 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about simple trigonometric identities or topologies and today's lecture is about law of science. As the previous lecture, law of cross-science, we are talking only about trigonometry of the triangles, actually trigonometry of the angles in the triangles. It's not like general trigonometric identity for any angle. These are only about triangles. But, however, we don't put any restrictions on the triangles. So what is the law of science? Actually, it looks quite, I would say, nicely and maybe unexpectedly. Actually, when I saw this particular identity, I was surprised how beautiful it looks, if you wish. Here is what it is. Let's say you have a triangle. A, B, C sides lowercase a across the vertex a, lowercase b across b, and lowercase c against c, and angles alpha into a, beta at b, and gamma at c. Now, here is the law of science, which, again, in my personal view, looks really very cool. A divided by sin alpha equals b divided by sin of beta equals a of c divided by alpha, sin beta alpha, alpha, beta, and sin gamma. Again, a divided by sin of a, a divided by sin of a equals b divided by sin of b, b and sin of b, and c divided by sin of gamma. So it's always opposite angle, so that's the point. Okay, warning number one. Well, we have something in the denominator. Maybe it's equal to zero. Well, no. You remember we are talking about triangles, and triangles have angles from zero to 180 degrees, zero to pi, not including zero and not including 180, obviously. And sin, as a function, equals to zero on sin x equals to zero when x is equal to zero, pi, two pi, minus pi, etc. So it's pi k. So if my angle is between zero and pi, not including the endpoints, then sin is not equal to zero. So this is valid for any triangle, no questions about that. Now let's talk about the proof. All right, the proof is actually very easy. So let me drop an altitude from b to ac to the point h. Now what can I say about this bh? bh is a catatus in two right triangles, abh and cbh. Well, they're right because this is the perpendicular. It's an altitude, right? So in the right triangle, catatus is equal to, if you remember, the side of the hypotenuse actually multiplied by a sine or cosine, depending on whether the catatus is adjacent or opposite to an angle. So let's consider abh, bh is a catatus, alpha is an angle and cab equals to c is a hypotenuse. Now let's call the length of this low vector h. So what can we say about h? h from abh is equal to c times sine of alpha. Since h is an opposite catatus to the acute angle alpha, then it's hypotenuse times sine. Now let's consider the triangle bh. Same thing, h is opposite to gamma. So it's equal to hypotenuse, which is a times sine of gamma. So forget about h now. And let's rewrite this as we can divide by sine alpha times sine gamma. They're not equal to zero, so we can safely divide it. And what will be c? Sine alpha divided by sine alpha sine gamma equals a sine gamma divided by sine alpha sine gamma. And what do we have? c over sine gamma equals a over sine alpha. Sine alpha and sine gamma. Now where is the b? Well, I mean obviously all you have to do is to draw another altitude. If you will draw an altitude to, let's say, where is my red? If this is h, same thing. Consider triangle ahc now. Ah is opposite to gamma. This is hypotenuse b. So b times sine gamma. That's what h is. Now from ahb triangle, the angle on the top is beta. Hypotenuse is c, so it's equal to c times sine of beta. Again, forget about this. Divide by sine gamma times sine beta. And what we will have? b over sine beta equals c over sine gamma. Okay? If we divide this by sine gamma times sine beta, sine gamma would be reduced. In this case, I will have only sine beta and sine beta would be reduced here. And I will have only sine gamma. Now these two actually is basically what this is written. Right? Because this is c over sine gamma and this is c over sine gamma. So I can just continue the whole thing. Have we proven this particular theorem really rigorously? Well, not exactly because, again, we have to consider different cases when the base of the altitude h falls not in between these points b and c. And we kind of dependent on this, right? So let's just think about what will be in case we have an obtuse angle. All right. So let's say we have this particular triangle c, a, b. Sorry. This is c. And this is b. And this is alpha. And this is beta. And this is gamma. And our altitude comes outside. Let's do exactly the same. We still have b, h, a, the right triangle and b, h, c, the right triangle. Okay? From the triangle b, h, c, b, h, c, h is equal to hypotenuse a times gamma is an opposite angle to the b, h, so it's sine of gamma. Now, from the triangle b, h, a, h is equal to c hypotenuse times sine of, oops, this is not alpha. This is pi minus alpha. So it slightly changes the proof because now that we have to really do and obviously I have to analyze what this is all about. Now, it's simple, basically. Again, consider unit circle, sine is ordinate. So if this is alpha, then this is pi minus alpha. But ordinates are obviously the same. We have proved many times, actually, that sine of pi minus alpha is equal to sine of alpha. So using the trigonometric properties of a sine, I can say that this is equal to c sine of alpha, which is exactly the same as before when the triangle did not have this obtuse angle. So all I'm saying is that formula remains absolutely the same, a sine gamma equals c sine alpha. And again, dividing this by sine alpha times sine gamma, we will have a over sine alpha equals to c sine over sine gamma. So no matter where this h actually falls, it falls on this side or on that side, we will have either one of these angles alpha beta gamma or one of these angles, pi minus alpha or pi minus gamma, et cetera, depending on which kind it's tilted this particular triangle. But no matter what it is, no matter where this h falls outside on the left or outside on the right, we will always deal with exactly the same type of identity. And the only difference is that in case we have pi minus some angle, and we're talking about sines only here, we just replace sine of pi minus some angle to sine of this angle. So the theorem actually is proven. Well, is it all? Not exactly. Look, if a and h coincide, we really don't have a triangle a, b, h, we have just one line, right? So the case of right triangle, we have to really consider separately. So, well, let's do it again. You know, we have to cover all the cases, right? So let's consider this particular line to be perpendicular. So it's a perpendicular. So it's a right triangle. Is the theorem true in this particular case? Well, let's just think about it. In this particular case, c, as a catheter of a right triangle, is equal to a times sine of gamma. Okay? Sine of 90 degree is 1, right? Sine of 90 degree is equal to 1. We know that. Sine of pi over 2 in radians, if you wish. So from this particular equation, c over sine gamma is equal to a over, and that will put 1, basically, sine of 90 degree, which is alpha. So as we see, we still have exactly the same equality. And obviously, for b, it looks exactly the same. So basically, that's what it is. And all I can say that this particular law of sines is really universal for any triangle. There are two important facts here. Well, number one is obviously the law of sines, which looks really beautiful in my particular viewpoint. From another point, I would like to mention that the proof which we were using heavily dependent on the picture which we draw, what kind of triangle. And it's very important to basically draw all kinds of relationship between the sides, angles, etc., so we cover all the cases because otherwise, the proof would not be really universal. So these are two points. I would recommend you to go notes for this particular lecture at Unisor.com. Also, I would strongly suggest to register you as a student and either your parent or supervisor as a supervisor, as a parent or a supervisor, because then you will be able to go through exams. And your supervisor will really see how you perform, what's your results from your exams. And then you can get enrolled in some other course, etc., so this would be like an educational process rather than just lectures. That's it for today. Thank you very much and good luck.