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Video: Deriving the Sine Rule

David Arnold

You may have used the sine rule to calculate the measure of angles or length of sides in a nonright triangles. Learn how to use trigonometric ratios in right triangles to derive the sine rule for nonright triangles.

13:47

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, 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 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 β€œa 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 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 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 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 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 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 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 𝐴𝐡𝐢, and we’ve been given the side 𝐴𝐡 which is ten, the angle 𝐡𝐴𝐢 which is thirty, and the angle 𝐡𝐢𝐴 which is forty. Now if we just label this triangle, opposite the 𝐴 we’re gonna call little π‘Ž, opposite the 𝐢 we’re gonna call little 𝑐, and opposite the 𝐡 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 π‘₯. So just like before if I drop the perpendicular line from 𝐡 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 β€” so that’s 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 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 β€” let’s just call that I don’t know let’s call that 𝑦 β€” 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 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 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 β€” so purely algebra. Okay, so as you can see I’ve drawn out another triangle 𝐴𝐡𝐢. 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 𝐢 is little 𝑐, opposite the angle 𝐴 is little π‘Ž. And I don’t need the 𝐡 because I don’t- we’re not trying to find anything about it. So this is angle 𝐴 and this is angle 𝐢. So our first objective is to use the left-hand triangle to work out that length, and we could call it π‘₯. 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 π‘₯, divided by hypotenuse, which is 𝑐. So if we want- if we can make π‘₯ the subject, we would first times through by 𝑐. So we get 𝑐 times sine 𝐴, which equals at π‘₯.

Now I’m gonna leave that there and then now, I’m gonna use this π‘₯, this π‘Ž, and this 𝐢. So opposite this 𝐢, 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 𝐢 is equal to our opposite, which is π‘₯, and divided by a hypotenuse, which is π‘Ž. So if we times this time both sides through by π‘Ž, we have π‘Ž Sin 𝐢 is equal to π‘₯. Now if I just change my pen color here, you’ll 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€” I’ll get back to my blue pen β€” so we have on the left-hand side β€” I’ll just go up here β€” that π‘₯ is equal to 𝐢 Sin 𝐴 and it is also equal to π‘Ž Sin 𝐢. 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 𝐢 equals π‘Ž Sin 𝐢 over Sin 𝐴. And then we’re gonna divide both sides through by the Sin 𝐢. So then we get 𝑐 over Sin 𝐢 equals little π‘Ž 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 𝐢 equals π‘Ž 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 𝐢 equals π‘Ž 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 𝐴𝐡𝐢. 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 𝐡 and this is angle 𝐢. 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 π‘₯. So that’s our opposite; opposite the right angle is our hypotenuse. So going over here, we have Sin 𝐡; this time is equal to π‘₯ over 𝐢. And if we rearrange that by timesing through by that little 𝑐, we’ve got 𝑐 Sin 𝐡 equals π‘₯.

Just as before we’re gonna do the same thing with this bottom triangle, opposite the angle is 𝑂; opposite the right angle is 𝐡. So we have 𝐢; I need to be a little bit careful that this is capital 𝐢 equals opposite, which is our π‘₯, over our hypotenuse, which is 𝑏. Now I’m gonna times through by 𝑏, so we end up with 𝑏 Sin capital 𝐢 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 𝑐 Sin 𝐡 is equal to 𝑏 Sin 𝐢. And if we divide through that by the Sin just like we did before, we’ll get little 𝑐 over Sin capital 𝐢 is equal to little 𝑏 over Sin capital 𝐡. And you can see there look if I just use the green this time, I’ve got 𝑐 over Sin 𝐢 there, which is equal to π‘Ž over Sin 𝐴. I’ve also got 𝑐 over Sin 𝐢 here, which is equal to 𝑏 over Sin 𝐡.

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 π‘Ž over Sin 𝐴 equals 𝑏 over Sin 𝐡 equals 𝑐 over Sin 𝐢. 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 𝐴𝐡𝐢. This time like we do know the angles for we know angle 𝐴 is forty, angle 𝐢 is thirty-five, 𝐴𝐡 is- has got length twelve, and 𝐡𝐢 is π‘₯. So if we just labelled our triangle first, we have that opposite the side 𝐴 is little π‘Ž, opposite the side 𝐢 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 π‘Ž is π‘₯, we have capital 𝐴 is forty, and we have capital 𝐢 is thirty-five.

So we’re gonna use that part of the sine rule which includes the 𝑐 and the π‘Ž; so little 𝑐 over Sin 𝐢 is equal to little π‘Ž 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 π‘₯ over Sin forty. And then we just need to get the π‘₯ 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 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, 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.