 In this video we're going to introduce the so-called law of signs, which honestly trick students typically love the law of signs. It's a very simple formula, but it's very powerful to help us solve for the missing sides and angles of oblique triangles. So imagine we are given some triangles called ABC. So the angles are A, B, and C, and their corresponding opposite sides will be called lowercase A, B, and C, such as like a picture you see right here. We'll get more into that in a second. So the law of signs tells us that if we look at the ratios, if you take sine of A over A, this will be equal to sine of B over B and sine C over C. So this proportion where you take the sine of the angle and divide it by its corresponding side length, the opposite side length, this proportion, this ratio doesn't change depending on the angle. So the sine ratio of the angle divided by its opposite side will be the same whether you choose angle A, angle B, or angle C. And so in particular, when you look at one of these, this gives us an angle opposite side pair or AOS for short. So when you look at a triangle, if you have an angle measure and its opposite side, this is exactly the situation where the law of signs will be useful for solving oblique triangles. I mean, you can use it for right triangles as well, but it's much more, for right triangles, you can use, it's just so katoa to find this information here. The law of signs is very useful for oblique triangles here. Whenever you have an angle opposite side in AOS, because what this tells us, the law of signs tells us is that the ratio of an AOS does not change depending on the angle or the side. It's also equivalent to take the reciprocals of these things because if these ratios are equal to each other, the reciprocals are equal to each other. And so you get little A over sine A is equal to little B over sine B, which is equal to little C over sine C. And so this law of signs proportion here is extremely useful. And let's see why this fact holds. All right. So what we're going to do is we're going to consider the altitude between the angle C and the, its opposite side would be the, which would be the side length AB, right? Its length would be called little C right here. Now, depending upon the angle C, you get one of two pictures. If C is an acute angle, you'll get something that looks like this. So acute meaning it's less than 90 degrees. On the other hand, if C is an obtuse angle, the altitude would look something like this. In particular, for an obtuse angle, the altitude is exterior to the triangle. The altitude would be right here. For acute angles, the altitude would be interior to the triangle. Of course, for a right triangle, the altitude is a leg of the triangle, but I digress. We're not going to worry about right triangles in this situation. Recall that an altitude is a line segment that connects a vertex of a triangle, and it connects the line on the opposite side of the triangle in so much that it's perpendicular. It forms the right angle right here. For obtuse angle, again, this altitude would be exterior to the triangle. So we draw these dashed lines to kind of, you know, these, these lines that are exterior to the triangle. So there's two, there's two cases that have to be considered, but it turns out they're going to be very, very similar to this. The reason we introduce altitudes into this problem here is that the altitudes create right triangles. You'll notice, for example, a, what's called the foot of this altitude, we'll call it point D on the first triangle here. You'll notice that the triangle A, D, C is a right triangle. So so Catoa applies in that situation, but likewise C, B, D is likewise a right triangle. So we can use so Catoa in that situation as well. For the obtuse case, the right triangles we're going to consider is going to be A, C, D again. Right, that's a right triangle. Now this is outside of the triangle, but still a right triangle. You also have the triangle A, B, D is a right triangle. And so in both situations, A, D, C is right and A, excuse me, B, D, C is a right triangle. So even though the picture looks a little bit different, the label is going to be basically work the same. So let's consider the acute case for a moment. Let's consider the triangle right here where we take B, C, D. Well, we have this altitude right here, which H, right? If we if we think of opposite over hypotenuse with respect to angle B, right, H is the opposite side, A would be the hypotenuse for that right triangle. We would have the sine ratio, sine of B is equal to opposite over hypotenuse, H over A, we get the following. Now let's focus on this right triangle right here, the right triangle A, C, D with respect to angle A, right? A to be the opposite side, B would be the hypotenuse this time. And so we end up with sine of A is equal to H over B, like so. Let's play this same game for now the obtuse case over here, all right? So if we think of angle A, angle A is referring to this angle right here. And so let's let theta be its supplement. In particular, theta would equal 180 degrees or pi radians if you prefer minus A. So these things are supplementary angles. But that doesn't make much of a difference. If you look at the triangle B, C, D right here, the opposite side with respect to angle B would still be H and the hypotenuse would still be A. So you still get the same observation that sine of B is equal to H over A. That didn't change. Let's think of angle A this time, but we have to think in terms of theta looking at the triangle A, C, D in this case. With respect to theta, the opposite side is H, the hypotenuse would then be B. So you end up with H over B is equal to sine of theta. But theta is the supplement of A. So we get sine of 180 degrees minus A here. And this is a very important property about sine. Sine is going to be equal, the sine of an angle is equal to the sine of its supplement. That is, in the first or second quadrant, acute angles would terminate in the first quadrant, obtuse angles would terminate in the second quadrant. Sine doesn't give a care between these two. The sine of an acute angle is equivalent to the sine of its obtuse supplement because sine is both positive in the first and second quadrant. So guess what I'm trying to say here? Sine of 180 degrees minus A is just equal to sine of A. In the first and second quadrant, sine is going to be the same thing there. So notice here we get sine of B equals H over A. That's the same in both cases. And then we also get that sine of A is equal to H over B. Sine of A equals H over B. So whether we're the acute or obtuse case, you end up with the same conclusions. It took a little bit longer to get there for the obtuse case, but using the symmetry of the sine function we get there. So let's take these equations and solve them for H, okay? And it doesn't matter which case we're in right now. To solve H for the sine B case, we'll times both sides by A. So we get little A times sine of B is equal to H. If we do that for sine of A, we'll times both sides by little B. So we get B times sine of A is like so. They're both equal to the altitude here. These things are equal to each other. What we're then going to do is we're going to divide both sides by A. We're going to divide both sides by B. Like so, the A's cancel on the left, the B's cancel on the right. And so we end up with then sine of B over little B is equal to sine of A over little A. And so this then gives us the law of sine, sine A over A equals sine B over B. Well, what about B here? Recall that this argument, what about the angle C, right? Well, this argument came about by drawing the altitude associated to angle C. If we were to draw the altitude associated to angle B, we would end up with sine of A over little A is equal to sine of C over little C. That would be the same conclusion we get, similar conclusion by using the same argument here. But since sine of A over A is equal to sine C over C, it's also equal to sine B over B. And so we see all three of these ratios are equal to each other. And this test proves for us the law of sine. The law of science is extremely useful when you're trying to prove the AAS or the AAS case, so angle, angle side. It's also useful for proving angle, side, angle. If we know the two angles of a triangle and we know any side with that side's interior or exterior to the two angles, we can then solve for the whole triangle. And we'll do that in the following videos.