 Hi, I'm Zor. Welcome to Unizor Education. I would like to remind you that trigonometry was called trigonometry because it helps to solve different problems about triangles, right? I mean, that was the original purpose of trigonometry. So, after all these lectures about the properties of trigonometric functions, etc., well, let's try to do something which trigonometry was supposed to do, which is just solve a few problems related to geometry of triangles. Well, first of all, to state these problems, I will need certain symbols. So, let me just introduce some standard which I will try to adhere to wherever possible. And I think it's a good standard to call the same elements of the triangle with similar letters. And I do recommend you to do exactly the same. So, the way how I will use the symbols is the following. So, if I have a triangle, my usual way of calling the vertices is with capital letters. Now, the sides will be called with the corresponding lower case letters. The side which is opposite to vertex A will be called lower case A, lower case B, lower case C. And the angles, I will also use the sequence of the same symbol letters in Greek alphabet which is alpha, beta and gamma. Now, next, there are different elements of the triangle which I will also use. Now, these are, well, the first three major elements are median angle bisector and altitude. So, if this is an altitude, I will usually call it HA, the length of this altitude. Now, if H, because it's the height, basically, and A, because it falls on the side called A. Now, the median, let's say the median from the vertex B, it goes to the middle of this. The length of this median I will call lower case M with our answer is on the B side. So, it would be B, the suffix B. So, the median is MB corresponding to MA or MC. And the altitude is also, this is HA, and obviously this would be HD and this would be HC. And now the angle bisector, let's say, the angle bisector of this, I will use the letter lower case L. And in this case, suffix will be C because it falls on the side C. The radius of circumscribed circle I will use letter R and inscribed lower case R. Well, sometimes it's written like this, whatever you prefer. I prefer this. And other elements I will basically introduce on the ongoing basis. But these are standards which I would like to state once. And then I can basically present the problem, something like this. And this will be the first problem which I'm going to solve today is this one. Prove that in any triangle A minus B times A plus B times sin gamma equals to C squared sin A minus B. Sorry, alpha minus B. What does it mean? Obviously A and B are two sides. C is the third side. Gamma is an angle which is opposite to the vertex C. And alpha and beta are angles opposite to vertices A and B corresponding. So that's what I would like to prove. Now, there are already a couple of things which I have mentioned in the previous lecture, one of the previous lectures. The law of sine and law of cosine. Now, you remember if you don't then go to the corresponding lectures that the law of sine is this. And again I'm using my symbolics which I have just introduced. This is law of sines. Now, I think it was one of the lectures in the trigonometric identities chapter. So go to unizol.com and in the trigonometric identities you will find the lecture which is devoted to the law of sines. And there is another one which is devoted to the law of cosines. And so these two, the law of sines and the law of cosines, square of any side is equal to sum of the squares of other two sides minus double the product and cosine of the angle between them. So these two actually are the first usages of the trigonometry in the geometry of the triangles. And I will definitely use these two laws, these two theorems which I have proven in that other lecture which I just mentioned to you to solve this problem. So how can I, using this particular law of sines, because it looks like this is the sine, this is the sine, it looks like law of sines should be used in this case, how can I do it to prove this particular equation? Well, using the law of sines I can express A in terms of C and sine of alpha and sine of gamma, right? C sine of alpha divided by sine of gamma, right? And similarly B, it's sine of beta divided by sine of gamma. Now, using this, I can substitute it in these cases and that will basically transform my original equation which I have to prove into something simpler, obviously, because there are no extra variables here. Now, also I have to mention another interesting property. Now, you know that the sum of angles in any triangle is 180 degrees, right? Pi. So, sine of gamma is equal to sine of pi minus alpha minus beta. Let me write it differently, right? Which is the same thing. Now, what is a sine? Sine, as you remember, is ordinate of the point on the unit circle. This vertical thing is ordinate. And obviously, if this is some kind of angle phi, this would be angle, now, if you count it this way, it also would be phi, right? Now, this angle is, it's pi minus phi, right? And the ordinates are obviously the same. We proved, in many cases, that the sine of pi minus some angle is equal to sine of this angle. So, that's another thing which I'm going to use instead of sine of gamma. So, all the left part I will express in terms of C sine alpha and again instead of sine gamma I will use sine of alpha plus beta. So, all will be in terms of C alpha and beta. And let's see if I will get an identity in this case. Because everything seems to be just dependent on these three variables, two angles and the side. Geonetically, by the way, it's worth mentioning that if this is C, this is alpha and this is beta. This is gamma. This is A, B and C, right? So, you know that triangle can be constructed using some three elements like, for instance, a side and two angles which are adjacent to this side, right? So, that's actually three independent elements which basically define a triangle. So, it's not a surprise that I will need exactly three elements to express all other elements. So, I need alpha, beta and C to express all other elements in terms of these elements. So, that's what I'm going to do. I will take three independent variables, in this case, C alpha and beta and express everything in these terms. And then see if I will get an identity. Now, this expression is invariant. I don't lose any specific values because it's a triangle which means none of the angle is equal to zero and all angles are less than pi, right? So, it has geometric sense in this case. It's a triangle. So, there is nothing wrong with dividing by sine of gamma. It cannot be equal to zero. So, this is an invariant transformation. So, that's what I'm going to do. I'll substitute it over there. And here is what I have. Okay. Alpha minus beta and alpha plus beta. Well, obviously, C over sine of gamma, I can take out, factor out the parentheses and I will have C divided by sine of gamma, which now I will replace with sine of alpha plus beta, as you know, right? Since alpha plus beta is pi minus gamma, or gamma is equal to pi minus alpha plus beta and sin is what I would say. Inside, I will have sine alpha minus sine beta times sine alpha plus sine beta. Now, since I have A and B here and A and B here, it would be square and it would be square, right? So, sine of alpha over gamma would be from A and C over sine of gamma would be from A and C over sine of gamma would be from B and these I take out from the parentheses and these guys have exactly the same thing. So, I have the C square and sine square. So, that's on the left. And sine of gamma, which is basically another sine of alpha plus beta, so I can just completely wipe out this square, right? That would be fine too. So, this is my left part. After I substitute it instead of sine of gamma, sine of alpha plus beta, and instead of A and B, I substitute these values. That would be on the left. Now, sine of alpha plus beta again is not equal to zero because it's exactly the same as the sine of the third and the gamma, which is not equal to zero. So, everything is fine, everything has meaning. Okay, so I have to prove that this is equal to C square sine of alpha minus beta. Okay, how can I prove that? Well, obviously I can reduce this C square. It's not equal to zero, so no problem. And they have only trigonometric expression, which I have to prove, trigonometric quality, which I have to prove. So, is this an identity with any angles alpha and beta from zero to pi? Not equal to zero, not equal to pi. Well, obviously we can multiply both sides by sine of alpha plus beta, and it will be here. So, identity which I have to prove is this one. Now, I multiply sine minus sine and sine plus sine, sine square minus sine square, right? X square, X minus Y times X plus Y is X square minus Y square, right? So, I have this on the left. On the right, I have sine of alpha minus beta times sine of alpha plus beta, right? So, this is sine alpha cosine beta minus cosine alpha sine beta. That's this one. And this one will do the same thing with a plus. So, again, I have X minus Y and X plus Y. So, it's basically X square minus Y square. That would be indirect. Well, obviously I have to replace cosine beta with one minus sine. So, I will have everything and cosine alpha with one minus sine. So, I will have on the right sine square alpha times one minus sine square beta minus sine square beta. Now, instead of cosine square, I will have one minus sine square alpha. So, what exactly? Now, the left part will do the same. Now, here I have sine square times one, which is sine square alpha minus sine square alpha sine square beta minus sine square beta. And now, minus and minus, it would be plus sine square alpha and sine square beta. As you see, this is reducible, and I have sine square minus sine square exactly what I have to prove. And these all invariant transformations, it goes both ways without any problems, and that's the end of the proof to the first problem. So, what I did use here is a theorem which is called the law of sines. It helped me to express all elements of the triangle involved in this particular problem through three independent variables. And in this case, it's one side, which is C, and two angles adjacent to this side. And then, basic reverence and chaos is technicality. Next problem. Okay, express A, B, and C in terms of R, alpha, beta, and gamma. R is the radius of circumscribed circle, and alpha, beta, and gamma are angles, obviously. Okay, so if you have a circumscribed circle around the triangle, A, B, C, alpha, beta, gamma, this is the center, so let's use this one. So, this is R, this is R, and this is R, okay? Now, look at this. Angle, beta is supported by this R, right? Angle A, O, C is a central angle, which is supported by the same R. So, this is inscribed angle, this is the center angle. And you know that the center angle is twice as much as the inscribed angle, right? So, angle A, O, C is equal to two beta. This is two beta. Now, correspondingly, this is two gamma and this is two alpha, obviously, for the same reason. If I draw a perpendicular from the center to a point M, which is the middle of the AC, now this perpendicular would obviously divide my isosceles triangle A, O, C into two rectangle, into two right triangles, sorry, because this is perpendicular, right? So, these are right angles. So, my isosceles triangle by an altitude from the top would be divided into two right triangles. Obviously, congruent to each other, well, because the sides are the same and the hypotenuse are the same and the characteristics is common, they're sharing it. So, angle A, O, M is half of the angle A, O, C. So, this is beta and this is beta, right? Now, and A, M is obviously half of AC. So, angle A, O, M is equal to beta and A, M is equal to half of B, right? This is the side B, lower case B opposite to the perpendicular B. Well, obviously, we can calculate the lengths of A, M in terms of hypotenuse and an angle using the definition of the sign in this case. So, obviously, A, M divided by A, O is a sign of beta. So, A, M, which is one half of B is equal to hypotenuse, which is R, times sine of this angle A, O, M, which is beta. From here, B is equal to 2R sine beta. And very similarly, obviously, A is equal to 2R sine alpha and C is equal to 2R sine gamma. Exactly the same logic can be in this triangle or in this triangle. So, this is how the sides are expressed in terms of radius of the circumscribed circle and angles. Now, let me go back to the law of sines. Law of sines says that A over sine of A of alpha is equal to B over sine of beta equal to C over sine of gamma. Now, I can put equals to 2R, right? So, this ratio, which we had established before as being the same for all three sides and all three angles, is not just an abstract number. Whatever that number might be, that knows what it is. It's basically a diameter to R is a diameter of a circumscribed circle. So, it has a geometric sense. So, any side divided by an sine of the opposite angle gives you the diameter of the circumscribed circle. That's actually an interesting kind of unexpected geometrical interpretation of this ratio. Okay, next. I usually mention it at the very beginning of every lecture, but let me mention it in the middle. Most of the problems which I'm presenting to you don't make any kind of practical interpretation or sense or anything like that. They are only for your exercise of your ingenuity or creativity and basically the ability to solve something which you have not really faced before. So, again, I'm categorically against the education when the teacher is saying, okay, this is the problem, this is the solution, use it wherever you can, because most of these problems will never be occurring in the real practice. It's important for you to make some kind of a way to solve the problem. So, that's why I'm always encouraging you, before you're listening to the lectures which I'm presenting here, try to solve these problems yourself. That's very, very important, that's the most important actually, because if you don't, if you just listen to these lectures without trying to solve something yourself, it's basically useless. Okay, now, another problem. Here is your triangle, and here are three altitudes. Let's call this point H. Now, what's necessary to prove actually that the radiuses of circumscribed circles around triangles A, H, C, triangle B, H, C, and triangle A, H, B, so A, H, C triangle, B, H, C triangle, and A, H, B triangles. So, these three triangles have the same radius of a circumscribed circle. But it's obviously not, you know, just looking at the picture, you don't see it. So, one circle would be, you know, something like this, another would be here, and the third one would be there. Why are they supposed to have the same radius? It's not obvious at all. However, let's just think about it. Now, these are perpendicular, right? Now, let's consider, let's say, M, N, P. Let's consider a triangle B, N, H, and triangle A, P, H. So, this triangle and this triangle. Now, obviously, these angles are vertical. These angles are right. So, these angles are equal to each other. So, angle H, B, C equals H, A. So, this angle is equal to this angle. Now, let's consider now a triangle B, H, C, this one, and triangle H, A, H, C, and this one, these two triangles. So, it's this and this. These are two triangles which I have to prove this property, the radius of the circumscribed circle is the same. Now, they share the side H, C, right? And these triangles have the angle opposite to this side. Opposite to H, C, in this triangle is this, and opposite to the same side, but in this triangle is this, and the angles are the same. Now, we just proved in the previous problem. Remember, A is equal to 2R times sine of alpha for a triangle, right? Alpha is an opposite angle to A. Now, applied to these two triangles, this would play the role of A, this piece, this side. In this triangle, the opposite angle is this, and in this triangle, opposite angle is that, and they're the same. So, the side is shared, and the opposite angles are the same, and that's why, considering this formula, I have to have A is the same and sine is the same, so R is supposed to be the same. So, that's why R for both these triangles are the same, the radius of the circumscribed circle, because it's expressed in terms of this side is shared and the angles are the same. That's why the radius is the same. That's it. And the last problem which I would like to present in this introductory lecture about trigonometry applied to geometric properties of the triangle is how to express the area of a triangle in terms of radius of a circumscribed circle and angles. So, I have angles, triangles, and I have radius. Well, this is easy. Again, let's connect these guys. So, as before, if this is an angle data, this is an angle data, because this is the half of the central angle, which is supported by the same arc as this inscribed angle. Similarly here, if this is gamma, then this is gamma as well, and this opposite, and this is alpha. Now, the area of a triangle can be now divided into six small triangles, and pairs are exactly the same, and this is the right triangle. Each one of them is the right triangle. What do I know about this right triangle? I know hypotenuse, which is R. In all cases, hypotenuses are R. And I know the angles, basically, this angle, and angle is enough with a hypotenuse angle. It is enough to determine both categories, both categories, right? So, for instance, the triangle AOM has area, these categories, times these categories, divided by two, right? If this is R, and this is beta, then one category is R sine beta, another is R cosine beta, and divided by two. That's the area of one triangle, right? Now, the area of another triangle is exactly the same. So, both of them together have this area, right? This is AOC now. Similarly, for a triangle BOC, I have R sine alpha times R cosine alpha, and for triangle AOD, I have R sine gamma, R cosine gamma. Now, total area, I have to sum them together, which is R square times sine cosine, sine cosine, sine cosine. What is sine times cosine of an angle? Remember that sine of 2 of 2 phi is equal to 2 sine phi cosine of phi, right? So, sine times cosine is half of the sine of a double angle. So, I will put one half in front of it, and instead of sine cosine alpha, I will put sine of 2 alpha plus sine of 2 beta or sine of 2 gamma. So, that's the formula. This is total area. By the way, I'm using the letter S for area. You can add this to all the other symbolics which I'm going to use. I would like to use it more or less wherever I can. So, that's the result of expressing the area of a triangle in terms of the radius of circumscribed circle and angles. Now, there are many more problems in the same category, which I'm going to present to you. I'm not sure I'll present many problems, but as much as I can. And all of them, they are very, very educational qualities. So, I'm encouraging you as much as I can to spend some time and try to solve these problems yourself. They are very educational, so to speak. They are very well in developing your ingenuity and creativity. So, all these subsequent lectures in this particular chapter of the course related to trigonometry and geometry together, all of them are very, very important for you to be able to solve yourself. Now, after this lecture is completed, try to basically do exactly the same with these problems. So, it would just, you know, kind of remind you exactly how it can be solved if you didn't do it yourself before. As everything else, this lecture is on Unisor.com, which you are welcome to visit this site. And what's important is if you register with this website and there is also some supervisor or your parent actually registered as a supervisor for you, then you will be able to take exam. Your supervisor, your parent will be able to enroll you in different topics so you will be able to go through exam. And you will see the results of your education yourself by basically checking the score on the exam. And exams can be just taken any number of times. Don't worry about that. The site is completely free so everything is available. Basically, that's it for today. Thank you very much. And I'll try to put as many other problems in the future lectures. Thank you. Goodbye.