### 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 us 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 divide it by sin of angle 𝐵. And I also get the same result if I
take side 𝑐 and then divide it by sin 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 14
centimetres and the angle 52 degrees and then I have this length 8.1
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 sin of angle 𝐵 divided by 8.1 is equal to sin of 52
divided by 14. 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 8.1. So, I have that sin of 𝐵 is
equal to 8.1 multiplied by sin 52 over 14. Now, I’m gonna use the inverse
sine function in order to calculate angle 𝐵. And then, using my calculator
to evaluate this tells me that angle 𝐵 is equal to 27.124. I’m gonna round that then to 27
degrees. So if I have angle 𝐵 is 27
degrees and I already know angle 𝐴 is 52 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 180
subtract 52 subtract 27, and therefore it’s 101 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 101
is equal to 14 over sin 52. 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 101. And I have then that 𝑐 is
equal to 14 multiplied by sin of 101 divided by sin of 52, which is 17.43. Now, I’m asked to round this to
the nearest centimetre, so then I have 17 centimetres for side 𝑐. So, those three calculated
values together then, I have that side 𝑐 is 17 centimetres, angle 𝐵 is 27
degrees, and angle 𝐶 is 101 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 27 degrees. What we need to consider is that
there is in fact another possibility for angle 𝐵 which uses the fact that sin of an
angle is equal to sin of 180 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 27 degrees, it could also be 180 minus 27, which would
mean that 𝐵 could be 153 degrees. However, if we look at the
information that we already had at this stage, which was that angle 𝐴 was 52
degrees, that doesn’t work. Because if we add 𝐴 and 𝐵
together, that would take the angle sum above 180 degrees, and we know that, in a
triangle, 180 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 153 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 52 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 180 degrees in order to find the third angle in the
triangle.

This question tells us that
𝐴𝐵𝐶 is a triangle, angle 𝐴 is 55 degrees, 𝐵𝐶 is 13 centimetres, and 𝐴𝐶
is 28 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 55 degrees and 13 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 sin of
angle 𝐵 divided by 28 is equal to sin of 55 divided by 13. 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 28. In doing so, I get that sin of
𝐵 is equal to 28 sin 55 over 13.

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 sin of this ratio, 28 sin 55 over 13. 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 1.764. So, we have sin of 𝐵 is equal
to 1.764. 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 sin 𝐵 to be equal to this value of 1.76, 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 40
degrees. James then moves 30 metres
further back, and the angle of elevation is now 25 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 40 degrees. James now moves 30 metres
further back, and the angle of elevation is now 25 degrees, so we add this part
onto the diagram. So, we 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 40 degrees is
sitting on a straight line with the other angle. And therefore, this other angle
must be 140 degrees, using the fact that angles on a straight line sum to
180. So, this angle here, angle 𝐴,
must be 140 degrees. We can also work out angle 𝐵
because we know the angle sum in a triangle is 180 degrees, so angle 𝐵 must be
15.

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 25 is equal to 30 over sin 15, using the opposite pairs. I can solve this equation then
in order to work out 𝑐. I need to multiply both sides
by sin 25. This tells me that 𝑐 is equal
to 30 sin 25 over sin 15. And evaluating that, it tells
me it’s 48.9861. 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, 40
degrees, I have length 𝑐, or 𝐴𝐵, which is 48, 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
40. 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 in 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
90-degree angle and a sin of 90 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 sin of 40 is
equal to the opposite 𝐵𝐷 over 48.98. I want to solve this equation
to find the value 𝐵𝐷, so I need to multiply both sides by that value of
48.986.

That’s why I kept that value on
my calculator because now I can just press multiplied by sin 40 in order to get
an exact answer. So, evaluating that gives me
31.48. And then, rounding it to the
nearest tenth as requested, gives me an answer of 31.5 metres for the height of
this building. So, in this question, drawing
an appropriate diagram first of all was important. We then used the law of sines
in a non-right-angled triangle and then used 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.