So again, it’s gonna be the sine ratio that we need to use
for this question.

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 sixty-one is equal to forty-eight 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 sine sixty-one is equal to forty-eight. Now to work
out the value of 𝑦, we need to divide both sides of the equation by sin sixty-one. And this gives us 𝑦 is equal to forty-eight over sin sixty-one; this I’m gonna
evaluate using my calculator. And I have 𝑦 is equal to fifty-four point eight eight zero nine nine. 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 fifty-four point eight eight meters 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 forty-eight. So that gives us a little bit of confidence that we’ve used the
sine ratio the right way up at least.

So to summarize 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 visualize the situation.