### 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.