### Video Transcript

In this video, we’re going to look at the general principle behind the law of cosines to make sure that we’re comfortable applying it in situations where the letters that we’re given in a question don’t match up with the letters as they appear in the standard formula for the law of cosines. We’re going to look at the structure of the law of cosines and see how we can identify which values within a question or diagram should be substituted into which parts of the formula.

So, this is the law of cosines as it’s usually introduced or perhaps written in a text book or a formula sheet. You often have a triangle in which the three vertices are labelled as 𝐴, 𝐵, and 𝐶 using capital letters. And the three sides are labelled as 𝑎, 𝑏, and 𝑐 using lowercase letters. It’s always the case that side 𝑎 is opposite angle 𝐴, side 𝑏 is opposite angle 𝐵, and side 𝑐 is opposite angle 𝐶. The law of cosines is then written in this form here. 𝑎 squared is equal to 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴.

And this law of cosines enables you to calculate the length of side 𝑎 if you know side 𝑏, side 𝑐, and angle 𝐴. Now, that’s brilliant if the triangle that you’re given is labelled in exactly this way here, and if you were given exactly the right pieces of information. But what if actually I wanted to calculate side 𝑏? Or what if the triangle wasn’t labelled 𝐴𝐵𝐶 at all, but actually was labelled as 𝐸, 𝐹, and 𝐺? Or perhaps you’re not given any labels at all, perhaps you’re given some lengths and some angles in the triangle.

What we need to look at is the general structure behind the law of cosines so that we could apply it in any of these different settings. So, returning then to this standard form of the law of cosines, let’s look at the information we’re given. We’re looking to calculate the length of a side. And in the standard form, it’s the length of side 𝑎. The information that we would need in order to do this using the law of cosines is the length of side 𝑏, the length of side 𝑐, and the size of angle 𝐴.

Now, that is a very specific set of information within this triangle. It is two sides and then the included angle. That is, the angle between them. When you’re looking to use the law of cosines to calculate the length of a side, it’s always this exact same set of information that you need. If you look at the law of cosines, you see that actually it’s symmetrical in 𝑏 and 𝑐. They’re both squared and added together, and then they both appear in the second part where we multiply by two and multiply by cosine of 𝐴 and subtract this. Therefore, it doesn’t matter which side is 𝑏 and which is 𝑐 because you’re doing exactly the same thing with each of them.

So, in fact we can forget about the labels 𝑎, 𝑏, and 𝑐 entirely and just look at where sides and angles appear in this law of cosines. For example, in this question here, let’s suppose we’re given this set of information. The two sides are four centimetres and 12 centimetres. And the included angle is 40 degrees. And we’re looking to calculate the length of the third side here, which is referred to as 𝑥.

Now, there are no 𝑎s, 𝑏s, and 𝑐s in this diagram, but let’s look at how we would write down the law of cosines for this question. We’re looking to calculate 𝑥, so we’ll begin with 𝑥 squared. Now, the law of cosines then has 𝑏 squared plus 𝑐 squared. Remember, 𝑏 and 𝑐 just represent the other two sides. So, for this question that will be four squared and 12 squared. Then, we need to look at this second term. It’s minus two multiplied by 𝑏𝑐, first of all. And, again, 𝑏 and 𝑐 just represent the other two sides. So, that will be 4 and 12.

Finally, then, we need cos of angle 𝐴. Now, again, there are no 𝐴s in this triangle. But that angle is just the angle opposite the side we’re looking to calculate, the angle included between these two sides. So, it’s 40 degrees. So, just by considering the information that I’ve been given and the structure of the law of cosines, I’ve written down what it would be for this particular triangle without the need for any 𝑎s, 𝑏s, or 𝑐s anywhere at all.

So, just for completeness, we’ll work through the remainder of this question. Our 4 squared and 12 squared together make 160. And two times four times 12 is 96. So, I have 𝑥 squared is equal to 160 minus 96 cos 40. This tells me then that 𝑥 squared is equal to 86.45973346. And if I then square root both sides, I have that 𝑥 is equal to 9.298376. Now, I’ll round this perhaps to one decimal place. And therefore, we have that the final side of this triangle is 9.3 centimetres.

So, the 𝑎s, 𝑏s, and 𝑐s are useful for having a format that we can write the law of cosines down in, but it’s thinking about the structure behind it that enables us to actually apply it in general.

Calculate side 𝑏 to the nearest hundredth.

Now, within this question, we’re going to be using the law of cosines. So, I’ll just recall it using its standard definition if you were to look it up in a textbook. And it’s this definition here. Now, this is actually gonna be quite confusing in this example, because we’re not asked to calculate side 𝑎, we’re asked to calculate side 𝑏. Now, you may think, oh, that’s okay, I’ll just rearrange this formula so that I get 𝑏 squared equals.

And if you do that, you would have 𝑏 squared is equal to 𝑎 squared minus 𝑐 squared plus two 𝑏𝑐 cos 𝐴. Now, here’s the problem. This would require you to know sides 𝑎 and 𝑐, which we do. They’re the sides opposite angles 𝐴 and 𝐶. So, they’re nine centimetres and five centimetres. But the other piece of information we would need is angle 𝐴. And looking at the diagram, you can see that we haven’t been given angle 𝐴. We’ve been given angle 𝐵.

So, just rearranging the law of cosines from this standard form doesn’t work because we haven’t been given the right set of information in order to apply it. Instead, what we need to do is write our own version of the law of cosines, where we cycle the letters around so that we’re looking to calculate side 𝑏. So, here is the information we have, two sides and the included angle, which is exactly the setup that we need in order to use the law of cosines.

So, what I’m gonna do is I’m gonna write out the law of cosines again, but cycling the letters through. I want to calculate 𝑏, so I’m gonna begin with 𝑏 squared. Then, the law of cosines tells me that I square each of the other two sides. So, in this instance, that’s going to be 𝑎 squared plus 𝑐 squared. It then tells me that I do negative two multiplied by 𝑏 and 𝑐, which are the other two sides. Well, in this case, that’s gonna be negative two multiplied by 𝑎 and 𝑐. Finally, then, I do cos of the included angle, so in this case that’s going to be cos of angle 𝐵.

So, this isn’t a rearrangement of the law of cosines because you can see it includes cos of angle 𝐵 instead of cos of angle 𝐴. Instead, it’s a rewriting of the law of cosines using the letters in a different order. And now, I have a version that I can use in order to answer this question. So, I can substitute the relevant information. I have then that 𝑏 squared is equal to nine squared plus five squared, first of all, minus two times nine times five times cos of 120. And now, I can just work through the stages here.

So, I have 𝑏 squared is equal to 106 minus 90 cos 120. This tells me that 𝑏 squared is equal to 151 exactly. That’s because cos of 120 is an exact value. It’s just negative a half. If I then take the square root, I have that 𝑏 is equal to 12.288205. And the question has asked me for this value to the nearest hundredth, so I’ll round my answer. And we have then that 𝑏 is equal to 12.29 centimetres.

So, when answering a question like this, you can look up the law of cosines in the form it’s usually given in or perhaps you’ve committed that form to memory. But if the side you’re looking for isn’t side 𝑎, then you need to think about how you can swap the letters around in order to make it relevant for the side you’re looking to calculate. Remember, you do this by just considering the structure of the law of cosines and the fact that it includes the two other sides of the triangle and the included angle, which is the angle opposite the side you’re looking to calculate.

Calculate the measure of angle 𝑄 to the nearest degree.

So, in this question, we’ve been given the lengths of all three sides of a triangle and we’re asked to calculate one of the angles. Now, this is exactly the setup required in order to use the law of cosines. So, if you were to look up the standard law of cosines, it would probably look something like this. 𝑎 squared is equal to 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴.

If you’re looking to calculate the size of an angle, as we are in this question, then that can be rearranged in just a couple of steps to give you this formula here. Cos of 𝐴 is equal to 𝑏 squared plus 𝑐 squared minus 𝑎 squared all over two 𝑏𝑐. However, there are no 𝑎s, 𝑏s, and 𝑐s in this question. This question involves 𝑄s, 𝑅, and 𝑆s, so we need to think about how we can apply this version of the law of cosines within this question.

So, we need to look at the structure within this law of cosines. There are two sides that are always treated identically, 𝑏 and 𝑐 are both squared and added in the numerator, whereas 𝑎 squared but then subtracted. 𝑏 and 𝑐 also both appear in the denominator. So, these two sides that are treated identically are the two sides that enclose the angle we’re looking for. So, in this case, that would be the side of eight centimetres and 10 centimetres.

The third side, which is treated differently in the law of cosines, it only appears in the numerator and is subtracted rather than added. This is the side that is opposite the angle that we’re looking to calculate. So, I don’t need 𝑎s, 𝑏s, and 𝑐s. I can just write down the law of cosines by thinking about where the sides are in relation to this angle.

So, we have cos of 𝑄 is equal to, well, the two sides next to it, first of all, are eight and 10, so we’ll have 8 squared plus ten squared. Then, I need to subtract the square of the opposite side, so that will be minus 15 squared. Then, I need to divide this by two multiplied by the two sides that enclose this angle. So, this is two multiplied by eight multiplied by 10.

If I then evaluate all this, I have the cause of angle 𝑄 is equal to negative 61 over 160. In order to work out angle 𝑄, I need to use cosine inverse. So, 𝑄 is equal to cosine inverse of negative 61 over 160. This gives me a value of 112.411132. And I’ve been asked to give this to the nearest degree, so my final answer then is that angle 𝑄 is 112 degrees to the nearest degree. So, again, there were no 𝑎s, 𝑏s, or 𝑐s within our question. We just looked at the structure of the law of cosines in order to work out which value should be substituted where.

Now, suppose instead I was asked to calculate angle 𝑆 this time. So, again, I would look at the structure of the law of cosines. So, first of all, I need to square two sides and add them together. So, that needs to be the two sides that are adjacent to angle 𝑆, which is going to be the eight and the 15. Then, I need to square and subtract the remaining side, so I’ll have minus 10 squared.

In the denominator, I need two multiplied by the two sides that are adjacent to this angle 𝑆. So, again, that will be the eight and 15. And there we have an equation that we could go on to solve if we wanted to in order to calculate the measure of this angle 𝑆. So, just by looking at the structure of the law of cosines, we can apply it to any question like this. Just remember that whenever two sides are treated identically within the formula, those two sides have to be the sides that are enclosing the angle that we’re looking to calculate.

In summary then, it’s useful to learn the definition of the law of cosines in its standard form involving 𝑎s, 𝑏s, and 𝑐s. But in order to apply effectively to a range of problems, you just need to look at the structure of the law and you need to be comfortable with what the letters represent in terms of adjacent sides, opposite sides, and included angles.