### Video Transcript

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, weβre just gonna recap a little bit of right angle trigonometry with these two triangles that Iβve drawn here. So if we look at the triangle on the left and we have a hypotenuse of six and weβre gonna try and find this side π₯, which is actually that opposite angle. So if we just label those triangles and this side π₯ is our opposite and the one opposite the right angle is called βa hypotenuse.β 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 π₯ 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 π₯. Now if you worked that out on a calculator, you will find that π₯ 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 letβs just look at the triangle on the right, so opposite the side forty and is our π and opposite the right angle again is our hypotenuse. So itβs still sine. And this time, Iβm gonna work out the π₯. So if we fill in into our formula, we have Sin forty this time is equal to opposite which is three over the π₯ like so. So to get that π₯ on its own, we first need to multiply through by the π₯ and that will be π₯ Sin forty equals three. Now this time we canβt work that out yet, so we have to get rid of the Sin forty. So weβre gonna divide by the Sin forty, and that will leave us π₯ 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 weβre gonna use this method in order to solve some non-right-angled triangles. So letβs look at that now. Right, so here we have a triangle π΄π΅πΆ, and weβve been given the side π΄π΅ which is ten, the angle π΅π΄πΆ which is thirty, and the angle π΅πΆπ΄ which is forty. Now if we just label this triangle, opposite the π΄ weβre gonna call little π, opposite the πΆ weβre gonna call little π, and opposite the π΅ if we wanted to we could call that π- little π. But we donβt actually need to work out anything to do that for trying to find this side π₯. So just like before if I drop the perpendicular line from π΅ and then we could call that point π·, 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 β so thatβs a bit bent β so two separate right-angled triangles instead of one β 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 π₯.

So we can work out this length here. So letβs do that, opposite the thirty is π, opposite the right angle which should be here is π». So we would have Sin thirty equals opposite β letβs just call that I donβt know letβs call that π¦ β so π¦ over ten. And then just like before times through by the ten, and we would get ten times Sin thirty equals π¦. And if we put that in a calculator, then you would work out that π¦ 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 π, opposite this right angle is π».

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 π₯. And just like before, we then need to times through by the π₯. And we get π₯ times Sin forty equals five and then divide through by the Sin forty to get that π₯ on its own. So π₯ 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 thatβs 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 letβs do this now without sines given, but just letters β so purely algebra. Okay, so as you can see Iβve drawn out another triangle π΄π΅πΆ. And this time, weβre hopefully gonna derive the sine rule which is what weβre setting out to do in this video. So Iβm gonna follow exactly the same steps as before. Iβm first gonna drop that perpendicular. And then- so we can see Iβm gonna label up my side to opposite the angle πΆ is little π, opposite the angle π΄ is little π. And I donβt need the π΅ because I donβt- weβre not trying to find anything about it. So this is angle π΄ and this is angle πΆ. So our first objective is to use the left-hand triangle to work out that length, and we could call it π₯. So letβs use our right angle trigonometry again. So this is our opposite and this is a hypotenuse. So we would know that sine π΄ is equal to opposite, which is π₯, divided by hypotenuse, which is π. So if we want- if we can make π₯ the subject, we would first times through by π. So we get π times sine π΄, which equals at π₯.

Now Iβm gonna leave that there and then now, Iβm gonna use this π₯, this π, and this πΆ. So opposite this πΆ, 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 πΆ is equal to our opposite, which is π₯, and divided by a hypotenuse, which is π. So if we times this time both sides through by π, we have π Sin πΆ is equal to π₯. Now if I just change my pen color here, youβll notice we then have two expressions in terms of π₯ 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 β Iβll get back to my blue pen β so we have on the left-hand side β Iβll just go up here β that π₯ is equal to πΆ Sin π΄ and it is also equal to π Sin πΆ. Now this isnβt- this is actually part of the sine rule, but itβs not in the usual form. So weβve just gonna divide both sides through by the Sin π΄, which will give us πΆ equals π Sin πΆ over Sin π΄. And then weβre gonna divide both sides through by the Sin πΆ. So then we get π over Sin πΆ equals little π over Sin π΄. Now that- oops letβs 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 π over Sin πΆ equals π over Sin π΄. And it usually says equals something else. So letβs just look at that for a second. So as you can see, weβve got part of our sine rule π over Sin πΆ equals π over Sin π΄. But usually thereβs another bit that comes on next to that. So what Iβm gonna do is show you hopefully where that comes from. So Iβve drawn another triangle π΄π΅πΆ. And this time Iβm gonna drop a perpendicular from the point π΄ over to this line. And if you label our triangle exactly the same way as we did before, this is outside π; this is outside π. Okay, this will be our angle π΅ and this is angle πΆ. So if we use our right angle trigonometry, it seems a bit funny that weβve tossed our triangle round, but itβs the same process. Letβs call this bit our π₯. So thatβs our opposite; opposite the right angle is our hypotenuse. So going over here, we have Sin π΅; this time is equal to π₯ over πΆ. And if we rearrange that by timesing through by that little π, weβve got π Sin π΅ equals π₯.

Just as before weβre gonna do the same thing with this bottom triangle, opposite the angle is π; opposite the right angle is π΅. So we have πΆ; I need to be a little bit careful that this is capital πΆ equals opposite, which is our π₯, over our hypotenuse, which is π. Now Iβm gonna times through by π, so we end up with π Sin capital πΆ equals π₯ just as before. And there look we have our two expressions again in terms of π₯. So if we put those together and I just do that up here in this little box, we have π Sin π΅ is equal to π Sin πΆ. And if we divide through that by the Sin just like we did before, weβll get little π over Sin capital πΆ is equal to little π over Sin capital π΅. And you can see there look if I just use the green this time, Iβve got π over Sin πΆ there, which is equal to π over Sin π΄. Iβve also got π over Sin πΆ here, which is equal to π over Sin π΅.

So what I can do and Iβll do that on the final screen is write those expressions together. And letβs do it in alphabetical order. So I have π over Sin π΄ equals π over Sin π΅ equals π over Sin πΆ. Great! And that is our sine rule okay. And we can use this to solve problems. So letβs just do one quick question so you can see how this works. What we have here is a triangle π΄π΅πΆ. This time like we do know the angles for we know angle π΄ is forty, angle πΆ is thirty-five, π΄π΅ is- has got length twelve, and π΅πΆ is π₯. So if we just labelled our triangle first, we have that opposite the side π΄ is little π, opposite the side πΆ is little π, and we donβt have any information about little π; so Iβm not gonna label it. So there we are, we have π is twelve and we have little π is π₯, we have capital π΄ is forty, and we have capital πΆ is thirty-five.

So weβre gonna use that part of the sine rule which includes the π and the π; so little π over Sin πΆ is equal to little π over Sin π΄. And weβre just gonna fill in to that and then solve a quick equation. So we have little π which is twelve, so twelve over Sin thirty-five is equal to little π₯ over Sin forty. And then we just need to get the π₯ on its own, just like weβve been doing before, so weβre 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 π₯. And if we put that in a calculator, we can work out that our π₯ to two decimal places equals thirteen point four five. And here you go, thatβs the relation of the sine rule and a quick example just for you to see how itβs used and thank you very much.