Video: AQA GCSE Mathematics Higher Tier Pack 2 • Paper 3 • Question 23

Find the value of 𝑥.


Video Transcript

Find the value of 𝑥.

Whenever I get a question like this, what I first will do is see if there’s any extra information that I can add on before I try and find the value of 𝑥. So if we take a look at our diagram, we can see that there’s an angle that we can add on. And we can add that on because we know that the angles inside a triangle sum to 180 degrees.

So we’re gonna call the angle 𝑦. And what I can say is that 𝑦 is gonna be equal to 180 minus 21 minus 30 which is gonna be equal to 129 degrees which I’ve now marked onto our diagram. So now what we need to do is decide what method or what part of math we’re gonna use to help us to find the value of 𝑥.

Well, what I do is I ask myself a couple of questions. And the first one of these being “is there a right angle?” Well, no, there isn’t a right angle. So we know that we’re not going to use Pythagoras and we’re also not going to use the standard trig ratios.

The second question I ask myself is “is there a matching pair?” And what I mean by a matching pair is an angle and the side opposite it. So in this case, yes, there is because we can see that there is 30 degrees and 11 centimetres opposite. So therefore, we know that there is a matching pair. And the reason we ask this question is cause it lets us know whether we can use the sine rule or we have to go on and use the cosine rule.

And because there is a matching pair, it means that we can use the sine rule. And what the sin rule tells us is that if we have a triangle where we have the sides 𝑎, 𝑏, and 𝑐 and we have the angles which are opposite these: capital 𝐴, capital 𝐵, and capital 𝐶, then 𝑎 over sin 𝐴 is equal to 𝑏 over sin 𝐵 is equal to 𝑐 over sin 𝐶.

We can also have this the other way around. So we can swap it so we have sin 𝐴 over 𝑎 is equal to sin 𝐵 over 𝑏 is equal to sin 𝐶 over 𝑐. And we would use it that way around if we were looking to find an angle. But in this question, we’re looking to find a side.

So therefore, using the sine rule, we can say that 𝑥 over sin 129 and that’s because 129 is the angle opposite our side 𝑥 is equal to 11 over sin 30. So now, what we need to do is rearrange and solve to find 𝑥. So therefore, what we do is we multiply each side of the equation by sin 129. And when we do that, we get 𝑥 is equal to 11 sin 129 over sin 30.

And we got that because as we said we multiplied the right-hand side by sin 129. So what we can do is we’re gonna put that onto the numerator. So we’ve got 11 sin 129 as our numerator. So this is gonna give us 𝑥 is equal to 11 sin 129 over 0.5. And that’s because sin 30 is equal to 0.5, so one of the values that we should know and we should memorize. So this gives us that 𝑥 is equal to 17.09721 et cetera.

So now, what we need to do is have a look at is it sensible to round this. Well, the answer is yes because we want to round it with a sensible degree of accuracy. So I’m going to round it is to two decimal places. And when we do that, we get 𝑥 is equal to 17.10 and that’s to — as we said — two decimal places.

And the way I did that, well, I take a look at the second decimal place which was a nine. Then, I’ve underlined next to it the digit to the right which is a seven. And this is the deciding number or deciding digit because this is gonna tell us whether to round our nine up or keep it the same. Well, because it’s five or above, it means we round our nine up. So our nine rounds to a 10. So we get 17.10 and that’s to two decimal places.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.