### Video Transcript

In this video, we are going to introduce the law of sines and then see how to apply it to some mixed problems. So this is what the law of sines is all about. It’s really useful because it enables us to do trigonometry and calculate lengths and angles in triangles that are not right angled.

So I have here a diagram of a triangle which has no right angle in it, and I’ve labelled it in a particular way. I’ve labelled the three vertices of the triangle as 𝐴, 𝐵, 𝐶 in capitals and then I’ve labelled the side opposite those with the same letter but in lowercase, so side 𝑎 is opposite angle 𝐴 and so on.

What the law of sines tells ask is that the ratio between the length of a side and the sine of its opposite angle is constant within a particular triangle. So if I take side 𝑎 and then I divide it by sine of the opposite angle 𝐴, I get the same result as if I take side 𝑏 and then divided by sine of angle 𝐵, and I also get the same result if I take side 𝑐 and then divide it by sine of the opposite angle 𝐶.

Now this is one way of specifying the relationship and this format is particularly useful if we’re looking to calculate the length of a side. But you can also specify it using the reciprocals, so I can invert each of those fractions. So it can also be written in this format here, where each of those fractions is just the other way up. This format is particularly useful if you’re being asked to calculate the size of an angle.

So when do we use the law of sines? Well as I’ve already said, we use it in non-right -angled triangles, but more specifically we use it when the information we’re given on what we want to work out is made up of opposite pairs. So for example, I might know the lengths of sides a and 𝑏 and I might know angle 𝐴 and want to calculate angle 𝐵. And so because those are opposite pairs, this would be the opportunity to use the law of sines.

This question says find all the possible values for the other lengths and angles in triangle 𝐴𝐵𝐶, and we’re asked to give lengths to the nearest centimetre and angles to the nearest degree. So we’ll discuss what is meant by all the possible values a little bit later, but let’s start off by recalling the definition of the law of sines that we’re going to need within this question.

And remember it was this ratio here, and of course we could have the reciprocal of that so we could have it written the other way up. So I have three things that I need to calculate here: two missing angles and then one missing length. The reason I know that I can use the sine ratio is because I have opposite pairs.

So I have that length fourteen centimetres and the angle fifty-two degrees and then I have this length eight point one centimetres, which means I have enough information to calculate angle 𝐵 first of all. So what I’m gonna do is I’m going to write down the law of sines just using angle 𝐴, side 𝑎, angle 𝐵, and side 𝑏.

Now as I’m calculating an angle first of all, I’m gonna use the reciprocal form of that relationship. So using all the known information, I have that sine of angle 𝐵 divided by eight point one is equal to sine of fifty-two divided by fourteen. And now I have an equation that I can solve in order to work out angle 𝐵. The first step is I’m gonna multiply both sides of this equation by eight point one.

So I have that sine of 𝐵 is equal to eight point one multiplied by sin fifty-two over fourteen. Now I’m gonna use the inverse sine function in order to calculate angle 𝐵. And then using my calculator to evaluate this, it tells me that angle 𝐵 is equal to twenty-seven point one two four. I’m gonna round that then to twenty-seven degrees.

So if I have angle 𝐵 is twenty-seven degrees and I already know angle 𝐴 is fifty-two degrees, I can work out angle 𝐶 straight away not using the law of sines but just using the angle sum in a triangle. So I have angle 𝐶 is one hundred and eighty subtract fifty-two subtract twenty-seven, and therefore it’s one hundred and one degrees.

So now I have all the angles in the triangle; I just need to work out the final side. I’m going to apply the law of sines again then. And as I’m calculating the length of a side this time, I’m going to use the first version where the sides are in the numerator. Now I only need to use one of the other pairs, so either pair A or pair B. I’m going to choose to use pair A.

So I have that 𝑐 over sin one hundred and one is equal to fourteen over sin fifty-two. Now be careful to distinguish between lowercase and uppercase letters here. Remember lower case letters represent sides, so that is a lowercase letter 𝑐. To solve this equation for side 𝑐 then, I need to multiply both sides of the equation by sin one hundred and one.

And I have then that 𝑐 is equal to fourteen multiplied by sine of a hundred and one divided by sine of fifty-two, which is seventeen point four three. Now I’m asked to round this to the nearest centimetre, so then I have seventeen centimetres for side 𝑐. So those three calculated values together then, I have that side 𝑐 is seventeen centimetres, angle 𝐵 is twenty-seven degrees, and angle 𝐶 is a hundred and one degrees.

Now let’s come back to that part of the question where it asked us for all the possible values for the other lengths and angles. And what we need to consider is when we worked out angle 𝐵, we saw that it was equal to twenty-seven degrees. What we need to consider is that there is in fact another possibility for angle 𝐵 which uses the fact that sine of an angle is equal to sine of a hundred and eighty minus that angle.

That’s just one of the properties of the sine ratio. So what this tells us is that although angle 𝐵 could be twenty-seven degrees, it could also be one hundred and eighty minus twenty-seven, which would mean that 𝐵 could be a hundred and fifty-three degrees. However, if we look at the information that we already had at this stage, which was that angle 𝐴 was fifty-two degrees, that doesn’t work. Because if we add 𝐴 and 𝐵 together, that would take the angle sum above a hundred and eighty degrees, and we know that in a triangle a hundred and eighty degrees is the sum of the angles.

This tells us that in fact there isn’t another possibility for angle 𝐵. Because if it were a hundred and fifty-three degrees, it wouldn’t be possible to incorporate that in a triangle with the information we already know. This is an important check though. And if it had been the case that this angle could be included in a triangle with the fifty-two degrees, then we would have another set of possibilities and we would need to go through the process of calculating angle 𝐶 and side 𝑐 again using the second value of 𝐵.

So in this question, we’ve applied the law of sines twice, once to calculate a side and once to calculate an angle. And then we’ve used the fact that angles in a triangle sum to a hundred and eighty degrees in order to find the third angle in the triangle.

This question tells us that 𝐴𝐵𝐶 is a triangle, angle 𝐴 is fifty-five degrees, 𝐵𝐶 is thirteen centimetres, and 𝐴𝐶 is twenty-eight centimetres. We’re asked if the triangle exists, find all the possible values for the other lengths and angles and then we’re told how to round our answers. Now it’s interesting that the question says if the triangle exists. So what we’ll do is we’ll assume the triangle does exist, and we’ll go through all the working out. And if it works perfectly fine, then the triangle does exist. And if we come up against a problem, then we’ll see that the triangle doesn’t exist.

So I’m gonna assume it exists first of all, and I’m gonna draw a sketch of this triangle. So here is my diagram, with all the information that I was given put onto it.

Now we’re asked to calculate lengths and angles, and I can recognise that I need to use the law of sines here because I’ve got an opposite pair there of fifty-five degrees and thirteen centimetres. So let’s recall the law of sines. And it’s this here: that the ratio between the sine of an angle and the length of the opposite side is constant throughout the triangle. Remember, lowercase 𝑎, 𝑏, and 𝑐 there are representing the sides opposite angles 𝐴, 𝐵, and 𝐶.

So I just include them on the diagram there. Now I’ve chosen to use the law of sines in this format, where the angles are in the numerator because I’m gonna try and calculate angle 𝐵 first of all. And it just requires less rearranging if I start off with the angle being in the numerator. I could use the other version with the reciprocals of each these fraction, but it would just require a slightly more complicated rearrangement.

So what I’m going to do then is I’m gonna write out this law of sines using the pair that I know so that’s pair A and using the pair I want to calculate which is pair B. And I have then that sine of angle 𝐵 divided by twenty-eight is equal to sine of fifty-five divided by thirteen.

So this gives me an equation that I’m looking to solve then to work out angle 𝐵. And the first step is to multiply both sides of this equation by twenty-eight. In doing so, I get that sine of 𝐵 is equal to twenty-eight sin fifty-five over thirteen. Now in order to work out angle 𝐵, I need to use the inverse sine function.

And I have then that 𝐵 is equal to the inverse sine of this ratio: twenty-eight sin fifty-five over thirteen. Now if you try to type that into your calculator, you’ll see that you get some kind of error and you can’t actually work out angle 𝐵. Let’s go back to the stage before to see why this is.

If I actually evaluate the fraction at this stage here, you’ll see that it’s equal to one point seven six four. So we have sine of 𝐵 is equal to one point seven six four, and that’s why we get an error. If you recall the value of the sine of any angle is always between negative one and one. And in the case of a positive angle as we’d have in a triangle, it’s always between zero and one. And therefore it’s not possible for sine 𝐵 to be equal to this value of one point seven six, which exceeds one.

This tells us then that we can’t calculate angle 𝐵 and therefore our assumption that this triangle exists must be false. Our answer to the question then is that we can’t calculate the lengths of any of these sides or the sizes of any of the angles because the triangle does not exist.

Let’s look at a worded problem. It tells us that James wants to calculate the height of a tall building. He looks at the building from the same horizontal plane, and the angle of elevation to the top is forty degrees. James then moves thirty metres further back, and the angle of elevation is now twenty-five degrees. We’re asked to calculate the height of this building to the nearest tenth.

So we aren’t given a diagram, and it’s always a sensible idea to draw our own. So we start off with a tall building. James is standing a certain distance away from it. We don’t know how far. And the angle of elevation to the top is forty degrees. James now moves thirty metres further back, and the angle of elevation is now twenty-five degrees, so we add this part onto the diagram.

So add some letters onto the diagram, and it’s 𝐵𝐷 that we’re looking to calculate, the height of the building. Now 𝐵𝐷 is a right-angled triangle, so in theory we can use normal trigonometry — sine, cosine, and tangent ratios — in this triangle. But we only know one angle at the moment, so we need some other information, preferably a length, in order to work out the length of 𝐵𝐷.

We also have a non-right-angled triangle, triangle 𝐴𝐵𝐶 in which we have a bit more information. We know an angle and a side. You’ll also notice that side 𝐴𝐵 is common to both of these triangles. So perhaps we can use the non-right-angled triangle to work out 𝐴𝐵 and then use trigonometry in triangle 𝐴𝐵𝐷 in order to find the height of this building.

So let’s look at the non-right-angled triangle, triangle 𝐴𝐵𝐶, first. And actually we can work out all three of the angles in this triangle because that angle of forty degrees is sitting on a straight line with the other angle, and therefore this other angle must be one hundred and forty degrees using the fact that angles on a straight line sum to a hundred and eighty.

So this angle here, angle 𝐴, must be a hundred and forty degrees. We can also work out angle 𝐵 because we know the angle sum in a triangle is a hundred and eighty degrees, so angle 𝐵 must be fifteen. So looking at this non-right-angled triangle, we know all three angles and we know one side, which means we can apply the law of sines in order to work out the length of either of the other two sides.

So I’ll give them the letters lowercase 𝑎, lowercase 𝑏, and lowercase 𝑐 corresponding to the angles they’re opposite. And let’s recall the law of sines. Here it is. I’m using it in this format with the lengths and the numerator because it’s a length that I’m looking to calculate.

So I’m looking to calculate the length of 𝐴𝐵, which is referred to here as side 𝑐, so I’m gonna use side 𝑐 and angle 𝐶. And I’m also gonna use side 𝑏 and angle 𝐵 because they’re the two that I know What I have then is that 𝑐 over sin twenty-five is equal to thirty over sin fifteen using the opposite pairs. I can solve this equation then in order to work out 𝑐; I need to multiply both sides by sin twenty-five.

This tells me that 𝑐 is equal to thirty sin twenty-five over sin fifteen. And evaluating that, it tells me it’s forty-eight point nine eight six one. Now I’m going to keep that value on my calculator so that I have it there exactly to use later in the calculation.

Now if I turn my attention to the right-angled triangle, triangle 𝐴𝐵𝐷, I have the size of an angle, forty degrees, I have length 𝑐 or 𝐴𝐵, which is forty-eight, and I want to calculate length of that side 𝐵𝐷. So I can use standard trigonometry. I’m going to begin by labelling the three sides of that triangle with their labels in relation to that angle of forty.

So I have the opposite, the adjacent, and the hypotenuse. Now I know the hypotenuse and I want to calculate the opposite. So that tells me it’s the sine ratio that I’m using, just the standard sine ratio and a right-angled triangle, not the law of sines. The sine ratio, remember, is the opposite divided by the hypotenuse.

Now you can actually use the law of sines in a right-angled triangle but it’s unnecessarily complicated it involves using that ninety-degree angle and a sine of ninety is just one. It’s more straightforward to just use the regular sine ratio. So I’m gonna write this ratio out for this triangle. So I’ll have sine of forty is equal to the opposite 𝐵𝐷 over forty-eight point nine eight.

I want to solve this equation to find the value 𝐵𝐷, so I need to multiply both sides by that value of forty-eight point nine eight six. That’s why I kept that value on my calculator because now I can just press multiply by sin forty in order to get an exact answer.

So evaluating that gives me thirty-one point four eight. And then rounding it to the nearest tenth as requested gives me an answer of thirty-one point five metres for the height of this building. So in this question, drawing an appropriate diagram first of all is important. We then use the law of sines in a non-right-angled triangle and then use the normal sine ratio in a right-angled triangle in order to answer the question.

In summary then, the law of sines allows us to calculate angles and sides in non-right-angled triangles. It tells us that the ratio between a side and the sine of the opposite angle is constant throughout the triangle, and we can use it in either of these two formats depending on whether we’re looking to calculate the length of a side or the size of an angle.