Hi guys, this video is about deriving the sine rule or the rule of sines as it is
sometimes called. Right, so before we start looking at non-right-angled triangles, we’re just
gonna recap a little bit of right angle trigonometry with these two triangles that I’ve drawn here.
So if we look at the triangle on the left and we have a hypotenuse of six and we’re gonna try
and find this side 𝑥, which is actually that opposite angle. So if we just label those
triangles and this side 𝑥 is our opposite and the one opposite the right angle is called “a
hypotenuse.” So if you remember your trigonometric ratios, this is actually the ratio for
sine. So we can fill that in; sine of an angle Sin thirty is equal to 𝑥 over six. And to
solve that, we would need to multiply through by six and that would give us six times Sin
thirty, which equals 𝑥. Now if you worked that out on a calculator, you will find that 𝑥
equals three. So notice here the triangle on the left is the same height as the triangle on
the right; so they are both three and that was not accidental.
So let’s just look at the triangle on the right, so opposite the side forty and is our 𝑂 and opposite the right angle
again is our hypotenuse. So it’s still sine. And this time, I’m gonna work out the 𝑥. So if we
fill in into our formula, we have Sin forty this time is equal to opposite which is three
over the 𝑥 like so. So to get that 𝑥 on its own, we first need to multiply through by the 𝑥 and that
will be 𝑥 Sin forty equals three. Now this time we can’t work that out yet, so we have to get
rid of the Sin forty. So we’re gonna divide by the Sin forty, and that will leave us 𝑥
equals three divided by Sin forty. And if we put that out in a calculator, that comes out as
four point six seven to two decimal places.
Great, so we’re gonna use this method in order to
solve some non-right-angled triangles. So let’s look at that now. Right, so here we have a
triangle 𝐴𝐵𝐶, and we’ve been given the side 𝐴𝐵 which is ten, the angle 𝐵𝐴𝐶 which is thirty, and
the angle 𝐵𝐶𝐴 which is forty. Now if we just label this triangle, opposite the 𝐴 we’re
gonna call little 𝑎, opposite the 𝐶 we’re gonna call little 𝑐, and opposite the 𝐵 if we wanted
to we could call that 𝑏- little 𝑏. But we don’t actually need to work out anything to do that
for trying to find this side 𝑥. So just like before if I drop the perpendicular line from 𝐵 and
then we could call that point 𝐷, then we could use the method that we just did-did with the
two right-angled triangles in order to work out the length of this one and then use that to
work out the length of this one. So if I just quickly draw out two separate right-angled
triangles — so that’s a bit bent — so two separate right-angled triangles instead of one — and fill in
those lengths, so this one is ten, this angle is thirty, and this angle is forty, and this angle
is this length of the side is 𝑥.
So we can work out this length here. So let’s do that, opposite
the thirty is 𝑂, opposite the right angle which should be here is 𝐻. So we would have Sin
thirty equals opposite — let’s just call that I don’t know let’s call that 𝑦 — so 𝑦 over ten. And
then just like before times through by the ten, and we would get ten times Sin thirty equals
𝑦. And if we put that in a calculator, then you would work out that 𝑦 then equals five. So we
also know cause these have the same height that that length is then five. Okay, so if we do
exactly the same thing, at opposite that forty is still 𝑂, opposite this right angle is 𝐻.
And then if we worked out using our sine ratio again, we have Sin forty equals opposite, which is
five, divided by the hypotenuse, which is 𝑥. And just like before, we then need to times through
by the 𝑥. And we get 𝑥 times Sin forty equals five and then divide through by the Sin forty
to get that 𝑥 on its own. So 𝑥 is equal to five divided by sine forty, which if you put in our
calculator to two decimal places, we get seven point seven eight to two decimal places. Okay, so
that’s how you can split a normal right-angled triangle up into two right-angled triangles in
order to work out a missing length.
Okay, so let’s do this now without sines given, but just
letters — so purely algebra. Okay, so as you can see I’ve drawn out another triangle 𝐴𝐵𝐶. And this
time, we’re hopefully gonna derive the sine rule which is what we’re setting out to do in this
video. So I’m gonna follow exactly the same steps as before. I’m first gonna drop that
perpendicular. And then- so we can see I’m gonna label up my side to opposite the angle 𝐶 is
little 𝑐, opposite the angle 𝐴 is little 𝑎. And I don’t need the 𝐵 because I don’t- we’re not
trying to find anything about it. So this is angle 𝐴 and this is angle 𝐶. So our first objective
is to use the left-hand triangle to work out that length, and we could call it 𝑥. So let’s use
our right angle trigonometry again. So this is our opposite and this is a hypotenuse. So
we would know that sine 𝐴 is equal to opposite, which is 𝑥, divided by hypotenuse, which is 𝑐. So
if we want- if we can make 𝑥 the subject, we would first times through by 𝑐. So we get 𝑐 times
sine 𝐴, which equals at 𝑥.
Now I’m gonna leave that there and then now, I’m gonna use this 𝑥, this
𝑎, and this 𝐶. So opposite this 𝐶, we have our opposite again. And opposite the right angle, we
have our hypotenuse. So if I fill in just like we did with the left-hand triangle into our sine
ratio, we have Sin 𝐶 is equal to our opposite, which is 𝑥, and divided by a hypotenuse, which is
𝑎. So if we times this time both sides through by 𝑎, we have 𝑎 Sin 𝐶 is equal to 𝑥. Now if I just
change my pen color here, you’ll notice we then have two expressions in terms of 𝑥 and they both must be
correct because this height must be the same for both right-angled triangles. So what we can do
is actually put those together.
So if I just clear or leave the screen there — I’ll get back to my
blue pen — so we have on the left-hand side — I’ll just go up here — that 𝑥 is equal to 𝐶 Sin 𝐴 and it is
also equal to 𝑎 Sin 𝐶. Now this isn’t- this is actually part of the sine rule, but it’s not in
the usual form. So we’ve just gonna divide both sides through by the Sin 𝐴, which will give us 𝐶
equals 𝑎 Sin 𝐶 over Sin 𝐴. And then we’re gonna divide both sides through by the Sin 𝐶. So then
we get 𝑐 over Sin 𝐶 equals little 𝑎 over Sin 𝐴. Now that- oops let’s just undo that; that is
part of our sine rule.
So-so you may have seen that before, but you might also have noticed I
used an extra bit here; we have 𝑐 over Sin 𝐶 equals 𝑎 over Sin 𝐴. And it usually says equals something
else. So let’s just look at that for a second. So as you can see, we’ve got part of our sine rule 𝑐
over Sin 𝐶 equals 𝑎 over Sin 𝐴. But usually there’s another bit that comes on next to that.
So what I’m gonna do is show you hopefully where that comes from. So I’ve drawn another
triangle 𝐴𝐵𝐶. And this time I’m gonna drop a perpendicular from the point 𝐴 over to this line.
And if you label our triangle exactly the same way as we did before, this is outside 𝑐; this is
outside 𝑏. Okay, this will be our angle 𝐵 and this is angle 𝐶. So if we use our right angle
trigonometry, it seems a bit funny that we’ve tossed our triangle round, but it’s the same process. Let’s
call this bit our 𝑥. So that’s our opposite; opposite the right angle is our hypotenuse. So going
over here, we have Sin 𝐵; this time is equal to 𝑥 over 𝐶. And if we rearrange that by timesing
through by that little 𝑐, we’ve got 𝑐 Sin 𝐵 equals 𝑥.
Just as before we’re gonna do the same
thing with this bottom triangle, opposite the angle is 𝑂; opposite the right angle is 𝐵. So we
have 𝐶; I need to be a little bit careful that this is capital 𝐶 equals opposite, which is our 𝑥, over our hypotenuse,
which is 𝑏. Now I’m gonna times through by 𝑏, so we end up with 𝑏 Sin capital 𝐶 equals 𝑥 just as before. And there look we have our two expressions again in terms of 𝑥. So if we put
those together and I just do that up here in this little box, we have 𝑐 Sin 𝐵 is equal to 𝑏
Sin 𝐶. And if we divide through that by the Sin just like we did before, we’ll get little 𝑐 over
Sin capital 𝐶 is equal to little 𝑏 over Sin capital 𝐵. And you can see there look if I just use
the green this time, I’ve got 𝑐 over Sin 𝐶 there, which is equal to 𝑎 over Sin 𝐴. I’ve also
got 𝑐 over Sin 𝐶 here, which is equal to 𝑏 over Sin 𝐵.
So what I can do and I’ll do that on
the final screen is write those expressions together. And let’s do it in alphabetical order. So I
have 𝑎 over Sin 𝐴 equals 𝑏 over Sin 𝐵 equals 𝑐 over Sin 𝐶. Great! And that is our sine rule okay.
And we can use this to solve problems. So let’s just do one quick question so you can see
how this works. What we have here is a triangle 𝐴𝐵𝐶. This time like we do know the angles for
we know angle 𝐴 is forty, angle 𝐶 is thirty-five, 𝐴𝐵 is- has got length twelve, and 𝐵𝐶 is 𝑥. So if we
just labelled our triangle first, we have that opposite the side 𝐴 is little 𝑎, opposite the
side 𝐶 is little 𝑐, and we don’t have any information about little 𝑏; so I’m not gonna label it.
So there we are, we have 𝑐 is twelve and we have little 𝑎 is 𝑥, we have capital 𝐴 is forty, and
we have capital 𝐶 is thirty-five.
So we’re gonna use that part of the sine rule which includes the 𝑐
and the 𝑎; so little 𝑐 over Sin 𝐶 is equal to little 𝑎 over Sin 𝐴. And we’re just gonna fill in
to that and then solve a quick equation. So we have little 𝑐 which is twelve, so twelve over
Sin thirty-five is equal to little 𝑥 over Sin forty. And then we just need to get the 𝑥 on
its own, just like we’ve been doing before, so we’re gonna remove that Sin forty, so times
through by Sin forty. And we will have twelve over Sin thirty-five times Sin forty equals
our 𝑥. And if we put that in a calculator, we can work out that our 𝑥 to two decimal places
equals thirteen point four five. And here you go, that’s the relation of the sine rule and a
quick example just for you to see how it’s used and thank you very much.