Okay, so let's start with properties of triangles. Now, most of you have already known the basics of a triangle. Okay, so let's say this is our triangle ABC. By the way, there is a convention in which we normally name the sites of the triangle. The site opposite to angle A is called small a. Okay, so this is a convention that we follow. If BC is written well and good, but if BC is not written, it is as good as small a. Okay, so if the question also says there is a small a whose value is that's a 5 centimeter. It means automatically it is to be considered as your length of BC. Okay, this is by conventions. So don't start calling your examiner and asking him, hey, sir, what is this A here? What is this B here and on? This is by convention that we follow that we name the sites opposite to a given vertex by using the same alphabet in small. So capital A is used for angle, small a will be used for the site opposite to that. Similarly, capital B is this angle and we use small b for the site opposite to that. Okay, and capital C is this angle and we use small c for representing the site opposite to that. Okay, so ABC are automatically as shown in the diagram, right? Now, most of the triangle inequalities that you have studied, I'll just recap it. So these are the basics that you should be knowing about the triangle. First of all, the sum of the angles of a triangle is 180 degree. Everybody knows it or pi. Okay, everybody is aware of it. The sum of two of the sites is always greater than the third. Please note this down. Please note this down. And you are most of you are only aware of it also. The sum of two sites of a triangle is always greater than the third. Okay, apart from it, please note that the difference of two of the sites is always lesser than the third. Okay, so I've written mod of a minus b because I just want to show you the difference. Okay, similarly, mod b minus c will be lesser than a and mod c minus a will be lesser than any questions. So this is something which is already known from your class 10 days. Okay, so you have it, you have learned this in your junior classes. Okay, so already known since ninth and 10th grades. So what are we going to study in this chapter in 11th grade? Many more things we are going to talk about several laws, several rules, which basically come from the properties of a triangle. Okay, so this simplest, you can say a close figure has a lot of properties and those properties are exhibited or expressed in terms of certain rules, certain laws. So we'll be talking about those rules and those laws in this chapter. Okay, so let's get started with the very first law, which is called the sign law. Sign law, many books will call it as the sign rule doesn't make any difference. Okay, so what is the sign law? So let me just make a triangle once again. Sign law says that, sign law says that in any triangle ABC, in any triangle ABC, the ratio of the side to the sign of the angle opposite to it is a constant. That means A by sine A is equal to B by sine B is equal to C by sine C. Okay, and this constant is actually 2R. Now what is R here? You will be wondering. Write down R here is nothing but the radius of the circum circle, or you can say the circum circle radius, circum circle radius of triangle ABC. So what's the circum circle? I think I had already discussed about it when I was doing with you the concept of review of Cartesian coordinate system. So what's the circum circle? Can anybody write it down on the chat box? What's the circum circle? Anybody? Circum circle, circum circle, come on. A circle which? Which? Yes, I'm waiting for somebody's response. Ah, a circle which passes through the vertices of a triangle. That is called the circum circle of the triangle. I hope you can see my screen. I'm drawing one of the case of circum circle. Okay, so this is the circum circle of the triangle ABC. So or you can say the triangle is inscribed within the circle, which is called the circum circle. Okay, so the radius of the circum circle is what is called as capital R. How many of you would you agree? Is it by convention? Yes, normally, normally, if nothing is mentioned about R and the question has used a capital R there, it is, you know, I mean, of course you should be doing a properties of triangle type of a question. Then that capital R obviously will represent the radius of the circum circle. Okay, now, the first thing is what is the proof for this? How does this result actually come out? So what's the proof? Okay, by the way, all of you put C from my figure. Let's say I call this as S, circum circle is normally called as S. So the distance of A, B and C from S, they're all equal to R, R, R each. Okay, this is also R, this is also R, this is also R. Of course, they signify the radii of the circum circle. Now, all of you, you would have definitely done this property in class 10th that if this is angle A, okay, what is this angle? If this is angle A, what is the angle B, SC? 2A, absolutely, very good Karthik. Okay, so this angle is 2A. Okay, great. So let me write 2A over here. Now, what I do is I'm dropping a perpendicular from S onto the chord VC. Okay, let's say this is M. This is M. Okay. Now, if you drop a perpendicular from the center of a circle onto a chord, what does it do to the chord? What does it do to the chord? It bisects the chord exactly. And since the side opposite to vertex A is called small A, can I say Bm will be A by 2 and MC will also be A by 2? Correct? Now, another important question I would like to ask you. Let me just erase this part. This whole thing was 2A. So how much is each part over here? How much do you think is this angle? And how much do you think is this angle? I mean, will it be split equally? A each? Why is that so sardly? Why do you think it will be split between both of them? Are these two triangles congruent? Exactly. Yes, they are congruent actually. Yes, what makes them congruent? By which law it is congruent? By which law it is congruent? RHS congruency. Correct. RHS is basically a special case here. So RHS congruency is coming up in the picture. Very good. Now, everybody focus in this triangle, focus on this triangle, SBM. Okay. In SBM, can I make this statement and correct me if I'm making a wrong statement that sine of A is opposite by hypotenuse? Correct? In other words, you're trying to say sine of A is A by 2R, which means you're trying to say A by sine A is actually 2R. Okay. So at least I proved 33.33% of this formula. So one third of the formula I proved. So can I prove the remaining 66.66%? Yes, of course, I can do the same activity by dropping a perpendicular from S on to chord VC. Okay, this will also bisect it. So let's say I call this as N. Okay, let's say I call this as N. So this will be B by 2, B by 2. Right. And again, you try, you try applying the basic trigonometric ratio on the triangle SAN. So in triangle SAN, if you just apply your basic trigonometry, by the way, this is B angle. Okay, because this whole thing is 2B. So this is B, B each, B, B each. Okay.  That means your 2R is equal to B by sine B. Okay. So similarly, we can also prove I'm not going to do all the proofs here because if one is understood, you can easily understand that other will also come in a similar way. Okay, therefore, in light of all the three results, in light of all the three results over here, 1, 2 and 3. Okay, we can claim that A by sine A will be equal to B by sine B is equal to C by sine C. And this ratio is actually twice of the circum-circle radius. By the way, many books will call it as a K. Okay, they just use a constant for it. They will not tell you that the constant is 2R. But it is for our internal understanding that now we know that it is twice of the circum-circle radius. Okay. First of all, make a note of this. And then we are going to take a quick problem on the same. Done everybody? So let's take a small question on the same. Let me pull out the chapter's name, properties of triangles. Where is it? Yeah, I'd say let's say this question first. I'm giving this as a proof that question, but it can always be given as an objective question also that find the value of this. Of course, it should also mention that there is a triangle involved over here. Else people will worry about what is this A and B and a C all of a sudden. Okay. So in a triangle ABC, prove that this expression will add up to give you a zero. Let me give you one minute time, one, one and a half minutes. I think it's good enough to solve this question. So let's wait for one and a half minutes for any one of you to solve this. Just like they're done on the chat box once you're done. So once again, wishing everybody here a very, very happy new year. A very, very, you can say I'm wishing you all that this year may prove to be very prosperous for you. And this is the year when you're going to enter your class 12 as well, which is going to be a year of huge, huge importance because board exams are coming your way. You will have to fill up your forms for many competitive exams. Of course, they are going to happen only next year for you. 2023 is the year that you will be writing these exams, but most of the form filling and applications will go this year itself. And of course, you'll be given a lot of projects and lab records and everything is going to happen this year. So this is going to be a tough year, no doubt. And year after year, it has been proven that people have to work extra in this year as compared to any other year. Again, I'm not scaring you, but just giving you a heads up that this is a year where you should pull up your socks and work really, really hard. And I'm just pay to God that all of you get that sense and motivation and energy to continue with that hard work, even for this year as well. Okay, anybody with any success? See, in this problem, all the terms are symmetrical, right? So just work with one of the terms. Let's say I start with the very first term, which is your a square sine B minus C by sine B plus sine C. Let's see what we can do here. By the way, when you have such kind of a figure that you are adding some things and it is coming to zero, that means there are there is some cancellation of terms happening. So some term that will come from here, some terms will come from here, some terms will come from here that would lead to canceling each other out, isn't it? So I need to convert this expression in such a way that looks very symmetric. And of course, it should have terms, which will actually get cancelled with some terms coming from some other, you know, let's say this is your first term, this is your second term, and this is your third term. So some set of terms coming from one, some set of terms will come from two, some set of terms will come from three, which would end up canceling each other out. That is our intention over here. Now a little while ago, we did our sine law, right? Where A by sine A is 2R, right? So from here, I can say A is 2R sine A. By the way, this is a very important outcome from the sine law that from sine law, we can express our side in terms of the circum-circle radius and the sine of the angles. Okay. So this is coming from directly from our sine law. So please remember this, this is going to be a lot of importance for us. Okay. Now, I'm planning to use this in simplifying or writing this term like this. So I want to write my A square as 2R sine A, the whole square. Okay. So I'm trying to convert everything in terms of angles here. And of course, R is definitely there, but let's see what happens. Okay. Now here also, I'll do Oswald. Here also, I'll do a small simplification. I will write this as 4R square. And this sine square A, I will write it as sine A into sine A. Okay. Now, please remember we are dealing with a triangle. We are dealing with a triangle. So in a triangle, in a triangle, we all know that A plus B plus C is pi. Okay. So A is pi minus B plus C. Correct.  What is sine pi minus B plus C? Sine B plus C only, isn't it? Sine pi minus theta is sine theta. So I will, I'm going to write this one of the sine A's. Let me choose this sine A and write it as sine B plus C. Okay. So I'm just giving a partial treatment to one of the sine A's and I'm writing it as sine of B plus C. Okay. Now, not long ago, we did a trigonometric identity where we had learned that sine B plus C into sine B minus C can be written as sine square B minus sine square C. Correct. So let's do that over here. So let's write these two terms product as sine square B minus sine square C. Okay. And of course, there is a sine B plus sine C sitting in the denominator. Mind you, this term is factorizable, right? It is factorizable as sine B plus sine C times sine B minus sine C. Correct. So one of the sine B plus sine C on the numerator that will get cancelled with the denominator sine B plus sine C. Okay. Now, here is a term which actually looks symmetrically poised. I'm just writing it, you know, in a very, very, in a crude format. Okay. So this is your first term, mind you. Okay.  Yes or no. So can I use this fact and write down the second term as well? Who will tell me how would the second term look like? Now, I don't want you to write that down, but at least we can know from the fact that if there was a B and C involved, we got sine A times sine B minus sine C.  Yes or no. Yes or no. Right. So it is sine B times sine C minus sine A. I've just expanded it and it and it. Okay. So you can just write these results similarly. You don't have to reinvent the wheel. You don't have to do the same activity for all the terms. Okay. Likewise, the third term, I can write it as four R square. Now, in third term, you have C A and B involved. Right. So you can write this as sine C sine A minus sine C sine B. Yes or no. Correct. Now just add the three. Just add the three. So one, two and three, what will happen when you add one, two and three, please take four R square common. And you can yourself see that this term will get cancelled with, let me just pair them up. Sine A, sine B is sitting over here and sine B minus sine A is sitting over here. So they will get cancelled. Okay. So these two will get cancelled. Similarly, sine B, sine C is sitting over here. And here's sine minus sine B, sine C sitting over here. So these two will also get cancelled. Correct.  And there is a minus sine C, sine A sitting over there. So these two terms will also get cancelled thereby leading to, so this will become zero. This will become zero. This will become zero. Thereby leading to four R square into zero. That's a zero. Okay. This is your RHS and hence two. Now, if I'm not mistaken, this topic has been put into the supplementary portion, right? What has your teacher told you about this topic? Is it going to be a part of your supplementary portion or the actual portions? Has this chapter been done in school at least? At least the basic sign law or sign law, et cetera. No, they haven't done. Oh, okay. No issues. Okay. This year in JEE Advanced, I remember three questions came from properties of Pringles, three questions in the paper number one. So it's a topic of huge importance, at least for people who are writing KBPUI, JEE Advanced type of exams. Of course, JEE may also, the questions are asked. So is it fine? Any questions, any concerns with this solution? Please do let me know. All set. Now, I'm sure you would like to solve one more question. So let's go to the next question. So here comes the next question for you based on the similar concept. Prove that. Prove that. Again, they have forgotten to mention that it's a triangle.  Of course, small A, small B, small C, capital A, capital B, capital C have their usual meaning. Okay. So please prove this. Again, I'm giving you around one and a half minutes. Anybody who's able to prove it, you can just reply with a done on the chat box. No need to give any kind of a justification. Just say a done. All right. Any success? Anybody? Okay. Let's start with the left hand side. Left hand side. We have cos A minus B. See, get the idea clear. This question is actually, you know, asking us to prove that this expression, this expression is equal to this expression, A square plus B square by A square plus C square. Now, if you try to write that in terms of angles, in short, they're asking you to prove that this is equal to this square, this square. Let me just write to prove that because we're not started the proof. I'm just rewriting the same stuff. This is what we are asked to prove actually, right? In other words, the question setter wants me to prove that this expression is actually equal to sin square A plus sin square B upon sin square A plus sin square C. This is what they're expecting us to prove. Of course, as you can all see, that 4r square is going to get cancelled anyhow from the numerator and denominator. So I have to convert this expression to this expression. So what we are going to do for that? Done, sir. Oh, very good. Awesome. Great. So let me start with my LHS. Okay, let's start with my LHS. So in LHS, now this is a deadlock unless and until I decide to write my C as pi minus A plus B. Okay, so that I can do some kind of, I can apply some kind of trigonometric identity between cos A minus B and something related to cos A plus B. Okay. So this is my idea over here. See, I may be unsuccessful also. I'm not claiming that whatever steps I take, I will definitely achieve my goal. In mathematics, we don't have that expectation that I should start with some method and I'll be successful. I'll meet the end result. It may not happen. You may fail and be ready to fail. See, please understand that the learning is hidden in the failure part of it, not in achieving it, not in getting the result. So when you try and it doesn't work, at least one learning will be there that such things will not such things are not going to work for such kind of portion. Okay. So many times it happens that you will try many things and it will not work. That is fine.  So can I say that this expression could be written as one minus cos A minus B times cos A plus B? Yes or no? Similarly, denominator also I can write it as one minus cos A minus C cos A plus C. Yes or no? Now, what is the formula for? Let me go to the side of the screen. Let me check your basic trigonometric identity. So who will tell me what is cos A minus B into cos A plus B? There were two formulas. There were two formulas. Yes. So one is cos square A minus sine square B and the other one is cos square B minus sine square A. Right. So please remember these and you can only remember them through practice. Okay. So good. So some of you definitely have retained this formula. I would not say remember retained this formula. So this is going to become one minus cos square A minus sine square B and this is one minus cos square A minus sine square C. Okay.  And here also if you open the brackets, you get one minus cos square A plus sine square C and it is obvious that one minus cos square is going to be sine square.  Is it fine? And this is what we actually wanted to prove. See here. Right. This is our right hand side and this is what we have achieved also. Okay. So this is our RHS hence proof. Is it fine? And all these questions which are proved that don't, don't try to underestimate such questions that oh, so this is a prove that question. It will never be asked. No, I can give which of the following is an expression for this for a triangle and this could be one of your options. Why not? Okay. So any proof that or any question which is subjective in nature can always be objectified. That is why many, you know, many questions of your, you know, the very past years that, let's say, you know, 90s and all of them are, they may have started to objectify those questions. And same trend has been seen in your CVSC paper as well. In your recently conducted semester one CVSC paper, they have objectified many previous year subjective questions because this year your seniors, they wrote an objective semester one paper. I hope you all know that. So it was an objective semester one paper. So the previous year subjective papers, they were objectified. Of course, you just have to put options. That's it. Any questions, any concerns here? Okay. So without much ado, we are now going to move towards the second law. Okay. So we already did sign law. So the second law that we are going to talk about is the cosine law also called as the cosine rule. And this rule is very much familiar to most of you the moment I write it down, you will be able to understand. So this rule says that in a triangle or in any triangle ABC, the following rules are going to be applicable. That is, cos a can be written as this. By the way, the way you have learned this in physics is somewhat different in physics. You have learned it as a square is B square per C square minus two BC cos a right in vectors maybe, isn't it? The reserve would have derived this result during the bridge course, isn't it? So the same thing is written in form of this. Okay. So cos of an angle a in a triangle can be written in terms of its sites as B square per C square minus a square by two BC. The way to remember this is whatever angle is involved, the side opposite to a is always in the minus. So minus this, the square of that side. Okay. The other two sides are in the positives and you divide by two times that these two sites which are there in the positives. So you can replicate this result and further write few more cosine laws here. That is cos B could be written as a square plus C square minus B square by two AC and cos C could be written as a square plus B square minus C square by two AB by two AB. Okay. So these three laws are together called the cosine law. They're called the cosine law. Now we will try to derive one of these. We'll not derive all of them because they are based on the similar type of proof. So we'll derive one of these laws. Let's try to derive the first one over here. Cos a is equal to B square per C square minus A square by two BC. Okay. Would anybody like to derive it first? I can give you minutes, few seconds if you want to attempt it. If not, then I can proceed with the derivation. Anybody wants to derive? Okay. Siddharth wants some time to derive. Okay. Think about it at least whether you get it or not. That's different thing, but at least you should make an honest attempt to derive it. Okay. So I'm giving you around one minute. Think about it and then we'll discuss it out. I have to prove the first of the cosine law here. Please, please keep this into your consideration that we have to derive this one. Okay. Okay. So let's let's try to prove it. So for that, I'm going to do a small construction over here. I'm going to drop a perpendicular from let's say B on to AC side. Okay. So we all know by the convention that this is your small A. Okay. This is the small B side and this is the small C side. Okay. Now let's call this foot of the perpendicular here as AM. Right. Now, if this is angle A and this is your small C, can I say AM is the base of the triangle ABM where A is considered to be your reference angle, right? So this is base. This is hypotenuse and this is opposite, isn't it? So in light of that, can I say AM will be C cos A. Am I right? Or you can say cos of A will be AM divided by C small C. So AM will be C cos A and BM will be what? BM will be C sin A. Okay. So this is C sin A. I'll just write it down on the on this expression C sin A and this is going to be C cos A. Right. So what is this remaining length? MC. So this remaining length MC would be nothing but B minus C cos A. Am I right? Yes or no? So let me write that down next to this else you will think that Sarah has written MC as a B. Okay. So this is C cos A. Is it fine? Now look at the triangle BMC. Look at the triangle BMC. Now can I say BMC is also right angle triangle? Yes or no? BMC is also right angle triangle. So can I apply Pythagoras theorem on BMC? That means can I say BM square plus MC square is equal to BC square. Okay. This is our Pythagoras theorem. Pythagoras theorem. Fine. So what is BM square? BM square is C sin A the whole square. What is MC square? MC is B minus C cos A the whole square. What is BC square? BC square is A square. So when you expand it you will see you will get C square sin square A. This will become B square C square cos square A minus 2 BC cos A equal to A square. Yes or no? So far so good. Now club these two terms. If you club these two terms by taking C square common it will become C square times sin square A plus cos square A which is a 1. So this two will become combinedly become a C square. So this is already a B square and minus 2 BC cos A. There you go. I think this is what we wanted to derive, isn't it? So A square becomes B square plus C square minus 2 BC cos A. Now if you want to make cos A the subject of the formula you can definitely do that. So cos A can be written as B square plus C square minus A square by 2 BC. Okay. Hence proof. In the same way you can prove the two other cosine laws as well. So I would not want to waste your time by doing it. The proof is more or less similar. Any question with respect to the proof please feel free to ask. Any question? With respect to the proof please feel free to ask. Now one of the most important use of cosine law is in finding the length of median of a triangle. So we'll take that as our question in the next slide. In case everybody has copied this can I go to the next slide or anybody is copying it? Okay. I'll stay for this on this slide for 30 more seconds. Okay. So let's take a question based on this. Length of medians. Length of medians of a triangle. No, no, no, no, no, no. It is always minus only. When does it become plus? Tell me. When does it become plus? It's always a minus 2 BC cos A. It's never a plus. Why would it become? But cos A is there to take care of it. Why you have to put an extra plus? If A is obtuse, cos A will automatically become negative. You don't have to change the formula for that. Formula still remains to be that.  Yes, right? Yeah. All right. So let's move on to the application of this cosine law, which I told you is in finding the length of the medians. So let me start with the triangle diagram over here. Let's say this is our triangle A, B, C. Okay. And these are our medians. Let me make them in different color. One I'll make in. Oh, so, so, sorry, so, sorry. I have to choose one line. This is a median. Let me call this as AP. Let me choose another one in orange color. Let's say, yeah, this is BQ. And let me choose another one in maybe a red color. Okay. CR. So as you can see, on your screen in front of you, I have shown a triangle with the three medians. Okay. So now we have to prove any one of them. Prove that. Question is proved that the length of the medians are as follows. Length of AP is half under root of 2B square plus 2C square minus A square. Length of BQ is half of under root of 2A square 2C square minus B square. And length of CR, which is the third median, is half of under root of 2A square plus 2B square minus C square. As you can see, there is a pattern in which I have written these results. So if you happen to remember one of them, you can easily get the other two as well. Okay. Meanwhile, I would request everybody to try approving this one out. Okay. Many people ask me, say, do we need to remember this result? Again, I would say yes, but through practice. Okay. It's not a Sanskrit stroke that you'll sit and mug it up. Okay. So remember it. Of course, you will remember it when you solve more and more questions. Yeah. So let's prove the length of AP is half under root 2A square plus 2C square minus A square. Okay. So in the interest of time, I'll be also pitching in so that things become slightly faster. So now let's say I call this length Bp as half of BC. Of course, it's a median. So we know it's going to be bisecting. Okay. So P is going to bisect the join of B and C, right? So each one of them is A by 2, correct? Can I use cosine law in the triangle ABP? Okay. Let's use cosine law in the triangle ABP by writing cos of B as, okay. So let me, for the sake of convenience, let's call it as X. Okay. So let ABB call X. Okay. So tell me how would I write the cosine law for this? Please note that this is your small C. Okay. This is your A by 2 and this is your X. So everybody focus on ABP triangle. This is C. This is A by 2 and this is X. So if I have to write a cosine law for this angle B, what would I be writing? So very simple. Just write the squares of the two sides, which are basically sandwiching this angle. So A by 2 square plus C square minus the side opposite to that angle B square, which is X square all divided by 2 into A by 2 into C. So when we were learning cosine law in the previous side, what did we learn? We learned that when you are writing a cosine of an angle formula, you write first square of that sum of the squares of the two sides, which are enclosing that angle minus the square of the side opposite to that angle divided by 2 times the product of the sides enclosing that angle. That is what I have done here precisely, isn't it? In short, it's nothing but A square by 4 plus C square minus X square by AC. Right. Any questions so far? Now, can I also write cosine law for the whole triangle? Okay. So let me write down the cosine law for the whole triangle A, B, C again, cos B only I will be writing. So can I say cos B for the whole triangle will be A square plus C square? I'm so sorry. I think this was a stray mark. Yeah. A square plus C square minus B square by 2 AC. So let's call it as 1. Let's call this as 2. Now remember, both of them are representing cos of B only. So since both 1 and 2 are representations for cos B only. So can I say from 1 and 2, I can equate the right hand sides. That means I can equate A square by 4 plus C square minus X square by 2 AC with A square plus C square minus B square by 2 AC. Oh, so sorry. There's no 2 here. My mistake. I wrote the 2. Yeah. 2 is here actually. Yeah. So AC, AC will go off. And of course, let's try to simplify and make extra subject of the formula. Send the 2 to the other side. Send the 2 to the other side. And this is what you're going to see. One of the C squares will also get cancelled off. Okay. I think this will give A square by 2. Okay. So here what we'll see is that you are going to get B square plus C square minus A square by 2 as 2 X square. Okay. So send this B square to the left. Send the 2 X square to the right. In short, you're trying to say 2 X square is equal to 2B square 2C square minus A square whole divided by 2. Now send this 2 down. This 2, send it down. Make it 4 and make it 2 square, whatever you want to call it. So your X will be under root of under root of this expression, which is nothing but which is nothing but half of under root of 2B square plus 2C square minus A square. Hence 2. Okay. So please remember this result. Many a times you would not require the proof. But if you don't do the proof, let me tell you, you'll forget it in some time. Because when you do the proof, you have actually dirtied your hands, you know, trying to get the result. Okay. So then you will remember that effort. That is why most of the things I derive in the class, because it helps you to retain the fact. Okay. Yes. Is this fine enough, Karthik or should I go further to the left? I think there's nothing further to the left. I think this is, is the screen at the right position for you? So whatever you want to see. Okay. Great. So Karthik is done. So in a similar way, you can also prove that the other medians, okay, would be given by these expressions. Is it fine? Using Heron's formula. Okay. Did it work? Okay. It became complex. Okay. Okay. Almost it worked. Okay. Yeah. There are many ways to do the same things, but we have to find out which is the most time efficient one. Anyways, let's resume a few more questions. Let's do a few more questions. So let's take, okay, let's take this question. With usual notations, if in a triangle ABC, B plus C by 11 is equal to C plus A by 12 is equal to A plus B by 13. Then do that. Cos A by 7 will be equal to cos B by 19 will be equal to cos C by 25 for that triangle. Of course, when you look at the problem, you realize that there is a cost, cost, cost sitting everywhere. And some use of cosine rule has to be there. Okay. So that could be like a tip for you to start working. I'm giving you our own one and a half minutes, one and a half, two minutes for this. Try it out. Yes. Let's proceed with the solution now. See, in order to get this kind of a ratio, we at least need to know the ratio of the cosines of these angles. Okay. But let me tell you, we have not been provided with the sides exactly. Of course, we know some kind of a ratio between them, but we don't know what's A exactly, what is B exactly, and what is C exactly. So I'm going to use this ratio to find at least some kind of a relationship between the sides. So let's say I call this expression to be equal to lambda. Okay.  Okay. You can take any variable name you want, alpha, beta, gamma, lambda, delta, t, whatever you want to take. So from here, I can see that B plus C is 11 lambda. C plus A is 12 lambda. And A plus B is 13 lambda. Correct. Now let's try to add them. If you add them, you'll end up getting twice of A plus B plus C as this is going to be 36 lambda if I'm not wrong. That means A plus B plus C is 18 lambda. Now, please note, if B plus C is 11 lambda and A plus B plus C is 18 lambda, what do you think will be your A? What do you think will be your A? Write down on the chat box. A will be, A will be, A will be, write down, write down, write down. What will be A is 7 lambda, right? Exactly, Shalini. Correct. Similarly, if C plus A is 12 lambda and A plus B plus C is 18 lambda, what is going to be P then? B is going to be 6 lambda. Similarly, if A plus B is 13 lambda and A plus B plus C is 18 lambda, then C is going to be 5 lambda. Now, let us use the cosine law formula to find our cos A. Cos A is nothing but B square plus C square minus A square by 2 BC. So by 2 BC, BC will be 6 into 5 lambda square. Ultimately, you realize that lambda squares will get vanished from everywhere. And this is nothing but 61 minus this, which is 12. 12 by, this is, again, this is 12 into 5, so you can cancel out the 12, 12. So that's going to be 1 by 5. So cos A is going to be 1 by 5. Please note that we don't need the exact value of A, B and C to get our cos A. Similarly, cos of B would be what? Cos B will be A square plus C square minus B square, which is 36 lambda square upon 2 AC. So this is going to give us, if I am not mistaken, 74. 74 minus 36, which is 38. Okay. 38 divided by this, so 19 by 35. Correct me if I am wrong. In a similar way, let's find out cos C. Cos C will be A square plus B square minus C square upon 2 AB. 2 AB. So here also, let's do the simplification. This is going to be 85. 85 minus 25, which is going to be 60. 60 divided by 84. In fact, 60 divided by 2 into 7 into 6. So let me just cancel off whatever terms are going to get cancelled. So it's going to be 5 upon 7. Okay. So now everybody please pay attention. Cos A is 1 by 5. Cos B is 19 by 35. And cos C is 5 by 7. Let's try to make the denominators of all of them same. Okay. So let's multiply this with 7. So that becomes 7 by 35. This is already 19 by 35. No need to do anything with this. And let's multiply this with the 5. So this will be 25 by 35. So can I say from here that cos of A by 7, which is 1 by 35 is same as cos B by 19, which is also 1 by 35, which is same as cos C by 25, which is also 1 by 35. That means we have achieved what we wanted to prove. This is what we wanted to prove. Cos A by 7 equal to cos B by 19 equal to cos C by 25. None. Is this fine? Any questions? They are simple. So in past few years, many questions have been framed very similar to what we just now did. Okay. So with this, we are now going to move towards the next law, which is called the projection law. Also called as the projection rule. Else, how would you find out, Karthik? Karthik has a question. How did you know that you have to start by equating it to lambda? See, in order to find cos A cos B cos C, I have to use the length of the triangle, A, B and C value. So unless and till I call that, I mean, of course, you can find A, B and C in terms of each other, right? So why not, you know, write all of them in terms of a independent unknown, which is lambda in this case. Of course, many things you will learn through problem practice also. So it's very hard to answer your question that how did you know that you have to start with this? Of course, a bit of, you know, you can say, you know, analysis where I needed the value of your sites and of course, a bit of experience when you solve more and more questions, you've come to know. That's the, that's the main problem with many of the math questions. You will not know how to start, right? So in order to know how to start, you need to start practicing, right? That's the, you can say that's the most convenient way of, you know, see, when you, when you practice many questions, you come to know many do's and don'ts. Okay. This might work. So that, that, that gives you a leap of faith. But again, leap of faith doesn't mean you will end up solving the problem. So there are some cases, there are some questions where you need to apply your learnings from the previous problems. Some of them will work. Some of them might not work. Okay. So let's talk about projection law now. What is the projection law and why does call projection law will discuss it in some time. So projection law says that in any triangle ABC, these three laws will always work. In any triangle ABC, these three laws will always work. One, your side A can be written as B cos C plus C cos B. Okay. Your side length B can be written as A cos C plus C cos A. And your side length C could be written as A cos B plus B cos A. Okay. So in any triangle, these three rules will definitely work. Now what is the proof for it? Maybe I'll prove one of them for you. Let's try to prove the first one. Okay. Now here I will be again dropping a perpendicular from A on to BC. By the way, very important. I will be helpful for you in vectors also. Please note that BM is called the projection of AB on BC. Okay. So this length that you see this length is actually called the projection of AB on BC. Projection means if you start dropping perpendicular from every point on that line. Okay. It will be basically all the points lying on on BM. You can also call BM to be like a shadow, shadow cast shadow cast of BM on BC. So if I put a light source over here. Okay. What would be the shadow of AB on BC? Okay. Of course, when you are casting a light exactly on top of it, every point will be having its image vertically down like this side. So if you start connecting all the points, yeah, it will become your BM length. Okay. So that is called the projection of AB on BC. So BM is called the projection of AB on BC. That's the reason why it has got its name projection long. Okay. Anyways, let's carry on with the proof to get more clarity. MC is called the projection of MC is called the projection of AC on BC. Okay. So just like BM was called projection of AB on BC. MC is called the projection of AC on BC. Okay. Anyways, now a very simple question, a very simple trigonometric question I will ask you if this is C and this is angle B, how much do you think is your BM length? You see it's a simple C cos B. Right. And if this is your small B and this is your angle C, how much do you think is your MC length? It's a simple B cos C. Okay. And you all know that AM, sorry, not AM, BC is nothing but BM plus MC. Okay. BC is a, BM is just now, we figured out C cos B, MC we just now figured out B cos C. Okay. And there you go, formula is proved. Okay. Now, any easy way to remember this formula, this chapter is going to be loaded with formulas by the way, right? So you have to find out a trick to remember the formula, at least one of the triads you have to remember others will automatically follow. So here if there's an A involved, you can see on the right side, there is no A or any angle related to A involved. So small n capital A are missing from the right. So it's only B and C. So once B becomes site, C becomes angle. In the next case, C becomes sites and B becomes active. Okay. So if you're talking about the expression for small a, it is completely in terms of small B, capital B, small C and capital C in this fashion. Okay. So you can see the same pattern is seen in the other two expressions as well. Okay. I hope you have noted this down. Any questions? Just to give you a connect between what we learned in our sign law and what is there on your screen right now? If you start writing something very interesting, let me write it like this, interesting. If you write, let's say the right hand side over here, B cos C and C cos B like this. B is 2R sin B. Okay. And C is 2R sin C. Take 2R common that leads to sin B cos C plus sin C cos B, which is sin of B plus C. Sin of B plus C is nothing but sin A because A plus B plus C is pi. So B plus C is pi minus A. So sin pi minus A is as good as sin A. And this is nothing but your A, which is left inside. Okay. So you started with this expression. You applied your sign law also and it ended up on getting A. So basically you're getting this result from multiple perspectives, from multiple starting points. Yes or no? Okay. Anyways, so many ways to keep things in mind. Let's take a problem on the projection law. Okay. Everybody's ready for a problem? Can we take a question? All right. Let's take a question. In a triangle A, B, C, if C cos square A by 2 and A cos square C by 2 is equal to 3B by 2, then show that A, B, C, that is small A, B, C are in arithmetic progression. Again, just write it down on the chat box only to give any justification to me. Just write it down on the chat box. Yes. Any success, anybody? Okay. Let's use this expression given to us. Now, please recall you have already done the formula for, in fact, let's multiply with 2 throughout. Let's multiply with 2 throughout. In your trigonometric ratios and identities, we have already learned the formula for 2 cos square x by 2. 2 cos square x by 2, if I'm not mistaken, is 1 plus cos x. Okay. So this expression I can write as 1 plus cos A. And this expression I can write it as 1 plus cos C. So when you express, when you expand it, this becomes A plus C.  And just now we learned that A cos C plus C cos A is going to be B. Isn't it? It's going to be B by your prediction law. So A plus C plus B is equal to 3B, which means A plus C is equal to 2B, which means A, B and C are in AP. Yes or no? Easy, right? Wasn't it easy? Okay. So please note this. We'll be discussing one more question, a simpler one. See, properties of triangles will use all the identities, all the trigonometric ratios and identities that you have learned is just that it is going to become a conditional case, conditional identity, where A plus B plus C is equal to 180 degrees. All right. Let's move on to the next question.  S is the semi-perimeter. I hope you all of you have come across the semi-perimeter expression, especially in the Heron's formula. This is a simple touch and go question. Simple touch and go should not take you more than 15 seconds. 15 seconds, not more than that. Done anybody? Not yet. Oh, wonderful. Prism is already done. Anybody else? Anybody else? Just one person. Okay. It's going to be such a easy question. See, just take the LHS. Okay. If you expand it, see how I'm going to expand it. B cos A, B cos A, I'm going to keep it along with B cos A, I'm going to keep it with A cos B, A cos B. I'm just canceling it out just to tell you that I have taken those terms. Similarly, C cos A, C cos A, I'm going to take along with A cos C. And similarly, C cos B, C cos B, I'm going to take along with B cos C. The moment you pair them up like this, as per projection law or from projection law, from your projection law, this expression becomes small C. This expression becomes small V. This question becomes small A, isn't it? And isn't this twice of A plus B plus C by 2? Isn't this twice of S? Yes, very much. That's your knowledge. Yes, absolutely. Is it fine? Yes. So this expression was just the sites written in terms of projection law. Okay. So with this, we are now cruising towards the next subtopic here, which is called the tangent law.  So now the fourth law coming your way is the tangent law. Tangent law is also sometimes called the Napier's analogy. So in many books, you'll find that they'll use the term Napier's analogy for it. Okay. Again, this law can be written in various shapes and sizes or various shapes and figures. So I'll write few of the ones which are normally used. Tangent of A minus B by 2, let me write down first in any triangle ABC, tan of A minus B by 2 can be written as A minus B by A plus B cot C by 2. Okay. Tan B minus C by 2 can be written as B minus C by B plus C cot A by 2, cot A by 2. By the way, if you want, I can put brackets around it and tan of C minus A by 2 is C minus A by C plus A cot B by 2. By the way, you can also, if you want to, you can also write it as tan B minus A. Okay. See, there are various shapes. That's what I was talking about. You can write this like this also. Tan B minus A by 2 is B minus A by B plus A cot C by 2. See, actually both are the same things just written in a different fashion. So many people ask me, sir, why A minus B by 2? Why not B minus A by 2? Can I write it for that as well? Yes, it doesn't make any difference. Okay. So similarly, for the second one also, you can write tan C minus B by 2 is equal to C minus B by C plus B cot A by 2. No issues. That will also work. Third one also, you can write tan A minus C by 2 is A minus C by A plus C cot B by 2. That will also work. No issues. So any of the forms you want to use, they're going to work. But the million dollar question here is how do you prove them? Or at least can we try proving one of them? Would you like to attempt the proof? Hint is sign law is going to help you to prove it. Yes, I'm giving you a hint. So let's prove one of them. The first one maybe and use sign law for the proof. Let's see who's able to get it. Just write it down if you're able to get it. Okay, any success anybody? Okay, let's discuss it out. See, let's start with this expression A minus B by A plus B. Let's start with A minus B by A plus B. What is A in terms of the angle and the circum-circle radius? Say sir 2R sign in, absolutely right. What is B in terms of the circum-circle radius and angle B? You say sir 2R sign B, absolutely right. Similarly here also I can write it. Okay, just get rid of your, just get rid of your R. So you'll have sign A minus sign B upon sign A plus sign B. Okay, now when you get such kind of an expression, what are the first thing that comes in your mind? When you see sign A minus sign B and sign A plus sign B kind of a thing. What are the first thing that comes to your mind? Using transformation formula, isn't it? So you feel like using sign A minus sign B formula that you have learned in your trigonometric identities and sign A plus sign B formula. So sign A minus sign B formula, we all know that it is cos A plus B by 2 into, into sign A minus B by 2. Okay, denominator will become 2 sign A plus B by 2 into cos A minus B by 2. Yes or no? Right? Just cancel off this 2 into, now all of you please pay attention. This term here, this term here is reciprocal of tan A plus B by 2. So can I write it like this? Okay.  So let me write it like this. Is everybody convinced of this? So this term is 1 by tan A plus B by 2 and this term is tan A minus B by 2. Right or no? So can I now say that tan A minus B by 2 that is your numerator of the right side can be written as A minus B by A plus B times tan A plus B by 2. Okay. Now keep telling yourself you are dealing with a triangle. You're dealing with a triangle. So A plus B plus C is pi. Right? So A plus B is pi minus C. So A plus B by 2 is pi by 2 minus C by 2. So tan A plus B by 2 is nothing but tan of 90 degree minus C by 2 which is actually, which is actually cot C by 2. Correct? So can I not write, can I not rewrite this whole stuff as, instead of writing tan A plus B by 2 I can as well write cot C by 2. Okay. Hence prove this is what you wanted to prove. Isn't it? So tan A minus B by 2, tan A minus B by 2 is equal to A minus B by A plus B cot C by 2. That is what was required. Okay. So in a similar way you can prove the other two results as well and this concept is going to be very useful when we are doing the concept of solution of triangles. Okay. When we are doing sort, solution of triangles will be using this concept. How did you get the last line? Okay. See here, Sharduli A minus B by A plus B is equal to this. So send the denominator to the numerator part of A minus B by A plus B. You got there. And then you know that you're dealing with a triangle. No. So A plus B by 2 is pi by 2 minus C by 2. So tan of A plus B by 2 is cot C by 2. So this guy, this guy, I replaced it with cot C by 2. Done. Okay. Great. Let's take a small question based on the tangent law. Let's take a small question based on the tangent law. Anybody wants to copy anything from this slide? Do let me know. I can wait for a few more seconds on this. Or if you want me to drag this slide left, right, up, down, do let me know. Okay. Great. Let's take a question. Okay. In a triangle A, B, C, if your side A is twice of B and mod of A minus B is pi by 3, mod of A minus B is pi by 3, what do you think is the angle C? Please leave your response on the chat box. Any question, any, any results? Anybody? Few hints I can see in this question. A minus B is given. So just now we did something which had A minus B by 2 kind of a thing, right? So that might, that may, you know, give you a kind of a hint to start with. Very good, Shalini. So Shalini has got one response, given one response. Very good. Anybody else? Okay, Prisham. Okay. Should we discuss it? All right. See, the first thing that I would like you to, okay. Okay, Siddharth. Okay. So the first thing I would like you to answer is, which is more? A is more or B is more? Which side is more? A is more or B is more? Obviously, A is more because A is twice of B. So which angle is more? A angle is more or B angle is more? The angle opposite to the longest side is more, isn't it? Yes, correct now. So if A is more than B, does it make sense to write mod around A minus B? Because A is anyway more than B. So it is as good as A minus B only, right? Okay. So the question center may have tried to confuse people by giving mod A, but it's as good as saying A minus B is 5 by 3. Nothing else, right? Now, let us recall our tangent law. Tangent law says tan A minus B by 2 is A minus B by A plus B cot C by 2. Correct. So let us use our given expression A minus B over here. So A minus B by 2 will be tan 5 by 6. A is already 2B. So 2B minus B by 2B plus B cot C by 2. I don't know cot C by 2 yet. So I don't know C yet. So let it be as it is. So this is 1 by root 3. And if I'm not mistaken, this is B by 3B, which is 1 by 3, which clearly means that cot C by 2 is root of 3. So cot C by 2 is root of 3. That means this is equal to pi by, this is actually cot C by 2 is equal to pi by, let me just remove this, this is equal to pi by 6, right? Because cot 5 by 6 is root 3. That means C value will be, this cancel of this 2 here, C value will be pi by. So those who have answered with 60 degrees or pi by 3, absolutely right. Is this fine? I think the first one to get this right was Shalini. Well done, Shalini. Any questions, any concerns with respect to this solution? Do let me know. All right. So with this, we are now going to some more set of, you can say formulas we can call them, which are actually called the half angle formulas. Half angle formulas. I'll say, sir, wait a minute. Didn't we do half angle identities in our trigonometry? Yes, definitely we did. And of course, we're going to make use of that. But these are half angles formulae pertinent to a triangle. Okay, pertinent to a triangle. So in a triangle, we have ABC angles. So there are some formulas, there are some trigonometric ratios based on the half of those angles. So these set of formulas is they are called auxiliary formulae. A auxiliary means a formula which basically helps you out in performing many tasks. Okay. So let's look into these formulas. They come in pair and become in triads. So I'll be sharing you the triads one by one. The first triad is where I'll be talking about these three formulas, sine a by two, sine b by two, sine c by two, of course, I will give you the similar formula for cos a by two, cos b by two, cos c by two, that's second of our formula list. And in the third of the formula list, I'll give you tan a by two, tan b by two, tan c by two also. Okay, meanwhile, note this out and we'll prove it. At least one of them will prove it. Prove that or let's sort it down first, we'll prove it in some time. In any triangle, in any triangle ABC, sine a by two is given by under root of s minus b, s minus c by bc. Where what is s? Where s is your semi perimeter as we already know it. Okay. Similarly, sine b by two is s minus a, s minus c by ac and sine c by two is s minus a, s minus b by ab. Okay, we will prove only one of them because the other two will follow from there on. Okay, so let's try to prove the first one. If you can prove this one, other two will be similarly can be done. Okay, so let me take you to the proof. First, note this down and please note the pattern in which they are written because that will help you to remember the result. Okay, let's go to the proof. The proof actually evolves from your cosine law. We all know that cos of a is b square plus c square minus a square by two bc. Now let's write down cos a in terms of sine a by two. So what is cos a in terms of sine a by two? You will say it's one minus two sine square a by two. I hope you all remember your identities. Let's try to make sine square a by two the subject of the formula. So I'm directly writing it. I don't feel like writing too many steps here one minus b square plus c square minus a square by two bc whole divided by two. Which is nothing but I'll just rewrite it just rephrase the same thing so that we can manipulate on it. So it's two bc minus b square minus c square plus a square. Let me write a square in front divided by four bc. Okay. This term that you see on your right hand side is actually nothing but a square minus b minus c the whole square. Please convince yourself first then we'll move further. Everybody convince yourself that whatever I have written you are happy with it. Okay. All right. I believe everybody's happy now. So now I'm going to write this term as something like x square minus y square.  Now please understand s is your semi perimeter. Right. So we know that two s is a plus b plus c. So can I say the first term over here which I'm showing with a tick mark which is a plus b minus c can be written as two s minus two c and the second term over here can be written as this term can be written as two s minus two b. Are you all convinced with this? So if you replace that back over here, you get sine square a by two as this is two s minus two. Let me write two common and this is two s minus two b. So let me take two common here as well divided by four b c. So this two, this two, this four gets cancelled off. So sine a by two can be written as plus minus under root s minus b s minus c by b c but there will be no minus. Can anybody tell me why is that so? Why won't it be plus minus? Why only plus? Can anybody tell me? In the formula also, I didn't write plus minus. I just sort of plus under root of this. Of course, if nothing is written, it's plus but why is it plus? Why is it plus? Why is it plus? Somebody can tell me. A by two is in the first quadrant. It has to be because any angle in a triangle has to be lesser than 180. So half of that will be lesser than 90. Lesser than 90 means you are in the first quadrant. In the first quadrant, all trigonometric ratios are positive. That is why no minus business, it will only be plus. Is it trying any questions? Okay. Now again, nobody is going to ask you the proof unless until you are a PU student. Okay. So if you are in note for a PU college and only the proof might be asked. Okay. So please note this down. But anyhow, I, as I told you, I'll be asking you to prove these formulas because proving helps you to retain the result. Yes, sir. Karthik, is it fine? I mean, have I stopped at the right place or should I move still further up? Great. Is it fine? Okay. Could you scroll down? Yes. Why not? Why not? Yeah. Good enough. Okay. Great. So let us look into a similar set of formulas for cosine as well. Okay. So just like we saw sine A by two, sine B by two, sine C by two, we'll also see for cosine as well. So in any triangle ABC, the following set of formulas will hold. Cos A by two is SS minus A by PC. Cos B by two is SS minus B by AC. And cos C by two is under root of SS minus C by AB. Okay. Again, we'll just prove only one of them. That is the first one. The process is almost similar to whatever we did for proving sine A by two formula. So again, I'll start with my favorite cosine law. Cos A is B square plus C square minus A square by two BC. Write cos A in terms of half angles in terms of cos. So that's going to be two cos square A by two minus one. Okay. We already know that. Okay. Let's make cos A by two or cos square A by two, do the subject of the formula. So that will be one plus B square plus C square minus A square by two BC divided by two. Let me just restructure this term. This term can also be written as B square plus C square plus two BC minus A square upon four BC. Correct me if I'm wrong. Convince yourself first. Convince everybody. Great. So we can write this as cos square A by two as B plus C the whole square minus A square by four BC, which is nothing but B plus C minus A and B plus C plus A as per our x square minus y square formula or identity. Now I have already discussed this with you while we were doing sine square A by two formula.  So let me write it in front. So two to four goes for a toss and cos A by two would be plus minus under root SS minus A by BC, but again, again, again, again, again, no minus. Why not? Because A by two is in the first quadrant and first quadrant all technometric ratios are positive. So here we prove the first of this formula. The other two can similarly follow up, which I'm not going to do.  It follows from the same process. Any questions? Any concerns? Is it fine? Great. So why to leave tan behind when we have already done sine and cos? Why to leave tan behind? So let's write down the formula for tan A by two, tan B by two, tan C by two as well. But here we don't have to start from scratch because we already know that tan is nothing but a ratio of sine by cos. So you can just use your sine A by two divided by cos A by two formula for it. I'll just do the first one and I will replicate the results for the others.  So VC, VC is going to get cancelled off. So you're going to get under root of S minus B, S minus C by SS minus A. In the same way, I can write the other result as well. This is going to be S minus A, S minus C by SS minus B. And the last one could be written as S minus A, S minus B by SS minus C. So if you happen to remember sine and cos result, tan, even if you're forgotten, you can actually figure it out in your examination hall as well. Any questions? Of course, there should be no question because just to use of our tan as sine by cos. So if you have noted it down, can you move on to a question based on this concept? Okay, great. So let's take a question. Okay, so this is a question which, okay, let's take this question. In, if in a triangle ABC, S minus A, S minus B is SS minus C, then which of the following is angle C? Okay, a single option correct question, I will launch the poll. After doing it, please put your response on the poll as well. Okay. CZ question, I think 90 seconds, sorry, one and a half minutes or two minutes is good enough. Okay, almost one minute, 15 seconds, four of you have responded. Okay, 90 seconds are gone, but I can still afford to give you a half a minute more. Okay, we'll discuss that. Five, three, two, one, go. Okay, just straight up you have responded out of which six of you say it's option number A. Okay, and one, one, the sponsor I've got for BNC as well. Okay, let's check this. If you use the formulas that we did in the previous slide, you can solve this question in less than five seconds. How? If you see, as per the question, it is given that this is equal to one. That means this is also one. And this is a formula of tan C by two. Right, so that means C by two is 45 degrees. So C is 90 degree option number A is correct, isn't it? Even if you'll not use the formula, it'll take you some time, but still you can solve it. So if you expand it, let me do alternative method over here. Even if you expand it, you end up getting S square plus AB minus SA plus B. Okay, this should be equal to S square minus CS. Okay, S square, S square will go off. So this could be written as correct me if I'm wrong. S is A plus B minus C equal to AB. Let's multiply our two on both the sides.  So that is nothing but A plus B whole square minus C square as two AB. Expand this, this becomes A square plus B square plus two AB. And this two AB and this two AB will get cancelled off, clearly giving you the Pythagoras theorem. And this suggests that your C is a right angle in the triangle. Okay, so this is nothing but 90 degree option number A. But don't you think the first approach is much faster as compared to the second one? Is it fine? Any questions? Any questions? Any concerns? Let's take the next question. Anybody wants to copy anything from here? I can wait on this slide for 15, 10, 15 seconds. So please copy anything that you would like to. Okay, so let's take another one. Let's take another question. In any triangle ABC, if 4A by 2, 4B by 2, 4C by 2 are in AP, then ABC are in which of the following? AP, GP, HP, none of these. I'm relaunching the poll. Okay, so I hope all of you can see the poll. Let me give you one and a half minutes for this as well. Your time starts now. All right, I can see four of you have already responded. Two minutes gone. I can afford to give you half a minute more. Okay, I think I've given you more than one and a half minutes. It's almost double the time I've given you. So let's discuss, guys and girls, if at all you want to pick an option, please pick up because I'm going to end the poll in the count of five. Five, four, three, two, one, go. Okay, nine of you have responded and out of nine, four of you have gone for option B. Okay, that's a GP. Let's check whether GP was right or not. See, we had already discussed the formula of tan A by 2, tan V by 2, tan C by 2. Okay, so we can use that in solving this question because I can clearly see cot A by 2, cot B by 2, cot C by 2 here. And as per the question, they are in AP. So can I say cot A by 2 plus cot C by 2 is 2 cot B by 2? This is nothing but 1 by tan A by 2, 1 by tan C by 2, 2 by tan B by 2. Yes or no? Yes or no? Now what is tan A by 2? Tan A by 2 is nothing but S minus B, S minus C by SS minus A under root. So can I write this term as this? I hope you don't mind me writing it like this. Similarly, this term, I can write it like this SS minus C by S minus A, S minus B. Okay, equal to twice of SS minus B by S minus A, S minus C. Yes or no? Yes or no? Yes or no? Yes or no? Yes or no? Okay, now here I'm going to do a small manipulation. All of you please watch it out very, very carefully. I'm going to multiply and divide within the under root with SS minus A in the first term. So when I multiply and divide with SS minus A in the first term, this term only I'm talking about. I'm going to get this. In the second term, I'm going to multiply and divide within the under root with SS minus C. So that's going to give me something like this. Oh, I'm so sorry. I think I should have written C over here. I don't agree by mistake. Okay, this is S minus B, S minus C. Right? And I can do a similar activity on the right hand side as well, multiply within the under root with SS minus B. Yes or no? Yeah. No, why at all am I doing that? Because this will lead to cancellation of the denominators from both the sides as well as getting rid of the under root sign because there is a presence of squares on the numerator. So this is as good as saying SS minus A plus SS minus C is 2SS minus B. Am I right? Yes or no? Correct? Let's get rid of your SSS also from everywhere. So the left hand side will become 2S minus A plus C and the right side is anyways, is anyways 2S minus 2B.  That means negative A plus C is negative 2B, which means A plus C is 2B, which clearly indicates to the fact that A, B and C are in AP. Oh, Ramakrishna, Janta, you were wrong. Why option B? I think this was a result of guesswork which happens, which results into option B getting more number of votes. Okay. Yeah, so it was actually option number A, which was correct. So these these three sides are in arithmetic progression. Is it fine? Any questions? Any concerns? Please immediately highlight. All set. Anything that you would like to ask me here? I hope the transition from this step to this step was clear to everybody. Okay, great. Next thing that we're going to talk about is area of a triangle. Sir, we already know area of a triangle. Why are you teaching us area of a triangle again? Yeah, see actually, you already know many formulas for area of a triangle. At least, you know, the Heron's formula and the basic half-basin to height formula, etc. But there are many more formulas which are going to be, you know, evolving. And those evolved formulas are going to help you depending upon what scenario is given when. Okay. So I'm going to show you at least at least 10 formulas for area of a triangle. Okay. Of course, all of them will be linked. It's not that they're all different formulas. So depending upon the situation, you can use any one of these formulas which I'm going to give you in the coming few minutes of our discussion. So let us start with, let us start with our very first formula that we know. Let me make a triangle first. Okay. So let's say this is our triangle. The very first formula that you would have all learned since your class eight and ninth days, maybe even earlier than that is half-basin to height, isn't it? Okay. So let's say AM is the height of the triangle and I consider BC to be the base. By the way, please let me tell you, you could consider any one of the sides to be the base. And accordingly you can choose the perpendicular from the opposite vertex to be the perpendicular or to the height. And also note that the height of the triangle need not be completely within the triangle. If it is an obtuse angle triangle, it can be outside the triangle as well. Okay. So please do not be under the impression that height of a triangle or altitude drawn from a opposite vertex will always be within the triangle. Okay. Anyways, so the first expression for area of a triangle, by the way, I'll be using delta, okay, to show area. So half-basin to height, this is known to everybody. Okay. Now I'll be just using the same formula in a different way and find an alternative expressions or alternative set of expressions for delta now. So this is the first formula. The second formula, the same formula, half base. Now this time I will use my A, B and C, the normal way of naming a size of a triangle. So let's say if this is A, this is B, this is C. And let's consider this to be angle B. Can I say height in this case? Height in this case would be written as C sin B. Am I right? So can I say I can write the area of this triangle as half base into height, which is C sin B. Right. So there you go. I got another formula here, which is half a C sin B. Right. In a similar way, I can write the same height as B sin C. So I can say this is half base into B sin C as well. That means I can write the area of the triangle as half a B sin C as well. Correct. And not only that, if you take the perpendicular from B on to AC, you'll end up getting one more formula, which I'm not deriving over here, but I'm just giving you the result. That is nothing but half BC sin A. Yes or no? Okay. So yet another formula, half BC sin A. So on this screen itself, I have given you four formulas for finding the area of the triangle. Yes or no? But let me tell you, we are not going to stop here. We are going to come across many such formulas in our subsequent discussion. First, make a note of these. Make a note of these. Okay. Now, third formula I'm going to write is the Heron's formula. I'm going to derive the Heron's formula from this half base into height or for that matter, any one of these three formulas. So let's use any one of these three formulas. Let's use the first one. Half AC sin B. Now instead of sin B, can I not write it like this? Using my sin log, sin B is B by two R. Correct. So this gives me one more formula for Delta, which is A, B, C by four R. Please note this down. Okay. So another formula for Delta, let me write it separately. Okay. A, B, C by four R. And sorry, I was about to derive the Heron's formula. Now I'm going to derive it. So half AC sin B. I'm not going to use, first of all, the half angle formula for sin B, which is two sin B by two cos B by two. Okay. So I've broken up sin B in terms of B by two, which is two sin B by two cos B by two. And then a little while ago, I was talking about the auxiliary formula. We are going to use them as well over here. By the way, two and two can be cancelled off. So you can just get rid of the two and two from here.  Sin B by two, as we have seen in the previous slides is S minus A, S minus C by AC, and cos B by two is SS minus B by AC. Correct. I hope your formulas are fresh in your mind, which we did in the afternoon. Okay. And the same expression. Now I'm going to write it as under root of SS minus A, S minus B, S minus C by A square C square. And this A square C square, when it comes out, it will come out as an AC and AC AC will go for a toss thereby leaving me with this expression, which is very much familiar to all of you. So this is nothing but your Heron's formula or the hero's formula. Both are, that you have already learned in your junior classes. Okay. So four formulas, not four formulas, in fact, there are many more. One, two, three, four, five, six formulas we have done already for area of a triangle. And stay tuned for more formulas, which we are going to take up right after the break. So we are going to take a break right now. And on the other side of the break, we'll be discussing few more unknown concepts about triangles. And yes, we are not done with the area of a triangle yet. So many more results to come that we write it down. Many more results to come. Results to come. Okay. Stay tuned. We'll take a break now. Yeah, sure, Prism. I'll just write down the break timings here. Right now it's six or five. We'll meet exactly at 6.20pm. Is this fine enough? Prisham, is this fine enough?  So I'll leave the screen at this position. All right. See you after the break. We're not talking about circles associated with a triangle. Okay. Circles associated with a triangle. Okay. With a triangle. The first type of circle, which we have already discussed in our sign law is your circum circle. Right. So we'll talk about circum circle first. Okay. A lot of things we'll discuss about it. We'll discuss about the circum circle radius. Yeah, we'll also discuss about the coordinates of the circum center. Okay. So and we'll also talk about distances of the circum circle from the sides of a triangle, etc. So we'll talk about all those aspects here.  So if you have a triangle, how do you make a circle with circumscribes the sides of this triangle? What will you do for that? How would you make a circum circle? Anybody? Anybody has an idea? Yes. Anybody? How do you make a circum circle? See, for circum circle, you need to know its center. Right. And of course you need to know its radius. So can I say in order to know the center, you need to perpendicularly bisect the sides of the triangle. Right. Why are you perpendicular by bisecting the sides of a triangle? Because you want to end up getting a point which is equidistant from the vertices, isn't it? So it should be equidistant from B and C. Any point which is equidistant from B and C will lie on its perpendicular bisector here. Similarly, you also want a point which is equidistant from A and C as well. So you'll be bisecting A and C as well. Okay. Please note you're literally bisecting it. Okay. Perpendicular bisector. Similarly, you also want to locate a point which is equidistant from A and B. So in short, what do you want? You want a point which is simultaneously satisfying the fact that it is equidistant from A, B, C. That means you are looking for a point which is common to all these three lines or all these three perpendicular bisectors, which actually happens when they are, which actually happens at their meeting point, which is this point. Okay. And as I already told you, we call this point as S. Many times, there are places they will call it as O also, but let's call it as S. Now with S as the center and any one of them, let's say S A or S B or S C as the radius, okay. They will all be equal to capital R. As I already told you, capital R we reserve for, you know, writing the radius of the circum circle. If you make a circle, that would be resulting into the circum circle of this triangle. So let me just fit our circum circle here. Yeah, there you go. Okay. So this is the circum circle of this triangle. Find any questions with respect to circum circle. So couple of points to be noted down. Number one, the distance of circum center from the vertices are all equal. The distance of circum center, okay. Circum center from any one of the vertices A, B or C is all equal to the radius of the circum circle. Okay. That is obvious fact. Okay. What are the distance of the circum center from the sides? Let us find it out. What are the distance of the circum center, circum center, circum center from, from let's say BC. Okay. We'll also write down distance of circum center from, from AC and distance of circum center from AB. This is the which I'm giving you. If you happen to remember it will save your time because it is useful in many situations. So what is the distance of, let's say I call this point as M. Okay. And let's say I call this point as N and I call this point as let's say M and P. Let's say okay. So what is your SM distance? How will you find that out? You'll say sir, very simple. If this is angle A, this whole thing is 2A. That means half of this is A. Correct. And this is already R. So can I say SM length will be R cos A. Note this down. So the distance of the circum center from the side BC is R cos A. Sorry. R cos A. I don't know why I wrote 2. Distance of the circum center from the side AC is R cos B. And distance of the circum center S from the side AB, side AB, which is your SP length, will be R cos C. Okay. It's good to note this down and keep it in your, you know, retain it in your mind so that wherever required you can use it. Okay. Fine. Now the third important thing that we are going to talk about is, is there any relation which helps you to get the circum center, circum circle radius? Yes. There are many situations. You have already learned the sign law from where we can get this. Okay. I'm just repeating the same thing, guys. There is no new thing in this particular formula. Of course, new thing is going to come next for you all. Okay. So please note down, you can write down the radius of the circum circle in terms of the side length and the angle opposite to that side. So your R can be written either as A by 2 sine A or B by 2 sine B or C by 2 sine C. And not only that, if you remember, we had done one formula for area of a triangle where we had learned that delta is ABC by 4R. So from there also, we get one expression for the circum circle radius, which is ABC by 4 delta. Please note this down also very, very important. Now, the fourth point that we're going to talk about is the coordinates of the circum center. Okay. The coordinates of the circum center. Let me write it down. Circum center coordinates. So circum center coordinates is given by X1 sine to, let me first make a diagram. So let me refer to that diagram while I'm discussing it. So let us say we have our triangle ABC whose vertices are X1, Y1. So A coordinates is X1, Y1. B coordinate is X2, Y2. C coordinate is X3, Y3. Okay. And I want to know, and I want to find out the coordinates of the circum center. Let's say this is our circum center. I'm just marking one position. So this is our circum center. So its coordinate is given by X1 sine to A, X2 sine to B, X3 sine to C upon, upon sine to A plus sine to B plus sine to C. And same will go for the Y coordinates as well. So let me repeat the same for Y coordinates. Y1 sine to A, Y2 sine to B, Y3 sine to C upon sine to A, sine to B, sine to C. Correct. Right. Now, if you recall, I had given this formula when we were doing review of Cartesian coordinate systems. And that time I did not prove it. And I told you, and I remember that class as well, that once I do properties of triangles with you, then I will be proving this formula for you or proving this coordinate for you. However, even I'm going to prove it, let me tell you beforehand only this formula is hardly used. Sir, why are you proving it then if it is hardly used? See, my interest is in the process. My interest is not in the result. Okay. So I enjoy the means, not the result. Okay. So what is the mean and what are we going to learn from that means, which we are using to get this end result. That is more important to me. Okay. So let's try to prove this. Okay. So let's try to prove it. So here I'll be making some constructions. Okay. So and I'll be asking you a few questions based on those constructions. Let me connect a line from S to A or A to S. Okay. Let's say extended such that it meets the BC line at D. Okay. My question to everybody here, let me see how many of you are able to answer that. What is BD is to DC? Can anybody answer this question? What is BD is to DC? Anybody? Don't say one is to one, by the way. So line connecting A to S is not the median. Okay. Line connecting A to the centroid, that's the median. This is not a centroid. This is the circumcentrum. Right. So let us find it out. See, even I don't know the result right now. Maybe I'll be finding it out along with you. Okay. So how will I find that out? Any way to find out BD is to DC. Okay. Let me help you out. Okay. Don't worry. Let's say I make a small construction. I connect B to S and I connect S to C also. Okay. Correct. Now, can I say, can I say ratio of area of triangle BSD? Okay. In fact, what's the area of triangle BSD? Let us write it down that first. Okay. I'll talk about ratio first. So it's half. Can I say product of two adjacent sides, which is BS into SD into sign of the angle BSD? Isn't it? Sign of the angle. Sign of the angle BSD. Isn't it? Do you all agree with me? Okay. Similarly, area of triangle CSD is nothing but half into, into CS, into SD, into sign of the angle CSD. Is everybody convinced with these two results? Is everybody happy with these two results? Then only I'll proceed further. Now, what am I going to do with the area? That's something which I'll be telling you in some time. But first of all, is everybody happy with this? Any questions? Okay. No questions. Fine. Now, here I have a question for you. What is this angle BSD? Who will tell me? What is this BSD angle? What is this BSD angle? Okay. Now, BSC angle is 2M, my dear, not BSD. Siddharth Kartik. Okay. No, that is not A. A is when you drop a perpendicular here. This is A. I'm not talking about, let's say I call this as M. I'm not talking about BSM. I'm talking about BSD. Okay. At least I've written it over here. If you've heard me wrong, you could have seen it also. Yeah. What is BSD? Now, nothing to worry. It's not a rocket science here. See, you already know that this angle is 2C. Isn't it? Of course, I've not drawn a circle, but if you draw a circle, you would realize that this is the center of the circle. So, angle subtended at the center is twice the angle subtended at the circumference. So, how much is this? How much is this then? Tell me. Tell me. 180 degree minus 2C. Great. So, let's slide it down here. So, this is pi minus 2C. Great. Okay. Similarly, I would like you to tell me what is CSD also because I need it over here. See, I need it over here. Why do I need it? That is a different story, but at least in this area of the figure formula, I need it. So, what is CSD? You'll say that similarly, like this is your 2B. So, this will be pi minus 2B. Correct? Any questions with BSD and CSD? Okay. No issues. Great. Now, all of you please also understand that if you take a ratio of these two, okay, it would lead to cancellation of these terms. But before that, can I say area of BSD is half base and let's say this is your height, which I had erased SM. Okay. So, area of BSD is half base, which is BD into height. Let's say I call it as SM. Okay. Similarly, area of triangle CSD, CSD will be again half base, which is CD into height, which is SM. Okay. Now, I think you should be able to connect the dots. Now, look at these two figures. We get to these two figures. Okay. So, these two figures will lead to PCF. Half-half will get cancelled. Okay. SDSD will get cancelled. You have BS, which is actually R. What is BS? What is BS? BS is actually R, isn't it? CS is also R. Correct? So, BS CS will also get cancelled. BS CS will also get cancelled because they are RH. So, you are left with sign of BSD, which is sign pi minus 2A. 2A only, no? BSD. Oh, sorry, 2C. Sorry, I wrote A for a C. My bad. Okay. Why? Sign CSD. Sign CSD is sign pi minus 2B. Here, if you see what happens, half-half goes, SMSM goes. So, it's BD is to CD. So, can I say from here that BD is to CD is nothing but sign 2C is to sign 2B. Okay. So, remember, I asked you this question. What is BD is to DC? Let me use DC only. You should not change the name, actually. Whatever you're using, you should continue using it. Doesn't matter. CD, DC is the same thing. So, in the beginning, when I asked you this question, what is BD is to DC? Now, I know the answer to that. That is nothing but sign 2C is to sign 2B. Okay. So, please note this down. This is us, information to us that if I draw a line connecting from a vertex to the circum-center and extend that forward, that is going to bias, that is going to cut the opposite side in the ratio of sign 2C is to sign 2B. That's why I told process is more important. So, I learned this through this particular thing. Okay. So, any time a question comes like that in your GE exam, directly, let's say. Okay. Then, you can answer that also. You can use this as if at all it is not required or helping you out. Now, if I know the ratio in which D divides B and C, which is nothing but sign 2C is to sign 2B. Can I write the coordinates of D? We'll say yes, sir. Section formula simple. So, it is X2 into sign 2B. Now, you must be realizing, are you now understand how the sign 2B, 2A, 2C are coming? Yes or no? Okay. But by the way, this is just half the work. We have to do much more work to get the final result. Okay. So, this is your coordinates of D. Correct me if I have missed out anything and note it down if you want to. This is just half the work. We are still have, we have still long way to go. We have to find S coordinate in fact. Right now, you have found out D's coordinate. Okay. Please note this down and any questions, please do let me know. Yes. Okay. Everybody is okay. Fine. Okay. Great. So now, let's go. Let's go to the next slide. I think this slide is completely full. So, I'll go to the next slide. Again, let me draw the figure once again. Okay. Sorry, I have to draw the figure once again here. Yeah. So, this was your S. This was your A. And when you draw a line passing through AS, it meets that D. Okay. Let's maybe get in white. So, you already know now the D coordinate. Okay. D coordinate is known. A coordinate is X1, Y1. Okay. Now, if I somehow, if I somehow, I don't know how, but somehow, if I know AS is to SD, right? If I know this, I'm asking you to get too many ratios. Am I not? So, let's say if I get AS is to SD, then my job will be done to find the coordinates of S because I already know A coordinate. I already know D coordinate. Just now we derived it. If you want, I can rewrite the result. So, this is X2 sin 2B, X3 sin 2C upon sin 2B plus sin 2C. Okay. Comma, Y2 sin 2B, Y3 sin 2C upon sin 2B plus sin 2C. Correct. So, if I somehow know this guy, AS is to SD, I can use section formula and get my, get my, get my S coordinate. How will I get AS is to SD? Now, again, this is not a rocket science. Very simple. AS is anyways R. This is that AS is anyways R. Isn't it? So, we know that AS is the circum-circle radius. So, it's anyways R. So, what is SD? Okay. Now, for SD, again, I will make use of the previous construction. Let's say I call this as SM, the perpendicular drop. You already know, you already know that this angle is how much? Write down in the chat box. How much is this angle BSM? Yeah, Shalini, absolutely. This is A. And this whole thing was how much? BSD. Isn't it? What was BSD? What was BSD? 5 minus 2C. Correct. So, can I say this leftover angle, which I'm showing with a white here, this leftover angle is nothing but 5 minus 2C minus A. Isn't it? Right? So, let me write that down over here in the corner. So, angle MSD, oh, minusing Dhoni, MSD. How many of you are MSD fans? Say me if you're MSD fans. Okay.  Now, I'm going to write this pi in a very unique way. Can I write pi as A plus B plus C? Because anyways, A plus B plus C is 180 degrees because you're dealing with a triangle. So, this A goes off and one of the C's will get adjusted. So, it'll become B minus C. Am I right? Oh, Siddharth is also a fan. Okay. Right? By the way, he was my neighbour long time back. Okay. I'm also from Jharkhand, Rachi. Okay. Anyways, so this is your B minus C angle. Okay. Or you can just write it as B minus C angle. Now, very simple thing I'm going to do here. See, we already know that this length SM, we already know that SM length, let me write it over here. We already know that SM length is R cos A. We had derived that result while I was giving you the distances of the circum-center from the sides of the triangle. So, distance of the circum-center from side BC will be R cos A. And if this angle is B minus C, so a simple trigonometry I'm requesting here for you to do. I'm just making that diagram over here. This was your S. This was your M. This was your D. SM is R cos A. Okay. This angle is B minus C. So, what is the SD length? Please tell me. SD, SD, SD. So, what is the SD length? Write it down on the chat box if I'm not asking you too much to write. Yeah. What is the SD length? This is the right angle triangle. This is B minus C. Base is R cos A. What is hypotenuse? Write down, write down, write down, write down. Fast. Oh, nobody. I'm waiting for you. I'm waiting for somebody to write it down because I'm not going to write it down. I'm just waiting for somebody to write it down. I write it down. Very. So, that, are you sure? So, it's R cos A divided by cos B minus C. Yeah. It cannot be multiplied with cos B minus C because if you're multiplying cosine with the term, that term will be becoming lesser than its original value. But this is more than that. Okay. Now, SD is known. SD is known. Just now you told me SD which is R cos A by cos B minus C. So, now what are we waiting for? Let's find AS is to SD. So, AS is to SD will become, if I'm not mistaken, it will become cos B minus C is to cos A, isn't it? Or just divide it AS by SD. So, just divide R divided by R cos A and this denominator will go to the numerator. So, this becomes this. Okay. Cos B minus C is to cos A. Now, a small activity I'm going to do here. Don't get scared. Very easy activity I'm going to do here. Okay. I'm going to multiply and divide with 2 sin A. Okay. Now, you must be wondering why? I got the ratio. Why is not doing the section formula here? Why he's doing this? See, our final goal is to write everything in terms of sin 2A, sin 2B, sin 2C. Do you remember that expression of the coordinate? So, I'm basically thinking in that line. Of course, this is also correct. I could have used it. But again, I may have to do further simplification later on to get to that desired result. So, I'm doing all my simplification here itself so that I don't have to redo the same thing. So, now see here, the denominator clearly becomes sin 2A. Correct. We already know that sin A and sin B plus C are same things because in a triangle, in a triangle, A is pi minus B plus C. So, sin A and sin B plus C are same things, isn't it? Now, 2 sin B plus C cos B plus C. Just remember your formula of, I'll just write it down. Recall your formula of 2 sin x cos y. What is the formula of 2 sin x cos y? You'll see it's a simple sin x plus y plus sin x minus y. Correct. So, in light of that, it will become 2 sin x plus y, x plus y or sorry, there will be no 2. Okay. Yeah, sin x plus y, sin x plus y will be 2B and sin x minus y will be 2C. Okay. So, as you can see now, we are inching towards a sin expression. Okay. So, in light of this, can I now say that this ratio, this ratio, the AS is to SD ratio will be sin sin 2B plus sin 2C is to sin 2A. Okay. So, this question mark which I had, now that answer is known. What is that answer? The answer is AS is to SD is sin 2B plus sin 2C is to sin 2A. Yes or no? Yes or no? Yes or no? Okay. Now, since I have known the ratio in which S is dividing the join of A and D and I already know my A coordinates and I already know my D coordinates, what are we waiting for? Let's use section formula and let's get our S. So, in light of that, I mean, I'm just writing the final result. Please remember, sin 2A will get multiplied with x1 and this will get multiplied with this guy. Now, you can imagine what is going to happen. It will give you x1 sin 2A and this term will get cancelled with the denominator of this term. Right? And of course, you have to divide that operation by some of these two ratios, like how we normally do it for the section formula. So, let me do it separately so that you can understand what is finally happening. So, let us say this was our, I mean, I'm just drawing part of the figure. I'm not drawing everything. So, this was your A which was x1, y1. This was your S whose coordinate is not known to me. This was your D which was x2 sin 2B x3 sin 2C upon, please bear with me. I just, I'm just copying the same expression once again. Sorry about re-doing the same thing. And this ratio is sin 2B plus sin 2C is to sin 2A. Right? Okay. So, now this multiplied with, so coordinate of S will be x1 multiplied with sin 2A and, there's no space here. Let me write it more closer. So, x1 multiplied with sin 2A. Okay? And this term multiplied with this. This term multiplied with this will lead to sin 2B sin 2C multiplied with x2 sin 2B x3 sin 2C upon sin 2B plus sin 2C. And of course, this is going to get cancelled, which I'm going to write it in the next step and the sum of the two ratios. Okay? So, this will get cancelled with this term. And same is going to happen, similar is going to happen with the y coordinates as well, which I'm not writing. I'm just writing it did too. So, when you see this expression, it just results into x1 sin 2A x2 sin 2B x3 sin 2C upon sin 2A sin 2B sin 2C. Okay? And a same expression, a similar expression for, for your y coordinate as well, hence proved. Is it fine? Any questions? Any concerns? Okay? So, that giant formula, and as I told you that giant formula is hardly going to be used, taught us so many things actually, right? It taught us in what ratio the circumcenter is dividing the join of A and D. Of course, it also taught us how is D dividing the join of B and C. That is more important. And of course, here and there, we learned certain simplifications or certain conditional identities as well. Okay? So, this is how this formula is getting derived. Is it fine? Any questions, any concerns with respect to the coordinates of the circumcenter? So, what all we learned with respect to circumcircle? First, we learned that the distance of the circumcenter from the vertices are all R and from the respective sides are R cos A, R cos B, R cos C. We learned the various expression for R. One is A by 2 sin A or B by 2 sin B or C by 2 sin C. We also learned it is ABC by 4 delta. And we also learned that the coordinates of the circumcenter is obtained by this entire process. Okay? So, I hope everything is clear so far. Next thing that we are going to talk about next thing that we are going to talk about is your in-circle. Let me quickly move on to in-circle. What's an in-circle? Anybody knows what's an in-circle? First of all, come on. I've already told this to you in your revision of Cartesian coordinate system. What is an in-circle? What is the in-circle of a triangle? First tell me what is in-circle, then we'll talk about how do we construct it. In-circle, what are the literal meaning of the word in-circle? Circle which is inside the triangle, isn't it? So, it's a circle which is going to be inside the triangle in a particular way. That means it is going to be in such a way that all the sides of the triangle are going to be tangents to the circle. In other words, basically I'm trying to say the center of this circle is going to be equidistant from all the three sides. Remember, in case of a circum-center, the circum-center was equidistant from all vertices. Here, i is equidistant from all sides. That means the distance of i from AB is same as distance of sides from AC, is same as distance of i from BC. Now, think a little bit in terms of locus and tell me, how would you find out the position of i if the triangle is known? If I give you the triangle, how do you get to the position of i? What construction will you do to get i? Tell me, tell me, tell me. How would I get to i? Let's say ABC is given to you. Okay, somebody has asked you to locate a point which is equidistant from all the three sides. What construction you are going to do? Tell me. Nobody or are you taking time to type it out? Which of the two cases is applicable? Are you taking time to type it out or you do not know the answer? I hope it's the former one, not the latter one. Okay, let me ask you a simple question. Forget about this. If I give you, just answer a simple question. Let's say this was AB and this was AC. Okay, AB and AC. I'm asking you to draw a construction or to draw the locus of such points which are equidistant from AB and AC. What will you do? You'll say, sir, bisect this angle, isn't it? If you bisect this angle, every point on this, let's say if I take any point, its distance from AC and AB will be equal. So it will be equidistant. Any point you take, isn't it? So now, here also I want a point which is equidistant from AB and AC. So it must lie on the internal angle bisector of A, correct? And at the same time, I want the point to be equidistant from AB and BC. So it must lie on the internal angle bisector of B. And at the same time, I want the point to be equidistant from AC and BC. So it is also lying on the internal angle bisector of, okay? By the way, when I am drawing an internal angle bisector, it's not perpendicular to the opposite side, okay? So this yellow line that you see, it is slightly visible from behind. So they may not be concurrent, okay? Sorry, they may not be coincidental. Yeah. So can I say the position of I can be known if you internally bisect the angles of a triangle? So A by 2, A by 2, B by 2, B by 2, and C by 2, C by 2. Now, please understand, the circumscentre was obtained by bisecting each sides. The in-circle centre, which we call as the in-centre. I is called the in-centre. The in-centre is obtained by the meeting point of the internal angle bisectors. Please note this now, okay? Is there any question with respect to the in-centre? Okay. Now we'll talk a lot of things about in-centre. Don't worry. One by one, we'll talk about it. The first thing that we are going to talk about is what is the coordinate of an in-centre? So again, let me make a triangle. Let me make a triangle. Let's use a green colour font here for making a triangle. Let me not make it appear like equilateral. Many times, we tend to draw triangles which look like an equilateral one and we start taking decisions based on that. So we are tricked by our own construction. Okay. So you internally bisected this. That's it. Okay. You internally bisected this. Okay. And of course, the internal bisector here also will pass through the same point, which I need not draw. Tell me, in terms of the length of these sites, let's call this as ABC. So this is a small b, small c. Okay. In terms of these small ABC and X1, Y1, X2, Y2, X3, Y3, what is the coordinate of I? In fact, I will give you the result and we'll prove it. So the coordinate of I is given by, note this down, AX1 BX2 CX3 by A plus B plus C, where your ABC have the usual meaning that we have discussed, comma same thing for Y as well, AY1, BY2, CY3 by A plus B plus C. Okay. Now the question is how do we get this result? Now, I don't know whether I have discussed this in the Cartesian coordinate review of Cartesian coordinate system chapter with you, but still I would request you to do it once again. So how would you get this result? So I'll be doing it very quickly for you in the interest of time. Let's say this point is again D point. So what is BD is to DC? At least this you answer, BD is to DC. I understand the previous one was a difficult one, but now what is BD is to DC? In what ratio does an internal angle bisector divide the opposite sides? What is BD is to DC? Who will tell me? Nobody. It's the length AB is to AC, absolutely right Siddharth. So BD is to DC is nothing but C is to B. Okay. So this issue is, this issue that we write it in different font, C is to B. So what is the coordinate of D? So this implies coordinates of D would be BX2 CX3 by B plus C, at least you are not inching towards that result which you want to achieve. Same with the Y. So I will not waste time writing that down. Okay. So this is your coordinate of D. Now, if I know AI is to ID, I can find out the I coordinate because now I'm ready with D and A was already given with me, given to me. So what is AI is to ID? Who will tell me? AI is to ID. Now, let's find AI is to ID. Tell me, tell me, tell me, tell me. You'll say, sir, same thing, sir. This angle is also bisected. So AI is to ID is same thing as, you can say BA is to BD or AB is to BD. Correct? AB is anyways a C and what is BD? What is BD? If this whole thing was A and this ratio was C is to B, what is BD? Can I say BD will be C by B plus C times A? Yes or no? In other words, that ratio will become B plus C is to A. Am I right? Correct me if I'm wrong. Have a good look at it, convince yourself, then I'll move forward. Can I convince everybody? So AI is to ID is as good as AB is to BD. AB is already small C and BD length will be C by B plus C times A. Okay. Anybody who knows ratio and proportion can easily get that. Okay. It's not a rocket science. They just need very simple question. This length is let's say 5. Okay. And let's say there's a point which divides it in the ratio 1 is to let's say 7. So what is this length? Tell me. 1 by 8 times 5, no? Same thing, it's C by B plus C times A. Simple. Okay. Anyways, so let's erase this fellow. Yeah. So now let me read all the figure once again. No, figure is already there. So I'll not paste my time. So this is already, this is already, please note down B plus C is to A. Right. So there is a line. There is a line. Okay. AID where A is X1, Y1. D is BX2, CX3 by B plus C comma BY2, CY3 by B plus C. And this ratio AI is to ID is B plus C is to A. So what's the coordinate of I? We'll say it's a simple section formula. So A into X1. So I coordinate will be, where should I write? Yeah, I'll write here. So I coordinate will be A into X1, B plus C into this. B plus C into this will be leading to cancellation of B plus C and of course some of A and B plus C. Same will be true for Y as well. Okay. I'm not writing everything down. Okay. Hence proved. Is it fine? Any questions? Any questions? Any concerns? Okay. So A coordinate is known, X1, Y1, D coordinate is known, which is this, this ratio is B plus C is to A. You can use your section formula and you will end up getting the same result that we had discussed. Is it fine? Any questions? Okay. Second important thing that we are going to discuss here. By the way, do you want to copy anything from this slide or should I move on to the next one? I hope you have all copied things, required things. Okay. So let's move on to the next one. The next thing that we are going to talk about is the length of the internal angle bisectors. So since we are talking about in-center, which is obtained by the meeting point of the internal angle bisectors, it's natural that we actually also learn the length of the internal angle bisectors, length of the internal angle bisectors. Okay. Let's try to find out the length of the internal angle bisectors. In fact, I'll give you the length expression and we'll derive one of them. Okay. That would be more fruitful exercise. So let's say this is an internal angle bisector of A. So it splits this angle A as A by 2, A by 2. Okay. Let me use a different color to show the internal angle bisector of B. Please note it might appear as a median, but it is not. Okay. So median is something else. This is an internal angle bisector. And let me show you the bisector from C. Okay. This is C by 2, C by 2. They're all concurrent at I. Okay. Anyways, we don't need that I. Let me name it. Let's say I call this as AK, BL and CM. Let's prove this that AK length will be 2 BC by B plus C cos A by 2. Okay. BL length will be 2 AC by A plus C cos B by 2. And CM length will be 2 AB by A plus B cos C by 2. Any one of them will prove it. We're not going to prove all the three. Let's prove the first one. Can you find out and can you prove that the length of the internal angle bisector from A is going to be 2 BC by B plus C cos A by 2. Okay. Would you like to try it or should I do it for you? What do you suggest? You want to give it a try? Okay. One minute you can try it out. Just try it out for one minute. Yes. Any success? Okay. See, I'll help you out with this. Let's find out AK length. Now just answer a few questions that I'm going to ask you. If let's say AK length BX. Okay. And a drop of perpendicular from K on to AB. Okay. Let's say, let's say I call this as KP. Let's say, okay. What do you think will be KP length? So you'll say very simple, sir. You have already called this as X. Okay.  So this is X and this is A by 2. So KP length will be X sin A by 2. Am I right? Everybody agrees with that. KP length will be X sin A by 2 if AK length is X. And if I'm going to drop a perpendicular from this on to let's say AC. Let's say I call this as Q. Then KQ length will also be the same. Am I right? Will there be any difference in the length of KP and KQ? Now wait, I'll tell you how am I planning to use those results?  All right. No issues? All are happy? Okay. Great. Now I'm proposing that the area of the triangle ABC, that is the whole triangle, is some of the area of the triangle ABK and area of the triangle AKC. You'll say sir, obviously yes. ABC whole area is some of this too. ABK and AKC. Okay. Now area of the whole triangle, let's say I plan to use half AB. Let's say I plan to use this formula, half BC. Half BC sin A. Remember I had given you several formulas for area of a triangle. One of them was half BC sin A. So I'm using half BC sin A because of course A angle is involved. That is why I want to involve the angle A in that. I could have used other formulas also, but in that case I would get a different expression. So if I involve this angle, I'll get half BC sin A because see, I want to get to this result. So angle A must be involved somewhere. Now what is the area of the triangle ABK? Let me show you the complete figure. ABK, it's half base into height.  So it is going to be half small c into height. What is height? Height is Kp. Kp, Kp, Kp, Kp. Kevin Peterson. Yes. So Kp will be x sin A by 2. Similarly for AKC, I can say half base. Base will be this length which is B for you and perpendicular will be Kq, Kq which is also x sin A by 2. Yes sir.  Half, half, half gone. Okay. Now you write sin A as 2 sin A by 2 cos A by 2. Let me just copy forth. In fact, I can do one small activity here. I can take x B plus C sin A by 2 common. Yeah. So I'm just taking x out and we'll have B sin A by 2, C sin A by 2. So B plus C also have taken common.  So it'll be 2 BC cos A by 2 divided by B plus C. There you go. Hence proof. Now sir, are we supposed to remember this result? Yes. Unfortunately, this is going to be useful in solving many questions.  So this derivation you will not do every time. It's very difficult to derive these results in the heat of the air. So please note this down and there is a pattern that you can actually remember. So if it is a perpendicular, if it is an internal angle bisector of A, you'll realize there is a B and C term involved here and this also many people learn it like this. This is the harmonic mean of B and C. Isn't it? 2 BC by B plus C is the harmonic mean of B and C and whichever angle is bisected, cos of half of that angle. That's how I can recommend you to remember it. Okay. In the same way, you can also find out the length of internal angle bisector of B, which is 2 AC by A plus C cos B by 2 and internal angle bisector of C, which is 2 AB by A plus B cos C by 2. Okay. So note this is down. I'm staying on this page for a few more seconds. Okay, done. Great. So the third thing that we are going, I think this was the length of the internal angle bisector. Now we are going to talk about expression for in-circle radius. In-circle radius is given this symbol, small r. Capital r is your circum-circle radius. In-circle radius is small r. Fine. So there are a lot of formulas for small r, which we are going to derive one by one. Okay. So the first formula that we are going to talk about is delta by s. Okay. So the in-circle radius is given by area of the triangle divided by, this is your area of the triangle, divided by its semi-perimeter, semi-perimeter of the triangle. Okay. How do we get this result? Let us derive that first. And by the way, this is yet another expression for area of a triangle. Area of a triangle is, as I told you, there are many more results. Right. So this is another one. Area of a triangle is given as the product of in-circle radius and its semi-perimeter. Okay. But anyways, let's derive this result anyhow. What is the proof for this? How do you claim that or how do we claim that in-circle radius is area of the triangle by its semi-perimeter? Okay. Let's derive that result.  Yeah. Let's say this is our triangle and this is our in-circle. Okay. I'm just showing the in-circle here. Better to draw a circle that would be more prudent way of showing it. Wow. Actually, it's not, you know, easy to fit a circle within a triangle. It's easy to draw a triangle around a circle though. Yeah. So this is not the center. This is the center. Yeah. By the way, the perpendiculars from the center eye to the sides are all the in-circle, are all, are the in-circle radius. Okay. So I hope everybody knows this. This is obvious actually. There's nothing to know here. Yeah. Now, since area of the triangle is involved, can I say one thing? Area of the triangle ABC is some of these three areas of the triangle. As you can see on your screen, I have divided the area or I've split ABC into three triangles, which is A, I, B, B, I, C and C, I, A. Okay. Isn't it? Obviously, area of the whole triangle is some of this area, which I'm showing with yellow, some of this area, which I'm showing with white and some of this area, which I'm showing in blue. Correct? Yes or no? So what is the area of the triangle, delta, which we already know? What is the area of A, I, B? Now, all of you, please pay attention. A, I, B, the perpendicular is R, base is C. So it is half C, R. Isn't it? What is the area of B, I, C triangle? Again, half base into height, which is half AR, and area of C, I, A will be half BR. In other words, if you take an R common, you have A plus B plus C by 2, which is nothing but the semi perimeter into R. In other words, from here, we can derive that your in circle radius is going to be delta upon S. Please note this down and commit to your memory because it is going to be very, very heavily used. Okay? If you want me to drag the screen to the right, yeah. Please copy this down. Good? Okay. So this is not the only formula for the in circle radius. There are many more formulas. In fact, there are two more formulas, which I'm going to discuss with you in this class itself. The next formula that we are going to talk about is this. R is S minus A tan A by 2. And it is also S minus B tan B by 2. And it is also S minus C tan C by 2. I must be wondering, are you three more formulas in one shot, sir, as given? Okay. So we'll prove them. We'll, we'll prove them very simple, actually. Let me make the construction once again. That's more important. So let's make let's make the construction. This is our, I normally draw the circle first.  So let's say this is our triangle ABC. And this is our in circle. And this is your I. Okay. Now, let's say this is the perpendicular is dropped from I. Okay. So we already know that this line, this angle, is A by 2. I'll just prove one of them. The first one, we already know this is A by 2. Okay. This is your R. Now, let us call this point as KLM, the meeting point of the, or you can say the normals, the perpendicular is also called normals. You will learn later on. The perpendicular dropped from I on to the sides are KLM respectively. Okay. Now, will you all agree with me if I say AM length and AL length will be equal to each other? Let's say I call them as XX. Why they will be equal? Why will AM and AL will be equal? Let's say X. Can anybody tell me why? I made one statement that AM and AL length will be equal. Why will they be equal? Right? They are tangents. No, their tangents don't form external point on to a circle. Remember that length of the tangents don't form any exterior point to a circle are equal. So AM and AL are equal. In the same way, can I say BM and BK will be equal? Let's say Y. Okay. Let's say Y. Let's say I'm just calling it as Y. Similarly, CK and CL will be equal. Let's say Z. Okay. Now, we all know that the perimeter of the triangle AB, BC, CA will add up to 2S. Right? Now, what is AB? AB is X plus Y. BC is Y plus Z. CA is Z plus X. So I'll end up getting 2X, 2Y, 2Z equal to 2S. Correct? Which is as good as saying this is equal to S. Yes or no? Now, let's look at, let's look at a very small, you know, let's say I talk about BC. Can I say BC is your Y plus Z? So this is your BC length. Correct? And BC length is actually your A length, which means X value is S minus A. Isn't it? So here, this X that I wrote, this was actually S minus A. Okay. This also is S minus E. By the way, just to tell you other results, this is S minus B. This is also S minus B. This is S minus C. And this is also S minus C. Okay. So if somebody asks you that what is the distance between a vertex to the point of contact of the in-circle with the side, then you know these, you know, values, S minus A, S minus B, S minus C. Now, focus on the triangle, focus on the triangle AIM. AIM. If this is R, and this is S minus A, can I say tan A by 2 will be opposite by base, which means R is S minus A tan A by 2. That's it. This is what we wanted to prove. Hence done. Okay. In a similar way, you can also get the result for R in terms of S minus B and tan B by 2 and similarly for S minus A tan C by 2. Isn't it? So please note this very, very important result. Okay. So the radius of the in-circle can be written in terms of the semi-perimeter and of course angles as these three. Is it fine? And just to, you know, wrap up the in-circle radius formulae, one more formula is there. I would not take it in the next class. I will take it right now. It's itself here. The third formula in-circle radius is also given in terms of the circum-circle radius like this. Very important formula. I will redo this concept in the next class also, but this is very, very, very, very, very important formula. I know we'll be discussing a few things about it also in our next class. But before I stop today's session, I'll be quickly deriving this result also. In fact, I'll be, you know, using this expression, a right side expression to come to the left side expression. Okay. So how do I prove this? Very simple. Let's start with our RHS. Okay. So RHS is for R sine A by 2, sine B by 2, sine C by 2. Let's use our half angle formula for sine A by 2, which is nothing but S minus B, S minus C by BC. Let's use half angle formula for sine B by 2, which is S minus A, S minus C by AC. And let's use half angle formula for sine C by 2, which is S minus A, S minus B by, by, by, by, by, by AB. Okay. If you simplify this, if you simplify this, you are going to get within the under root sine S minus A square, S minus B square, S minus C square. And you'll also get A square B square C square. Correct? Yes or no? I hope everybody is able to follow till here. Correct? No? Okay. Let's do a small manipulation. Let's multiply with S square within the under root and compensate by dividing by S outside. Okay. Now let's see, something interesting is going to come up. This is nothing but, this is nothing but square of this term, S, S minus A, S minus B, S minus C. And this is nothing but ABC because under root, within the under root, you had A square B square C square. So it'll come out as ABC. Now, what is 4R divided by ABC? If you remember, we had done an expression ABC by 4R as delta when I was doing the area of the triangle. So reciprocal would be 4R by ABC. So this is 1 by delta. Okay. There's already an S over here. And what is this? This is area of a triangle square. So delta and 1 delta will get cancelled off. Right? And what is delta by S? Just now, we figured out that is small r. There you go. So this is an alternate expression for in-circle radius in terms of circum-circle radius. That is why this formula is very important. We'll talk more about this formula in our next session as well, which will be the last session for properties of triangles. Okay. We'll try to wrap up that chapter in that session. Meanwhile, thank you. Bye-bye. Take care and get yourself vaccinated, because I think the government has already started vaccination for 15 to 18 years old. So please get yourself vaccinated and do take good care of yourself.