# Proof of the law of sines

CCSS Math: HSG.SRT.D.10

## Video transcript

I will now do a proof of the law of sines. So, let's see, let me draw an arbitrary triangle. That's one side right there. And then I've got another side here. I'll try to make it look a little strange so you realize it can apply to any triangle. And let's say we know the following information. We know this angle -- well, actually, I'm not going to say what we know or don't know, but the law of sines is just a relationship between different angles and different sides. Let's say that this angle right here is alpha. This side here is A. The length here is A. Let's say that this side here is beta, and that the length here is B. Beta is just B with a long end there. So let's see if we can find a relationship that connects A and B, and alpha and beta. So what can we do? And hopefully that relationship we find will be the law of sines. Otherwise, I would have to rename this video. So let me draw an altitude here. I think that's the proper term. If I just draw a line from this side coming straight down, and it's going to be perpendicular to this bottom side, which I haven't labeled, but I'll probably, if I have to label it, probably label it C, because that's A and B. And this is going to be a 90 degree angle. I don't know the length of that. I don't know anything about it. All I know is I went from this vertex and I dropped a line that's perpendicular to this other side. So what can we do with this line? Well let me just say that it has length x. The length of this line is x. Can we find a relationship between A, the length of this line x, and beta? Well, sure. Let's see. Let me find an appropriate color. OK. That's, I think, a good color. So what's the relationship? If we look at this angle right here, beta, x is opposite to it and A is the hypotenuse, if we look at this right triangle right here, right? So what deals with opposite and hypotenuse? Whenever we do trigonometry, we should always just right soh cah toa at the top of the page. Soh cah toa. So what deals with opposite of hypotenuse? Sine, right? Soh, and you should probably guess that, because I'm proving the law of sines. So the sine of beta is equal to the opposite over the hypotenuse. It's equal to this opposite, which is x, over the hypotenuse, which is A, in this case. And if we wanted to solve for x, and I'll just do that, because it'll be convenient later, we can multiply both sides of this equation by A and you get A sine of beta is equal to x. Fair enough. That got us someplace. Well, let's see if we can find a relationship between alpha, B, and x. Well, similarly, if we look at this right triangle, because this is also a right triangle, of course, x here, relative to alpha, is also the opposite side, and B now is the hypotenuse. So we can also write that sine of alpha -- let me do it in a different color -- is equal to opposite over hypotenuse. The opposite is x and the hypotenuse is B. And let's solve for x again, just to do it. Multiply both sides by B and you get B sine of alpha is equal to x. So now what do we have? We have two different ways that we solved for this thing that I dropped down from this side, this x, right? We have A sine of beta is equal to x. And then B sine of alpha is equal to x. Well, if they're both equal to x, then they're both equal to each other. So let me write that down. Let me write that down in a soothing color. So we know that A sine of beta is equal to x, which is also equal to B sine of beta -- sorry, B sine of alpha. If we divide both sides of this equation by A, what do we get? We get sine of beta, right, because the A on this side cancels out, is equal to B sine of alpha over A. And if we divide both sides of this equation by B, we get sine of beta over B is equal to sine of alpha over A. So this is the law of sines. The ratio between the sine of beta and its opposite side -- and it's the side that it corresponds to, this B -- is equal to the ratio of the sine of alpha and its opposite side. And a lot of times in the books, let's say, if this angle was theta, and this was C, then they would also write that's also equal to the sine of theta over C. And the proof of adding this here is identical. We've picked B arbitrarily, B as a side, we could have done the exact same thing with theta and C, but instead of dropping the altitude here, we would have had to drop one of the other altitudes. And I think you could figure out that part. But the important thing is we have this ratio. And of course, you could have written it -- since it's a ratio, you could flip both sides of the ratio -- you could write it B over the sine of B is equal to A over the sine of alpha. And this is useful, because if you know one side and its corresponding angle, the angle opposite it that kind of opens up into that side, and say you know the other side, then you could figure out the angle that opens up into it. If you know three of these things, you can figure out the fourth. And that's what's useful about the law of sines. So maybe now I will do a few law of sines word problems. I'll see you in the next video.