 So good morning, everyone. So I'll be starting with a quick revision on properties and solution of triangle So first I'll start with properties of triangle Properties of off triangle. What is solution of triangle solution? The word solution is used when you want to find out something like solution of an equation So you're finding out the roots of the equation in the same way solution of a triangle is when you're trying to figure out a Triangle from its certain known parameters That is something which we'll do in the second part of our discussion today So properties of triangle there are certain theorems laws which we need to know most of it you already know So I'll be slightly faster with respect to that The first thing is the sign rule or the sign formula Sign rule or the sign formula and do let me know if you want me to do I will of course derive a lot of formulas here But if you need the derivation of basic formulas also do let me know Okay, so sign rule says a by in any triangle a by sine a is equal to b by sine b is equal to c by sine c However, everybody here knows how a triangle is basically named So when you are trying to name a triangle that's a distort this The side opposite to angle a is called small a the side opposite to angle b is called small b side opposite to angle c is called small c So in any triangle this law is going to be true and this ratio is actually to our Okay, our is the circum circle radius circum circle radius Okay, the derivation of it is very simple. Let's say this is your circum center Okay, and if you I'm not drawing the circle over here, I hope you can imagine that there is a circle which is passing like this Okay, so if you drop a perpendicular remember this angle is a So this whole angle is to a so this angle is a each Okay, and this length. Let's say I call this as Ob is our OC is also are Okay, and we know that the circum circle center is Obtained by perpendicularly bisecting the sides. So this length and this length will be equal to a by 2 each Okay, so on this triangle, let me call this as obm. So in triangle obm if you apply a Basic trigonometry, you will realize sign of a becomes a by 2 by r So it's a by 2 r. So 2 r becomes a by sine a Similarly by dropping a perpendicular on AC and AB you can figure out the other two ratios as well. So this is your sign rule cosine rule cosine rule says that in any triangle These three formulas will be valid. So any of the three formulas Can be asked The again the proof for this is very simple. So I'll just make a triangle over here. Let's say this is your angle a b See, let me drop up perpendicular This is small c small b small a now if you look at the Triangle, let me call this as p if you look at the triangle AB p This side length would be actually C cos a Okay, and this side length will be C sine a Right so the CP length which will be left off will be B minus C cos a and Since this is also a right angle triangle So in triangle BPC you can apply Pythagoras theorem. So C sine a Square B minus C cos a square is equal to a square if You expand this if you expand this you'll get B square You'll get minus 2 BC cos a and you'll get C square cos square and C square sine square Which will be actually a C square Okay, so from here you can make cos a the subject of the formula You'll automatically get B square per C square minus a square by 2 BC So that proves the first part of the formula. Similarly, others can also be proven Okay, any question here because I'm slightly faster here if you feel that you want explanation for something Please feel free to stop me Especially HSR people can you drag up Okay Because of the sign rule you can write your side in terms of angles like this. So this is a connection between the side and Angles Connected by the circum circle radius So this is very important very highly used Many books will write k for 2R because sometimes they will get cancelled. So you don't need them exactly Okay, can I go to the next page? Yes, sir Next is your projection formula projection formula we said because they come in Triads so projection formula says in any triangle C is equal to a cos B plus B cos a B is equal to a cos C plus C cos a and A is equal to B cos C plus C cos B. It can be easily be proved. So let's say this is a Triangle ABC Drop-up or pendicular from a on to BC Okay, now you can see that You can see that this length BM Let's say this is angle B. This length BM is C cos B because of the triangle ABM and This length MC would be B cos C because this angle is C And if you use this triangle AMC, this will be B cos C So together if you add it that is BM plus MC It makes your and is makes your side a and hence a is nothing but B cos C plus C cos B Which groups the C part of your formula and similarly others can also be proven Okay Any questions here? Any questions in this case? Okay, so Let's say we take a problem So whatever we have done so far We'll take a question on this in a triangle angle is 55 degree angle B is 15 degree angle C is 110 degree Then C square minus a square is which of the following options I'm going to run the poll for this Yeah, I'm I get every morning. I have it Not a new thing for me Okay Coming cold also This is nice So quite an old concept for all of you trigonometry is something that you have done Almost one one-half years back But yes, you constantly deal with it While you're solving other topics also Alright, we'll not have we'll not give a lot of time to such questions. Two minutes is what is expected I've just got four words so far We'll close this in another 30 seconds As I've already discussed we have a lot of concepts to cover five four three two one Go okay Most of you have gone with option number a let's check See we already discussed that I can write a C as to our sign C and I can write a as to our sign a as per our sign rule, isn't it? So C square minus a square I can write it as for our square sign square C minus sign square a Correct, we already know that the sign square C minus sign square a formula is Basically nothing but sign C plus a into sign C minus a Okay, don't forget your basic identities Okay, so sign of C plus a would be a sign of C plus a will be 165 degrees 165 degrees and sign of C minus a would be sign of 55 degrees sign of 55 degrees Okay now sign of 65 65 degrees and sign of 55 degrees Can we write this in this way? sign of 15 degrees into sign of 55 degrees because sign of theta is sign 180 minus theta So you can split up to our sign 15 and to our sign 55 To our sign 15 will automatically become your B and to our sign 55 will automatically become your a Yes, I know so this becomes a B Option number a is correct Okay, is it fine any questions any concerns? Okay, let's take this all right. So this question is a Circle of area 20 square units is centered at O Suppose triangle ABC is inscribed in that circle and has an area of eight square units The central angles alpha beta gamma are as shown in the figure Find the value of sine alpha sine beta sine gamma sine alpha plus sine beta plus sine gamma Again, let's have a time limit of two two and a half minutes not more than that All is on good good Okay, our last 20 seconds five four three two One go All right, most of you have gone with option number B B for Mangalore Let's check See if you want to find the area of the triangle ABC or if you want to relate the area of the triangle ABC Let's figure out the area of each of these small small triangles which comprise that triangle So let's say this triangle if I talk about This triangle will have area half into Now in a triangle, we know that if I know the two sides And a sandwiched angle Okay, then area of such a triangle is given by half XY sine of the sandwiched angle, right? This is something which is supposed to be known to you So this area will be half R square sine beta correct similarly This area which I'm shading with orange that would be half R square sine gamma and This angle which I'm shading with Blue that would be half R square sine alpha Together they will make up your Together they will make up your bigger triangle ABC. So I can say half R square sine alpha sine beta sine gamma Will make up your triangle which is having a area of 20 square units So eight square units eight square units sorry eight square units And we have been given the area of the circle also so I can find my capital R square from there So pi capital R square is 20 So R square is 20 by pi So if you put it over here 10 by pi times sine alpha sine beta sine gamma is given to us as eight That clearly means sine alpha sine beta sine gamma has to be four pi by five Four pi by five Option number a is correct which did not get enough votes. I think a only got 17 percent of the votes Is this clear any questions? So with this yes, sir So one second Sure, then okay Next is your tangent law. So sine lock or sine law and now the tangent law or tangent rule Whatever you call it tangent rule also called as the napier's analogy napier's analogy and This is basically us not a very important concept, but you will find its application To a certain extent in the solution of triangle spot So as of now, I'll give you what are the rules So it says that tan of a minus b by two Is a minus b by a plus b cot c by two Similarly, you can say tan of Let's say if you want to write b minus c by two you can choose any difference of the angle over here by two But accordingly, this will change. So b minus c by b plus c and Cut off whatever angle is left off. So let's say a is left off. So a by two Similarly, if you can write a tan of a minus c by two So here you'll have a minus c by a plus c and whatever angle is left off by two Okay, this proof is very easy. Just one of them. I'll prove it very quickly. It comes from the sine rule only So if you just talk about If you just talk about the first formula a minus b by a plus b As for the sine rule, I can write it as two r sine a minus sine b By two r sine a plus sine b And you can use your Transformation formula sine a minus sine b is two cos a plus b by two into sine a minus b by two Similarly two sine a plus b by two cos a minus b by two If you cancel this off On the right-hand side You clearly see tan a minus b by two Divided by tan a plus b by two Okay, so if you make tan a minus b by two the subject of the formula it becomes a minus b by a plus b times tan a plus b by two Now we we are dealing with the triangles. So we know a plus b plus c is 180 degrees So a plus b by two is 90 degree minus c by two So tan of a plus b by two is going to be caught off c by two So this thing over here you can replace that with caught off c by two and there you go So this gives you the desired rule a minus b by a plus b into caught off c by So we'll not take any question on this we'll take questions when we are doing the solution of triangles Next thing that we are going to talk about which Probably most of you would not have done so far Is your concept of Let me take one question. I think one question I can take let's say Um In a triangle in a triangle a b c Side length a is two times side b and it is known that Mod of the difference of angle a and b is pi by three Question is find angle c question is find angle c You still remember that Sir, how could we forget those tortuous moments? Yeah, just a simple question on the tangent law See if the time permits will take a lot of questions, but my immediate concern is to let you know all the wherewithals Very good. Very good. Uh, shomic Good richer Yeah, so if you use your formula directly you can get this in one shot now see The the moment they say a is to be you know that The side length a is more than the side length b So in a triangle the side which has got the longest You can see the the greatest side will have the greatest angle opposite to it So angle a has to be greater than angle b because of this relation And because of this the mod of this will change just to a minus b Okay, so mod here will become redundant because side a is more than side b So angle a also has to be more than angle b, right? So now we can directly use our, uh, Napier's analogy here So this will become a minus b by a plus b cot of c by 2 This is already Pi by six And a is already 2b Okay So pi by six is one by root three and this guy is going to be one by three cot of c by two So from here we get tan of c by two is equal to One by root three So c by two has to be pi by six. So c has to be pi by three Is this fine any questions here any concerns? all right, now we go to the Tignometric ratios for half angles what we call as the auxiliary formula Why does call the auxiliary formula because it comes from your basic laws only which you have already studied The sine null cosine null etc so Let's have these formulas as well. So let me start with Half angles of sine of Angles a b and c Okay, so sine a by two is under root of s minus b s minus c by bc This is given as s minus a s minus c by ac so ac And this is given as s minus a s minus b by ab Uh easy way to remember this is in sine you realize whatever angle is here The side which is not corresponding to that angle is present over here bc bc Since angle b is here ac ac like this, okay Uh Let's prove this one of them. I'll prove it very quickly. Uh, let's say I want to prove the first one So it comes from your cosine formula So you all know that cos of a is b square plus c square minus a square by two bc Okay, cos square a is known to be one minus two sine square a by two correct So let us make sine square a by two the subject of the formula so it'll become It'll become one minus that is I'm Bringing it to the Left side and half of this and half of this Okay, so this becomes nothing but two bc minus b square plus c square plus a square By four bc. This is equal to sine square a by two Now this term over here is actually a square minus b minus c the whole square. Check it out correct This is sine square a by two So if you apply x square minus y square formula on the numerator It becomes a plus b minus c and it becomes a minus b plus c by four bc Sine square a by two now we all know that s is the semi perimeter of the triangle Okay, so two s is a plus b plus c So a plus b minus c is equal to two s minus two c So two s minus two c So this will be similarly two s minus two b by four bc sine square a by two Okay, so this will become s minus b s minus c by bc And take the under root without having to take plus minus because in any triangle sine a by two will be positive This will always be positive. So don't have to take plus minus Uh symbol over here. Just a plus will be sufficient. Okay Similar see why i'm telling you the derivation is because formula may ditch you some time and let's say in very dire situation You may need to derive it. So you should know from where it is coming Okay I can play the role of giving the formula for you, but then it will be no difference for you for you to attend this class and see uh Probably a Arihan formula that black book. So all formulas are there also Okay, so this is how this formula is derived Is this to be remembered? I would say yes because it will save a lot of time for you in the exam if at all you need it So make a note of this This is for the sign of the half angles Similarly, there will be a formula for the cost of the half angles, which I'll be giving you on the next page Have you copied this? Can I go to the next page? Okay So second set of formula is cost of uh A by two cost of b by two cost of c by two So cost of a by two is under root of s s minus a by bc Okay Now the way to remember it is It should have the same denominator as the sine a by two term and whatever angle is present is the same angle present over here Now most of you would be thinking that itself is a formula for me Okay, so similarly s s minus b and ac Or you can say whatever is not present Whichever angle is used the side which is corresponding to the other two angles that is has that has to be present in the denominator Similarly s s minus c by a b Now the proof is exactly similar as what we did for the previous one Instead of writing cost a as one minus two sine square a by two Use this formula. I'm not going to derive it. I'm just going to write the expression So in this formula All you need to do is change your cost a with two cost square a by two minus one Okay, and from here you proceed you will end up getting your cost a by two expression like this Okay, so please proceed as a homework question on this Is this fine and similarly we have a formula for tan of half angles as well First make a note of this all right, so tan half angle is You don't have to do anything extra over here If you know your sine half angle and if you know your cost half angle, this will be automatically be there So sine half angle was sine a by two was sine s minus b s minus c by bc and cost of a by two was s s minus a by bc Okay, so bc and bc will go off and you'll end up getting s minus b s minus c by s s minus a Now This is something which many people also remember it as delta by s s minus a Okay, how very simple all you need to do is within the brackets you multiply with You multiply with s s minus a so this will become s square Okay and we all know that We all know that this is the This expression is delta square By heron's formula So it becomes delta square by s square s minus a the whole square under root So this becomes delta by s s minus a Okay, from here itself you will get a formula which we'll do in in some time and that formula is tan a by two Delta by s is actually known to be small r That is your in circle radius So this formula comes From here. Okay, so they're all interlinked. This will do later in today's discussion. Don't worry Is this fine? Similarly without much adieu we'll write down the formula of tan b by two So tan b by two formula is under root of Under root of s minus a s minus c by s s minus b And tan of c by two formula is under root of s minus a s minus b by s s minus c Okay, so we'll take a few questions where we'll be learning how to apply them Sir, can you direct the right? Yeah, sure Uh, sir. What did you say r was here? Oh r r was the in circle radius in circle radius So basically when you have a triangle and you inscribe a circle within it that circle is called the in circle So this circle will have all the sides of the triangle being tangent to it And such a circle will have a radius which is called small r by convention. We call it as small r Okay, sir. This is that small Rajaji Nagar, uh, I'm sure I have done this chapter to a certain extent trippan. Do you remember? Trippan shomic Oh Some of you used to go off. Yeah Okay, anyways, we'll take some questions In a triangle if tan a by two tan b by two plus tan b by two tan c by two is two third Then a plus c is which of the following options Let me start the poll Has anybody joined late? Oh wow, that was fast Okay, okay Okay, Karthik. Okay, Anjali Sir, you are taking our names sir and telling everybody that we joined late Yes, which is absolutely correct. You're Richard Singh done. Yeah scamming is always welcome. What sir? You're corrupt corrupting us scamming is Always welcome Sir, in mains will questions be scammed? Ah, 10% question will definitely be scammable And I would suggest everybody to first start scamming only. Okay, because be smart I mean, there is no credit attached to solving a problem. They're rigorous with Scam at least you'll get to know which could be your one of your answers, right? So If you do it properly, you will not be able to ever score more than 200 marks. Okay Scamming only can make you touch to 30 to 20 to 40s like that Richard and Sukirti used to scam like anything Not Richard. Uh What is it? Bharat and Sukirt Huh, try the scam first. That is what I also suggest first thing is scamming Many questions like matrices and all will be like you can put values When you can solve the question in five 10 seconds by using scamming why why you need to solve it in a proper way? And knowing the way is okay. Good. You have that in your armory, but don't start using that initially itself Okay, guys, I'll stop the poll three minutes already done option B is what people say So most of you have gone with option B. See, uh, since I'm teaching you this I'll not scam here But I'll like to recall, uh, some conditional identities for you Some of them are very important for a triangle. So when you know that you're dealing with a triangle However, uh, this is true for any n pipe Okay, but you can always use it for a triangle as well. So in any triangle tan a tan b tan c Is nothing but tan a tan b tan c. This is a very important conditional identity, which is true for a true for a Uh case of a triangle And one more is there and think there are many more but these two are very important So tan a by two tan b by two tan b by two tan c by two tan c by two tan a by two What is this? Can anybody tell me? This is always equal to A1 yes Yeah, so this is your very important conditional identities which you need to keep in mind very important conditional conditional identities Because they're true under a condition Now from here These two information is already given to me as two third Okay, so this has to be one third Okay, so now if you know, this is one third you can use your tan c by two formula tan c by two formula will be what? s minus Tell me s minus a s minus b By s s minus c Similarly tan a by two also i'll include over here So s minus b s minus c by s s minus a Okay I believe a lot of terms will get cancelled off this will go with this this will go with this and you'll have And you'll have s square and s minus b. So we'll have s minus b by s equal to one third Equal to one third Okay, so three s minus three b is s. So two s is equal to three b Okay, two s was a plus b plus c is equal to three b So if you drop a b you end up getting the desired answer which is a plus c is equal to two b option number b is correct Okay, you can always guess an equilateral triangle and do it in a much faster way That is always welcome Is it fine any questions here? So you could do it without using these two identities Yes, yes, you can start applying the auxiliary formula right in the beginning only Yeah, it just saved a bit of time Absolutely correct you can do it without using the identities as well No problem. Yeah All right, let's take another one Okay, let's take this problem In a triangle if two delta square is given to be this Then which of the following Is the correct option? Yes, delta means the area of the triangle So there are certain symbols that you should be aware of Uh, delta is the area of the triangle s is the semi perimeter small a is the side bc small b is the side ac And small c is the side ab so these things will be uniformly used Okay, it may require you to know one small formula. Uh, if at all It is required abc by four delta is equal to r So your circum circle radius I'll speak about it officially in some time But I think it was required in this question or it may be required in this question. I'm not giving you a hint kind of a thing So abc by four area of the triangle is your r It's very easy to prove we'll we'll prove it in some time Sure good good seven of you have responded Yeah, just I think Only one option is correct Venkat. It's a single option Okay, we'll stop the poll in another 20 seconds five four three two one Okay, so most of you have gone with option number c Okay, let's check See first of all Can I write this as a square plus b square plus c square? as as a square b square c square by four delta square Correct Oh, sorry two delta square not four delta square two delta square Okay Now just now we discussed that abc by delta is equal to four r correct So abc By delta the whole square is 16 r square. So if I use it over here, I will end up getting Half of 16 r square, which is actually eight r square Right Now a square b square c square if I use my Sign rule I can easily write my left hand side like this Okay, this will go by a factor of two from here Okay Now what should I do with this? Okay Okay, let me write this in double angles of course. So this will be cos 2a 1 minus cos 2a by 2 1 minus cos 1 minus cos 2b by 2 and 1 minus cos 2c by 2 so 2 to all I'll send it to the other side in other words cos 2a Plus cos 2b plus cos 2c Is nothing but a minus one Now how many of you remember this conditional identity that cos 2a cos 2b Cos 2c is actually 1 minus 4 minus 1 minus 4 cos a cos b cos c very important Very important conditional identity all the conditional identities are actually important because you never know where you require them And they will save a lot of time. So this will be minus 1 minus 4 cos a cos b cos c Equal to minus 1 minus 1 minus 1 goes off which clearly implies that cos a cos b cos c has to be 0 Means that at least one angle here must be a zero That means at least Sorry, at least one angle must be a 90 degree One of a b and c Must be 90 degrees So what I can say with conviction that this can be a right angle triangle not very sure about isosceles So please do not mark something which you are not very sure of So you can say it is a right angle triangle may not be isosceles Okay, so isosceles is an extra restriction that you are putting on the system Which has not been very evident from whatever workings we have done so far No where it is evident that two of the sites have to be equal correct So option number d is correct, which I think Not many people only six of you marked option number d Is it fine any questions any concerns? Next is uh various formulas for the area of the triangle That may be required by us. So the next concept is area of a triangle There's so many formulas. You have already learned A few of them in your junior classes So half base into height Okay, half base into height. So if you have let's say Now based on this, we can write several other formulas also derived from air. For example, half base you can take it as a Okay, height you can take it in several ways For example, height height you can take as c cos b Okay, sorry c sin b. So it'll be c sin b It can also be taken half into a into This is b and this is angle c b sin c Okay, so two formulas already known half ac sin b half ab Sin c and you can also write it as half Half bc sin a as well. So this is already used in one of the questions which we saw today So it says nothing, but if you have been given a triangle Right, so all you need to do know is two sides and the sandwiched angle between them So if you know bc and a you can get this formula if you know b c and sorry, this is your a not b So if you know ac and b you can get this formula And if you know ab and c you can get this formula Okay, now using this only we'll get one more formula which is abc by 4r Area of a triangle also known as abc by 4r How will I get this formula? Let's take let's take the third one So if you use your third formula Which is half bc sin a sin a you can write it as a by 2r Okay, so when you do that you end up getting abc by 4r This is what I was discussing a little while ago when I gave you the formula delta is equal to abc by 4r So this is a very important relation Okay, so this is another formula for the area of the triangle Apart from it, you already know your heron's formula that is s s minus a s minus b s minus c You can prove this also from the very same fact For example, if I want to use the let's say The formula half bc sin a Okay, so sin a you can split it up as 2 sin a by 2 cos a by 2 Okay, so this will become bc sin a formula. We all know is under root of s minus b Sine a by 2 is s minus b s minus c by bc cos a is under root of s s minus a by bc So in the under root you'll have bc square. So that will come out and cancel out with the The outside bc and inside the under root you'll have s s minus a s minus b s minus c So this is how you can prove your heron's formula as well Okay, in due course of time, we'll see more formulas Like delta is equal to small r by s. We can also write it in terms of x x Ascribe circle radius all those things will come to those formula in some time Okay, please note this down because I'm going to go to the next page Mehul, I don't know how much you have lost in the previous page. I think we had discussed about One of the questions. I believe right one questions. We are discussed in the previous page So this is a theory that we are discussing on this page. So I would request you to note this down So we have discussed around one two three four Five six six formula for area of a triangle and there are many more to come There are many more to come Yeah, sure done Yes, sir Okay, let's Now talk about circles connected with the triangle Circles connected with the triangle So there there'll be a lot of type of circles that we'll be talking about of course the circum circle in circle The ascribe circle, etc So let's start start talking about them We all know how to make a Circum circle, right? So if there's any triangle How do you make a circum circle? We basically find the Find the center of that circle by perpendicularly bisecting the sides of the triangle Okay, and it will be Concurrent at some point and that point will be called as the circum circle center, which we normally write it as s Okay, so with this as your center and you can take any one of the Distance from the Vertex as your radius so they will all be equal Oh, sorry So s a s b s c they will all be equal to r r each Okay, you can draw a circle and that circle is called the circum circle Okay, we have already discussed a lot of things We have already discussed that your circum circle radius is nothing but a by two sine a We have also discussed that it is a b c by four delta Okay, a few things that you should also remember it is used in a lot of questions The distance of s from the vertices and the sides Now from the vertices it is no Difficult work it is all equal to r only but from the sides, let's say I call this as p q r p q r Okay, so what is sp? What is s q and what is s r? Now there's no need to remember this because you can easily find it out So if you know that this angle is angle a Okay, and this side is side s s b is r You can easily get the base Base will be nothing but r cos a Right, so s r is r cos a Sp will be similarly r cos b And s q will be r cos c Okay, no need to remember but just Unnoted down and you will automatically remember this to practice There's no separate, you know requirement for you to remember these formulae Done noted down Yeah, okay Next is your this was your circum circle This was your circum circle. Let's talk about in circle in circle. We have a lot of things to discuss in circle is basically a Circle which is made By internally bisecting the angles of the triangle So if you bisect the angle a internally That means this angle is equal to this angle. You bisect this angle b internally That means this angle is equal to this angle and you bisect this angle c internally They will all be concurrent at a point And that point will be called as the in-center that is denoted by the Alphabet i So what is great about the in-center as you have found out this by internal bisector of the angles This point is equidistant from all the three sides. So this length This length And this length They will all be equal and hence with that as a radius. You can always sketch a circle like this This circle. Sorry for crossing this this circle is going to Have all the sides of the triangle as tangents to it Okay The radius of such a circle is called small r Now there are several formulas for finding small r small r First of all is given by delta by s Now what is the proof for this proof for this is very simple You probably would have done this in your Class 10th also Area of a triangle will be made of area of triangle a i b area of triangle b i c And area of triangle c i a Area of triangle c i a Okay, so area of triangle a i b will be half base base will be c height is r so half c r Okay, let me know Yeah, similarly area of b i c will be half base, which is a height is r and again This is half b into r. So if you take r common it will give you this And we all know a plus b plus c is 2 s so 2 s by 2 will be s So this is another formula that you have for the area of a triangle. You can note this down also very important Delta is equal to s r or r is equal to delta by s Okay, i'm sure you've done this type of question in your class 10th. It is there in your r s sigarwal book of class 10th Couple of more formulas that we need to note down over here So please copy this i'll be Going to the next slide Yes, one minute one minute Okay With hsr batch i had not done this at all right hsr. Thank you, sir. Yes So we only did it to find and cosine rule Oh, okay Okay, anyways, uh, there are a few more formulas for r This So all of them will give you the r expression only so note down I'll prove one of them I can prove let's say the first one in two ways One is by using a half angle formula. So we all know that tan a by two Is nothing but s minus b s minus c by s s minus a right? Okay, so if you apply here this will become under root of s minus a s minus b s minus c By s multiply and divide with an s The numerator will become delta square by s square Okay, numerator will become delta square denominator s square. So this will give you delta by s which is actually an expression for Which is actually an expression for r as we had proved earlier. Okay. This is one way to prove it Other way to prove it is Let's say I bisect this angle Okay, and just let me draw this in circle in between You should actually always draw a circle before drawing this Anyways, I'll try to fit one inside Maybe distorted a bit. Yeah Anyway, so let's call this as your point of contacts a b See, okay, and let's say this is your point of Actually one more thing I would like to tell you that This line that when is extended it need not pass through where the circle is touching the Uh, the bc line. Okay, so this need not pass through the same point Let me call this as p q and r. Okay. Now if I call this as Let me drop a perpendicular from the center. Let's say this is your i Okay Now if I call this as x can I say this will also be x a p and a q correct Because they are tangents from an external point to the in circle If I call this as y that is a p Br will also be y Similarly c q is z Cr will also be z correct in other words to x to y to z Is equal to your semi perimeter Sorry is equal to a perimeter. So it is twice of the semi perimeter. So x plus y plus z is equal to s Correct. Yes or no. And y plus z is already a Because this length is your a that is your side Opposite to the angle a right So x is nothing but s minus a So this length is s minus a Any ways you should remember this result because it is required. Okay. So this is s minus a this is also s minus a s minus a S minus a Similarly, this will be s minus b s minus b s minus c s minus c like that So if you focus on this triangle triangle api If you focus on the triangle api You know already that this angle is half of a so you can easily say tan of a by two. Let me write from triangle api tan of a by two would be r by Ap ap is s minus a So from here you easily get r as s minus a tan a by two. There's another way to prove the same formula Just by using pure geometry No auxiliary formula used here. Okay. So either ways you can remember it So all of these will give you the expressions for your in-circle a radius Done noted down Can I go to the next formula? Any questions? Okay The next set of formula for r is Yeah, the next set of formula for r is r is given by a sin b by two sin c by two upon cos a by two Okay, it is also given by b sin c by two sin a by two by cos b by two And is also given by c sin a by two sin b by two By cos c by two Okay Uh, can you prove this? Again, very easy to prove if you use your auxiliary formula Yeah, so if you use your auxiliary formula in many one of them you will end up getting your desired result And each side will give you delta bias. Okay, so just prove it for homework Prove for homework And the most important expression for r that you should know and it is also a relationship between capital r and small r Which is this formula A very heavily used one This formula is not only an expression for small r, but it also relates Small r with capital r. So this gives you the relationship between the small r and the capital r Again, you can use any one of the formula over here to prove it For example, if you use the first one So if you replace your a as two r sin a And sin a itself can be written as two sin a by two cos a by two So you would realize a lot of terms will start getting cancelled off For example, cos a by two and cos a by two will go off Two r into two will give you four r And you'll end up getting the desired formula Is this fine? All right, so from here a very important inequality arises So, uh, you would have all done this inequality in your, uh, trigonometry chapter that in any Triangle sin a by two Sin b by two sin c by two is always less than equal to one by eight Okay, equality will exist when your triangle is a equilateral triangle So now using this condition I can say r by four r Because your sin a by two sin b by two sin c by two is r by four r. This is less than equal to one by eight Which clearly means r is greater than equal to two r Remember your circum circle radius is greater than twice the in-circle radius Equality will hold when the triangle is Equilateral is this fine Equality holds when the triangle is Equilateral So, uh, if you see we have taken at least four formulas four set of formulas for your small, uh, r I think, uh, all of them are interrelated So if you remember one of them, I think you'll be able to produce the other one in case it is required Okay, uh, three of them is absolutely important. One is the one which I have shown on the screen right now And the other one is delta by s And s minus a tan a by two s minus b tan by two s minus c tan c by two This one is not that important The one which is on the top of the screen. This is not that important So I've not seen much use of this formula. I've seen mostly the second the third and the fourth one being used More frequently Okay, we'll take up questions on this little later on. Uh, first I would like to do ascribed circles and their radii So let me go to the next page ascribed circles So we have done circum circle We have done in circle Now we'll be talking about ascribed circles because they are more than one So what are ascribed circles? Let me just, uh, draw one for you In fact, I should first draw a circle for you for myself Then I think making the side becomes Oh, sorry Somebody has a doubt Oh Venkat Yeah, Venkat will talk about it So let's say A, B, C is my triangle So I have purposely drawn the circle first because then it becomes convenient for me Now this circle is basically obtained by internally bisecting angle a So center of the circle is obtained by internally bisecting angle a externally bisecting angle b And angle c Okay, so this point is what we call as i one Okay, i one or i a also many books write it Okay, i a or i one just to signify that it is a center of a scribe circle opposite to the vertex a of the triangle So this is the actual triangle Okay, so what I did I just internally bisected this angle I externally bisected b and c Okay, and I obtain a point of concurrency and from this point The perpendicular distance from the three sides whether Extended or not They are all equal So These three lengths will be all equal and they will be called as r one Or many books will call it as r a also doesn't make a difference So r one is basically the radius of the circle and i one let me stick to i one only I one is nothing but your center of the circle Now a similar thing can also be drawn had I Extended b a and b c internally bisected b externally bisected a externally bisected c and drawn a circle like this Okay, so that circle would be having a center of i two And radius will be r two Similarly, when I do the same thing on this side, I'll get a circle with center as i three And radius as r three so there can be three ascribed circles. They can be three ascribed circles Ascribed outside aditya inscribed means inside there's a name given to it So Yes, sir So the extended sides are always perpendicular like tangent Yes, yes, yes This circle will have tangents as side bc Let me write it down tangents would be bc ab extended ab extended and ac extended Okay, so they will always be tangent. So I can say let's say I call this as uh point p q r so you can say c q is same as c p b q is same as br In fact, you can say ar is same as ap Are you getting my point? So we'll do some geometry on this. Don't worry. So as of now the step of construction I'll again repeat First you have a triangle given to you correct Uh extend let's say if you want to make an ascribed circle opposite to a Join ab and extend it of course ab is already joined so extend ab further extend ac further Okay, internally bisect a externally bisect b and c Right, so I'll write it down Internal bisector of a and external bisector of or bisectors of b and c So they meet at I1 So I1 is the point of concurrency of internal bisector of a and external bisector of b and c Again, let me tell you the perpendicular dropped from I1 to bc Need not need not meet this internal bisector Now once you've got I1 The perpendicular distances of I1 from each one of these will be equal. It is because of the locus condition Right, we know that if you are bisecting the angle of a ray or of a pair of ray The the any point on that bisector will be equidistant So this point is equidistant from ab extended and ac extended. It is equidistant from bc And again ab ac extended and is equidistant from ab extended and bc So this will be having equal distances from all the three sides Okay, and that distance is called r1 by convention. We call it as r1 Now many people ask me sir will In the combative exams will they use the same kind of uh, uh, nomenclature In jee, they will define it because jee, uh, these national exams don't assume people to know all these names Okay, uh, but yes certain things they will not define. They'll not say a is called this and delta is called this They will assume you to know all those things Now what are the values of r1 etc. We'll figure it out uh r1 The first formula that we'll talk about r1 is r1 is delta by s minus a Okay It is also equal to s tan a by two Okay, it is also equal to a cos b by two cos c by two by cos a by two And it is also equal to Also equal to 4 r sin a by two cos b by two cos c by two if you remember this r2 r3 expressions Will automatically come out by the trend that you will see in the formula, but before that will prove it Because you'd know from where are these uh facts coming up So for the first one, okay, let's prove That r1 is equal to delta by s minus a. This is very easy proof It is done in a similar way as what we had done for finding r is equal to delta by s. That means you're dealing with the area So the proof is we know that area of the triangle Is Just watch out this figure Can I say it is area of a i1 b Plus area of A i1 c minus area of b i1 c Can I say that? I would request all of you to scan this figure In your mind properly So area of this triangle is just watch out my uh white marker area of a i1 b Plus area of a i1 c Minus area of b i1 c because b i1 c is coming extra in that this thing. So I only wanted this triangle abc area Correct, that's my subtraction Correct now. What is area of a i1 b? So, you know the base You know the height Correct So I can say it's half cr1 So this is half cr1 Similarly area of a i1 c is you know the base base is b Height is again r1 Okay, so it's half b r1 An area of the triangle which you need to subtract That is half base base is a and perpendicular is r1. So I can say half. Let me write it in yellow half a r1 So if you take half r1 common you get b plus c minus a Okay, and b plus c minus a you can write this as 2 s minus 2 a so 2 2 will get cancelled. So r1 s minus a is equal to delta and You should not shy away from noting this down also as one of the another Formula for the area of a triangle. So delta is also given as r1 s minus a So from here, I can say r1 is delta by s minus a Is this fine? Any questions here in the proof See nobody is going to ask you the proof but see what you learn while you're proving that is more important Okay Next, how do you prove that it is s tan a by 2 now for that it is very easy to prove it. I'll again explain it from this figure See We know that Bq length is same as br length Right. So if I call this as x This will also be x correct We know that cq length is cp length. So if you call it as y, this will also be y. Am I right? Yeah, now now ab Sorry for writing on top over it ab Plus x plus y plus ac is equal to the perimeter of the triangle correct and remember ab plus x is ar And ac plus y is ap And both are equal by the way Both are equal. Why both are equal? why ap is ar Because again, they are tangents drawn from an external point a to the ascribe circle So if I call this as let's say What name should I give z z? Okay. So 2 z is equal to 2 s. That means z is equal to s That means this whole length is actually an s Okay, so if you see this is your a by 2 right And this is your r1 So if you focus on your triangle if you focus on your triangle a i1 r a i1 r Can I say tan a by 2 can be written as r1 by s And for me here we get another formula, which is what you wanted to prove s is equal to sorry r1 is equal to s tan a by 2 So this formula is problem. Is that fine? Yes, sir. Any questions do let me know Could you scroll to the right? Sure Yes Let me know once you're done so that we can move on Okay, so rest of the formula. I think you can easily prove by using your you know, uh, basic Conversion basic auxiliary formula. So I'm not going into details of that But yes, I will definitely write r2 and r3 expression similarly so that you know What is this? You know analogous expression for r1 r2 r3? So Let me do a comparison chart here So a very summary of your ascribe and the inscribe circle radius See we already have seen Small r is delta by s Okay, it is s minus a tan a by 2 It is s minus b tan b by 2 And i'm just writing the important ones the ones which will be very very useful for you For sine a by 2 sine b by 2 sine c by 2 correct in the same way r1 The small change that will happen from this formula instead of s you will have s minus a Instead of s minus a you will have an s Okay, uh rest formula I will not copy. Oh by the way here is one more formula I can write s minus c tan a by 2 And in this formula what will happen only Sine a by 2 will remain rest all will get converted to cos cos Let's both will get converted to cos cos Okay r2 You can easily predict from here r2 will be what tell me delta by S minus b yeah, and it will be s tan b by 2 Correct and what will happen here This one will become sine b by 2 rest both will be cos cos So this is what I said if you remember one you will be able to produce the other one in the exam very easily Is this fine Any questions? Now what is more important is to know the relationship between these small r's so how is our small r Small r1 small r2 small r3 connected. This is more important for us because uh, you know many questions will be directly based on that However, I will uh give you few of them And we'll derive also few of them So the process of derivation should be known to all of us Okay, I will not be able to take up all the questions Okay, so let us start with this particular formula And this is very important The formula is 1 by r1 1 by r2 1 by r3 is equal to One second square. I forgot Yeah, plus 1 by r square is equal to a square plus b square plus c square by s square s in fact I should write delta square r1 r2 r3 is delta square There are a lot of properties. I can just give you a few of them This is a square one of them is very important r1 Uh plus r2 plus r3 minus r is equal to 4 r. This is very important The many there are actually tons of formula which relate your small r to capital r Meanwhile, uh, can you try proving one of them? Let's say try proving third So you just write it as s tan a by 2 and then we know tan a by 2 tan b by 2 plus tan b by 2 tan b that's 1 Okay, square you can pull out common Is that clear everybody, please try done a couple of more we can add here, uh cos a cos b cos c is 1 plus r by r Okay A cos a b cos b c cos c By a plus b plus c A plus b plus c is equal to small r by capital r There are many actually there are many. I mean I cannot give you all of them But yes, we should know how to uh tackle, uh, no these proofs Okay, these proofs are very important Uh one one I wanted to give it in the beginning. I think I changed my mind when I was giving it. So This one, okay Okay, so You don't have to remember them, but you should know how to you know deal with them how to prove them Now which one we were trying to prove The third one, right? Right, so how do you prove the third one? So, uh, if you know this formula If you know this formula and if you know the last one We can prove it very easily So if I if I able to prove this one first so r1 r2 r3 r1 r2 r3 We all know it's delta by s minus a delta by s minus b delta by s minus c Correct. So it's delta cube by by By s minus a s minus b s minus c Correct Oh once again, I think I missed out an r over it Yeah, so delta by s Delta by s. Yeah, so this will become delta to the power four and this itself is the heron's formula delta square Okay, so if you're able to see this then you are able to also Figure out that r1 r2 r3 is delta square by r correct Okay, and here here In the left hand side, let me write it down over here In the left hand side if I take r1 r2 r3 common, I get 1 by r1 1 by r2 1 by r3 Okay, this is already delta square by R And this is actually 1 by r So it's delta square by r square And delta square by r square is known as the square of the semi-perimeter because small delta delta is small R into small s Okay, so they're all very interlinked So This you can find it in any standard book. Okay, so try practicing a few of these questions. Meanwhile, I'll take some direct questions Which can be uh based on these particular uh properties So let's take a few questions on them Scroll down. Yeah here this side Let's say this question we want to do In a triangle small r1 plus r2 plus r3 is 9 r Then the triangle is necessarily Which of the following Then the triangle is necessarily which of the following Pole is on Very good to cheer Good good Good speed Normally in the crash course and all the speed is very good. I mean Roughly students will take around 1 minute 15 seconds to somewhere around 1 minute 45 seconds to solve questions But we have to come to that speed Awesome, all of you have voted for the same option. Good I think the first thing that everybody tries is taking an equilateral triangle So b option is most obvious for many of them. Okay, end of poll will check which option is correct Most of you have in fact, all of you have voted for option b See If you if you know the formula r1 plus r2 plus r3 minus r is equal to 4r Which is one of the well-known formula You can automatically say this is 4r plus r and as per the question, this is 9r That means 4r is equal to 8r. That means r is equal to 2r little while ago I had discussed that equality between r and 2r will exist when the triangle is equilateral So that further proves that option number b is the right option Okay, you may have your own way of doing it not a issue easy Next In triangle abc small r is capital r by 6 and r1 is 7r find the measure of the angle a Should I put the pole? All right, so let the pole be on Let the pole be on Okay, good. I've got a few responses So give me one more minute Yeah, yeah, take your time Nice See a few have responded and all of you have voted for the same option So chances that it's correct is good enough High enough Okay, let's discuss enough time. Let's close this in another 20 seconds five four three two one go Okay, so most of you have voted for d d for deli Let's check See first of all, let's do r minus r1 or let's say r1 minus r r1 we all know it's 4r Sine a by 2 cos b by 2 cos c by 2 And small r is 4r sine a by 2 sine b by 2 sine c by 2 correct If you take 4r sine a by 2 common You'll end up getting cos b by 2 cos c by 2 minus sine b by 2 sine c by 2 which is conveniently written as cos of b plus c by 2 Am I right? Now we know that we are dealing with a triangle. So b plus c by 2 is pi by 2 minus a by 2 So cos of b plus c by 2 is actually sine of a by 2 So this reduces to 4r sine square a by 2 Now again, this is that I wanted to share with you, but I thought, you know, let not everything be shared Because you'll not be able to remember it But yes, you should know how to work around this. This really makes your life very easy in solving this question because You know your small r1 is 7 small r Or your r1 is 7r. So this is small r minus r, which is actually your 6r And 6r is your capital r Okay, in other words sine square a by 2 is equal to 1 fourth. That means sine of a by 2 is half That means a by 2 is 30 degrees So a has to be 60 degrees Option number d is correct. Is that clear any questions? Any questions on this? Okay We'll take one more Before we start a new Yeah, let's take this one If abc is the right angle triangle at a cos inverse capital r by r2 plus r3 Where capital r r2 r3 mean the normal things that we have discussed so far This is equal to which of the following options All is on Good one minute gone two people responded so far Okay, we'll stop it in count of five five four three two one Most of you have gone with option number b Let's check See again, there's a lot of hint given to us in the question. Let's find r2 r3 R2 as we all know is 4r cos a by 2 Sin b by 2 cos c by 2 And r3 is 4r Cos a by 2 cos b by 2 sin c by 2 Okay, if you take 4r cos a by 2 common Okay, you will end up getting sin b by 2 cos c by 2 and cos b by 2 sin c by 2 which is clearly sin b plus c by 2 Okay, since we're dealing with a triangle we can all say that Oh, sorry We can all say that sin b plus c by 2 is actually cos a by 2 Right Isn't it So r2 plus r3 is this So clearly the expression within the cos inverse which is r by r2 plus r3 Is actually 1 by 1 by 4 cos square a by 2 am i correct And a is 90 degrees. So a by 2 is 45 degrees. So it's 1 by 4 1 by root 2 square So it's actually 1 by 4 into half, which is actually 1 by 2 So cos inverse of this Is as good as Doing cos inverse of half and cos inverse of half is 5 by 3 Option number c Is that fine? Okay Now a couple of things that we would like to link up to whatever we have done so far number one is the length of the Median Okay, this is very important length of the median of a triangle So let me just make a triangle quickly so length of Median coming from vertex a Okay Does anybody remember this formula? We had done this. I think yashfandpur will remember it Anybody from yashfandpur Length of median Length of a median coming from vertex a right sure Okay, it's half under root of 2b square 2c square minus a square This is your let's say I call it as am Okay, can you prove this quickly? Let's say even if you don't remember it Can you prove this? Can anybody prove this? hint Use cosine law Use cosine law. You'll be able to prove it. Okay Let's say am is x Okay, this is a by 2 a by 2 correct focus on this angle b Okay, if I write a cosine rule for b What will I write a square plus c square minus b square by 2 ac correct. So you're doing cosine for abm sir I'm going. No, no, I'm doing cosine for the whole triangle. Oh, okay, and I will also do it for abm Correct. So abm if I write it, it'll be a by 2 square. By the way, this side is c c square minus x square by ac correct And both of them Basically will give you the same value because they are cost of the same angles So if you compare 1 with 2 What do you get you get a square plus c square minus b square by 2 ac is equal to a square by 4 plus c square minus x square by ac ac ac gone And if I multiply with the 2 I'll get a square plus c square minus b square as a square by 2 plus 2 c square minus 2 x square correct So 2 x square is nothing but c square plus b square minus a square by 2 That means 2 x square is equal to 2 b square 2 c square minus a square by 2 If you bring this 2 down, it will become a 4 And if you take the square root of it This is what you're going to see Square 4 will come out and it'll become 2 times 2 b square 2 c square minus a square Okay, this result should be kept in mind. It's half off under root 2 b square 2 c square minus a square That is the length of the median coming from the vertex a In a similar way if I ask you what is the length of the median coming from vertex b and c respectively I'm sure you'll be able to figure that out also. Let's I call it as n and let's say I call this as p So what is b n then? Take a clue from this Tell me the other word median b n so half This will be 2 a square 2 c square minus b square And the other median cp I hope I have called it cp only. Yeah cp would be half under root of 2 a square 2 b square minus c square So these three Length of the median are very important So you know how it is derived and you should just note down the formula Similarly, we'll derive the length of the internal angle bisectors also Done noted down Can I go to the next page? So could you go down a derivation part down? Yes, sir, okay Good enough Yes, sir. Yes, sir. Thank you Any qualities is already there in your trigonometric in equations Aditya to answer your question. You should go for trigonometric in equations I don't know which book you are following But that is a very explicit chapter for you in Arihant If you follow Amit Agrawal's Arihant They have given As a separate chapter trigonometric inequalities Yes, sir Okay, so length of Angle bisectors Okay, so let's say if you bisect the angle internal angle of a Okay, let me call this as a p Okay, please note that a p length is given as 2 bc by b plus c cos a by 2 Now I can give you other formulas also, but let me derive this first of all So the derivation part of it is very simple. Uh, just drop a perpendicular Just drop a perpendicular, uh from these Vertices on this So let's say I call this as an x Now we all know that area of a triangle is half bc sine a Okay, so let's say angle is known and this is your a by 2 a by 2 Okay, and it is also equal to area of the triangle a b p Plus area of the triangle a p c Okay Now since Since This is 90 degree and this is a by 2 Okay, and we know these set side lengths Okay, what is the perpendicular here? Base is x only What is the perpendicular height over here? It may not be a very accurate diagram So perpendicular will be b sine a by 2 Okay Similarly for the other also this will be c Sine a by 2 Okay Now sine a here I can make it up as sine a by 2 cos a by 2 half Half gone sine a by 2 sine a by 2 sine a by 2 gone So you'll end up getting x b plus c is equal to b c cos In fact 2 b c cos a by 2 So from here x can be figured out as 2 b c by b plus c cos a by 2 This is a very very important formula. You must remember it Okay This is the length of your internal angle bisector coming from a Okay, similarly, let's say if I make an internal angle bisector coming from B as Bq and internal angle bisector coming from c as cr So b q will be b q will be take a clue from this and tell me what should be b q to a c a plus c cos b by 2 Okay, and c r will be 2 a b by a plus b cos of c by 2 So just like you noted down the formula for your medians, please note down the length of the angle bisectors Is it fine any questions? Okay Next is your mn theorem mn theorem Are not very important. Basically it comes from your sine rule only So mn theorem is basically a formula Which links The following I'll just Write it down over here Let's say this is your triangle abc Okay And let's say there is a point on this side d. Let's say okay, and I connected to a Okay Such that this angle is theta This angle is alpha This angle is beta and d divides vd into dc In the ratio of m is to n Okay Then there's a relationship which connects all these things that you see over here mn alpha beta and theta and that relation is m plus n cot theta is m cot alpha minus n cot beta Okay There's one more version of it Basically, they say the same thing m plus n cot theta is n cot b b is the angle of the triangle minus m cot c Uh, let's try to prove the first one second one will automatically come from there So everybody, please try to prove the first one. This may come as a direct question After doing this we can take a small break on the other side of the break. I'll talk about the pedil triangle and how Pedil triangle is related to the x-central triangle All those things will take up We'll also have to talk about the concept of distance between the critical points We'll also have to talk about the concept of quadrilaterals Uh, cyclic quadrilateral Solution of triangles. So we have a lot of grounds to cover. In fact, I should start doing this Uh, if you look at abd Okay, apply sign rule there So can I say bd by sine alpha Is equal to ad by ad by In fact, yeah ad by This angle now. What is this angle? This angle I can write it as theta minus alpha correct because theta is The sum of the two interior angles over here, right? So this is theta. This is alpha. This has to be theta minus alpha Very good Richard. Very good Okay, so let me call this as one of the equations Similarly, if I apply sine rule on triangle adc Can I say dc by Sine of beta is equal to ad by ad by now ad by this angle What is this angle? This angle will be Okay, this is 180 minus theta, right Right, so can I say this angle is Pi minus Theta minus beta So it is sine of theta plus beta because sine pi minus something is sine of the same angle Is this clear that this is pi minus theta? And this is an external angle So these two sum should be equal to this external angle correct Yes, or no, let me call this as second equation Now take the ratio so bd by sine alpha Okay correct Now bd by dc is already given to us as m is to n. So m is to n sine beta Sine alpha will be equal to this This you can expand sine theta cos beta cos theta sine beta by Sine theta cos alpha minus sine alpha cos theta cross multiply Just a second Yeah, sorry. Yeah, so cross multiply. So if you cross multiply you get m sine beta sine theta cos alpha We can also write it as m sine beta sine alpha cos theta Equal to n sine alpha sine theta cos beta plus n sine alpha cos theta sine beta Okay, let's remove this bracket unnecessary bracket Now let's do one thing. Let's divide both sides Let's divide both the sides Divide both sides by sine alpha sine beta And sine theta Okay, so let's divide through throughout with sine alpha sine beta sine theta. So what will happen to this term? This term will become m cot alpha Correct, what will happen to this term? This will become m cot theta What will happen to this term? This will become n cot beta And this will become again n cot theta Okay, so let us bring your uh theta terms to one side. So m plus n So I'm bringing this term to this side and this term to this side So m plus n cot theta will be m cot alpha minus n cot beta This is called the m n rule. I have not seen many questions on this in j But yes, this is a very favorite topic of uh olympiad people So people who write rmo p rmo Okay, so please try to prove the other one similarly. That's very easy. It can be proved from in a similar way Okay, so this is called the m n theorem This is called the m n theorem Okay Uh, let's have a break right now Because on the other side of the break we'll talk about uh, a new concept I don't want to take that concept right now. It'll take At least 15 20 minutes So next concept that we are going to talk about is the pedal triangle Pedal triangle Okay, so what is this triangle? Uh, this triangle is basically obtained by Uh, connecting the feet of the altitudes drawn onto the sides of the triangle from the opposite vertex So let me just draw a triangle for you Okay So if I start dropping perpendicular If I start dropping perpendicular from The opposite vertices On to the opposite sides. So they are all perpendicular Okay And the triangle which I connect And the triangle which I connect by joining these feet Okay, so let's say this is a b c And this is d e f Okay, so this triangle is basically called as the pedal triangle The one which you see in the blue color The one which you see in the blue color Now there are certain things that we need to understand about pedal triangle Number one, what is the side length of the pedal triangle? What is the angle uh Of these of this triangle, for example, what is angle d? What is angle e and what is angle f etc? So we need to understand that thing We need to also understand what is the uh, circum circle radius of this triangle Right, so let's try to you know discuss that First thing first. What is the side length of this pedal triangle? Let's try to figure that out Yeah, so Let's try to prove the following Okay, let's try to find out e f length Let's try to find out f d length And let's try to find out d length Okay, e f length is given by a cos a a is the length of bc And angle a is the normal angle of the triangle abc f d is given as b cos b And d e is given as c cos c Okay Now, how do we prove this? How do we find out these lengths? The proof of this is very simple Uh, if you focus on this triangle small triangle a fe What do you think is the length of ae It's very simple since this is 90 degree and this is angle a Okay, then this angle has to be 90 degree minus a correct So this length that is ae is like the opposite. Sorry, uh Yeah, this is your hypotenuse. This is your perpendicular. So I'll just draw this diagram again So this will be something like this So this is going to be 90 minus a Okay, and this side length is c We have to find ae length ae is this length Okay, so I'm just drawing the same structure over here Okay, so tell me what is a length if this is c and this is 90 minus a It's very obvious that this length will be c cos a Correct So this is c cos a What will be this length a f length? Can I say a f length will be b cos a am I right? No, no, let us use our cosine formula on a e f cosine rule on On triangle a e f So let's say I call this as x for the timing. So can I say, uh b cos a square c cos a square minus x square by 2 b c cos square a is equal to cos a itself Is it fine? Have I written everything correctly? Just confirm this Correct So I can say x square is nothing but uh b square plus c square cos square a minus 2 b c cos q a Am I correct? Okay now here if you see If you take cos square a you will end up getting b square plus c square minus 2 b c minus 2 b c cos a What are this formula actually this formula is actually a cosine rule Which gives you a square isn't it? So what we end up getting is x value, which is a cos a This is to be kept in mind. You will not sit and derive the pedal triangle length every time So this length is a cos a in the same way you can prove that this is b cos b and this is c cos c So these are the lengths of the pedal triangle So how did we construct the pedal triangle foot of the perpendicular we combined? No, that is the first thing I said What your internet went blank that time No, sir. I came Drop perpendicular is from your vertices on to the opposite side connect the feet of the perpendicular that triangle is called the pedal triangle Okay. Now, how do I find these angles? How do I find these angles? Angle fde Sir one more time. Yes. Yes. Tell me. So if the angle is like Opti use then or when you drop a perpendicular won't it come outside? Yeah See, uh, we draw it for an acute angle triangle Okay, so we are making this pedal triangle. That's a very good thing that you pointed out So we are making this for an acute angle triangle Okay, in this case, we are making it for the acute angle triangle. However in reality Your pedal triangle can be drawn for any triangle in this case in the figure I have drawn an acute angle triangle just to make it clear to you Because if you have, uh, you know a different kind of a triangle and if you are looking at an acute Opti use angle triangle some of these angles may be more than 90 degree and in that case your length of the sides will become a negative Okay, so we are assuming that our triangle drawn here is an pedal triangle. There's something called orthic triangle That's that's the similar concept Okay, uh, normally orthic triangle and pedal triangle are basically You know interchanged, but they are not actually Okay, we'll not talk about orthic triangle. That's not a part of our syllabus However, this is also not that important, but indirect questions can be framed on it. That's why we are discussing it out Okay Now Next is the angle of this pedal triangle The angle of these pedal triangle is basically Let me write it down here It is this part Sir Yes, sir, trippin So so for obtuse triangles also pedal triangle is defined, but not in this way not in the sidelines and all like this. Yes Okay So pedal triangle angles Let me just Remove this. Yeah So angle f de Is pi minus 2a angle def Is pi minus 2b an angle e fd Is pi minus 2c So these are your angles of the pedal triangle. Now, how do I find that out? How do I find that out? Now one thing you have to understand here is that In the pedal triangle the ortho center point, which is your edge point is actually the in center of the pedal triangle Very important note it down ortho center of Ortho center of Triangle abc is your in center Of your pedal triangle, which is triangle def That means This angle Is bisected here. This angle is bisected here And this angle is bisected here Okay Now let us try to figure out From this information what should be your angle f de Now if you see very closely, this is 90 degrees And this is 90 degrees That means triangle Triangle h f b d I hope you can see the figure Triangle, sorry, quadrilateral h f b d that would become a cyclic quadrilateral That means you can always pass a circle through these four points Because in a cyclic quadrilateral the property is the opposite Vertices are supplementary. So which is happening in this case So this angle And this angle adds up to 180 degrees So it's a it can be considered to be a cyclic quadrilateral If it is a cyclic quadrilateral You can all see that this is your one of the you can say segments on a circle And this angle which is your 90 minus a is lying on this segment. Correct. So As per the circle property This angle and this angle should be the same because they are Angles lying on the same arc Correct. So if this is 90 minus a this should also be 90 minus a And just now we said that This is a in-center. So this is basically an internal angle bisector So you can say this also is 90 minus a So 90 minus a 90 minus a the total angle will become 180 minus 2a and that's how you end up getting this angle Okay, similarly you can prove the other two as well So these two things should be kept in mind that The sides of a pedal triangle is given by this And the angle is given by this one pi minus 2a pi minus 2b pi minus 2c Now what about the circum circle of a pedal triangle? Let's say that is given to you as a question How is the circum circle of the pedal triangle related to the circum circle of abc? So if I draw a circle, let me draw fresh diagrams. This is all cluttered up Let me go to the Oh my god, what is all these opening in my Okay, let me go to the pedal triangle diagram once again. Okay So This was our pedal triangle If you draw a circle which circumscribes the pedal triangle I mean my circle is not that good. Let me just Make it from a tool Yeah, so let's say this circle has a radius of r dash And thus the circum circle radius of abc is r Note that r dash is r by 2 Now how very easy to prove it see in any triangle This condition which is your sign rule will be satisfied, right? So for this triangle also, can I say? 2 r dash will be a Now what is a for this a for this is e f Or fe whatever you have written earlier also fe By sign of the angle opposite to it. This is nothing but pi minus 2 a right So if you start writing this in terms of your capital a and small a this is what you end up writing Correct And this is nothing but a sin a sorry a cos a By 2 sin a cos a Right, so cancel this off Now we know that a by 2 sin a is capital r So you can say that 2 r dash is equal to r. So your r dash is equal to r by 2. So this is an important result Which says that the circum circle radius of the pedal triangle Is half the radius of the circum circle of the original triangle Now one interesting thing that I would like to share with you is Do you remember we had discussed about the scribe circles? Okay So let me show you something very interesting So let us say this triangle is made by connecting the Scribe circles center I1 I2 I3 I hope you remember I1 I2 I3 where the centers of the scribe circle Please be aware that The scribe circles pedal triangle is your original triangle abc That means if I drop a perpendicular If I drop a perpendicular on the pedal triangle and start connecting the basis Or start connecting the feet of the altitudes Then this will be your original triangle abc. Are you getting my point? The original triangle abc Whose scribe circle had ready centers I1 I2 I3 Okay So I1 I2 I3's pedal triangle is abc So let me write it down triangle abc is the pedal triangle For Triangle I1 I2 I3 It's a very interesting fact This triangle is known as the x-central triangle This triangle is known as the x-central triangle That means The in-center of abc Will be ortho center of I1 I2 I3 so that also I will write it down That is The in-center of Triangle abc Is the ortho center of ortho center of Triangle I1 I2 I3 Okay Now on the basis of this known fact I have a question to all of you. Can you give me The length of The sides of the x-central triangle What is I1 I2 what is I2 I3 and what is I1 I3 I1 can you give me in terms of In terms of the dimensions of abc Assuming that this is your normal a Okay, this is your b This is your sorry. This is your c. This is your b Okay, and you know all your angles also. This is angle a this is angle b. This is angle c Can you give me I1 I2 I3 in terms of The dimensions of abc as you can see on the screen How will you give me I1 I2 length? Okay, it's very simple Now if you try to write this length ab Correct. Can I say it is I1 I2 cause of Angle a Correct, which is this angle? Am I right? Yes I1 I2 cause of this angle now. How much is this angle can somebody tell me? I1 I2 c Okay, let's say I call this theta for the timing Right, can I say this angle which is angle c Is actually supposed to be pi minus 2 theta Am I right? Am I correct? Yes, sir So 2 theta is going to be pi minus c. So theta is going to be pi by 2 minus c by 2 Okay, so that also gave me an opportunity to know these angles of the x-central triangle So this is pi by 2 minus c by 2 Okay, similarly, can I say this angle will be pi by 2 minus a by 2 And this angle will be this angle will be pi by 2 minus b by 2 Okay, now having known this let us complete our pending work Okay, so I1 I2 I1 I2 cause pi by 2 minus c by 2 is actually your side length small c Am I right? Yes or no? Oh, yes, sir Correct So I1 I2 would be nothing but c by sine c by 2 Correct Yes or no? This c we write it as 2r sine c by sine c by 2 Okay, so when you expand it it becomes I mean when you break it in half angles It becomes 4r sine c by 2 cos c by 2 divided by sine c by 2 This will get cancelled and you'll get the length of the petal triangle as 4r You can write this also not a issue But let me let us write it as 4r cos c by 2. So this is your length of I1 I2 This is your length of I1 I2 Okay So I1 I2 is 4r cos c by 2 So can you take a clue from this and tell me what is I2 I3? 4r cos a by 2 And this will be 4r cos b by 2 Is that fine? Any questions with respect to this derivation? Sir I had a doubt on why the orthocenter of the outer triangle is the center of the petal triangle Okay, this is a very beautiful question you asked and I'll give this as a proof for all of you to figure out why is the why is the in center of the petal triangle is the orthocenter of the You can say the original outside triangle or for that matter here Why is the in center of abc orthocenter of the extra x-central triangle I1 I2 I3? Okay It's a very simple thing. You just figure it out. It's not going to take much of a time You have to make three cyclic quadrilateral Three cyclic quadrilateral in fact in fact four will be formed So one cyclic will be this guy Other cyclic will be the one which I already used this guy other will be This guy, okay. So from there you can figure out very easily. So I'm giving you as a hint I try to figure it out. You have to make cyclic quadrilateral to figure that out You'll automatically realize that this angle. What do you get for the? This part the same thing you'll get for this part also Okay, automatically will come to know from there. Don't worry about it Okay, excuse me, sir. Yes, Shamik So we can simply say that the line a I1 is bisecting angle a and similarly I3 she's bisecting angle c so That point of Intersection is basically the point of intersection of all the angle bisectors. So it's the in center No, it is by the construction actually that I have already incorporated it, but his question is more you can say More basic. Why is it happening like that? He's asking He's asking why is the internal angle bisector of abc meeting at the same point as the altitudes drawn from I1, I2, I3 That is that is his question actually Got the point So did you get my approach here? See, I'll give you an idea about it See, this angle is how much? What is this angle? Let's say this angle was I1 angle. Can I say this angle is 90 degree minus I1 Right no Yes, sir. Can I say this angle will also be 90 degree minus I1 because they can be a cyclic quadrilateral fitted over here Yes Similarly, this angle is also 90 degree minus I1 Correct and this is equal to this because you can fit a cyclic quadrilateral like this Correct. Yes. I got what I wanted you to figure it out, but okay fine now. I've given it out. So it's fine So that's why this angle and this angle become same this angle and this angle become same and this angle and this angle become Same and hence it is along the internal angle bisector and hence this point is the in center of abc Which is also the coinciding point of the altitudes of the x-central triangle Okay Got it. Is this fine? All right Okay, now my question to you is Can we find a relationship between Can we find a relationship between The circum-circle radius of the x-central triangle and the circum-circle radius of abc I'll say come on sir two hours Is that fine It is basically dependent on the previous question that I had asked you where the pedal triangle Circle radius was half the circum-circle radius of this So if this guy has a circum-circle radius of r x-central triangle will have a circum-circle radius of 2 r Okay, so both are interlinked So both our concepts are interlinked. So just keep this in mind. It may help you in solving No, some questions Now comes the critical distances between Couple of Critical distances we'll talk about. Let's talk about some important distances Okay So the first thing that I would like you to Tell me in center We know all that in center is obtained by Internally bisecting these angles Okay All right What is ai bi and ci length What is what are these lengths? It's very important to know these lengths because they will be very much used in finding You know a lot of other important distances which we'll take up later in this chapter So what is ai bi and ci? Who will tell me? So you'll say simple if I drop a perpendicular from i Let me just use a If I drop a perpendicular from i on to the side ab. This is going to be r Okay, and you already know this is a by two so can I say ai Will be the hypotenuse of this triangle. Let me name it as m So am i hypotenuse is ai Correct. So r by ai is sine a by two. So ai is r by sine a by two. So please keep this in mind a very important result Which can further be written as four r sine b by two sine c by two Now don't ask me why because your small r can be written as four r sine a by two sine b by two sine c by two So sine a by two will get captured off So this is your ai Similarly bi will be r by sine b by two Or it can be written as four r sine a by two sine c by two And ci can be written as r by Sine c by two which is four r Sine a by two sine b by two Okay Now it is needless to talk about the distance from the sites because you know, it's all r So we do not talk about it Next it is important to know the same thing for the ortho center as well So let me so ortho center we all know is obtained by the altitudes Okay, and we normally write ortho center as h So who will tell me a h bh And c h distances Also, let us call it as p q r. Let us also figure out h p h q h r Okay, so for finding a h can I say this angle is 90 minus c Because this is 90 and this whole angle is c. So this is 90 minus c. Am I right? Okay Second thing that I can say is that this whole length is actually a small c. Isn't it this whole length is actually a small c And this angle is 90 minus a correct Because this is whole thing is a and this is 90 degrees. So this angle is 90 minus a So now you have a right angle triangle a b q This angle is 90 minus a this is c. So this can be written as c Cos a correct me if I'm wrong a q can be written as c cos a Am I right? correct me if I'm wrong Now this angle is 90 degree minus c and a h is known a h is unknown to me. So let us find that out So can I say c cos a by a h Will be cos of 90 minus c right Which is actually sin c So c cos a By sin c is equal to a h And c by sin c is 2 r A very important result if you remember in one of our Vector question we had used this result and many of you did not know about it Okay, now, you know now, you know this result Yes, which is 2 r cos a Yes, sir So I have a different method There can be so many methods. Yes, you can share with us Like we can find a p because we know the angle and we know the length Okay And then we can subtract the radius of the Incent in center That you're saying hp is the uh, uh, uh, in center radius Yes, no, no It is not In circle radius Okay Similarly bh Can I say from the same thing bh will be 2 r cos b and ch will be 2 r cos a cos c This result is I have felt it is very commonly used so better to keep it in mind Okay Now from the same thing I can also get the hq hp and hr also So hq if I just look at the diagram again a hq triangle hq will be what So hq by c cos a will be tan of 90 degree minus c Hq by c cos a is tan of this angle Which is cot c So hq will be c Cos a cot c Okay, right Now c by sine c again will become a 2 r. So you can say this 2 r cos a cos c So hq length is 2 r cos a cos c Can you take a clue from this and tell me the others? Oh, sorry. I wrote it against the wrong one hq 2 r cos a cos c Now remember hq was the one which is opposite to your vertex b. So that's why there was no b Hp is the one opposite to a so there will be no a there. So it'll be 2 r cos b cos c And hr will be similarly 2 r cos a cos b. Is that fine? You could have also multiplied you could also use this length also For getting your hq hp and its hr etc This is not very important. No need to remember Not very important Not important. I should write But this is these are important because they are used. I've seen many problems which directly Require you to use these results Okay, so so far what have we covered we have already covered. How far is your i from a b c How far is your i from the sides? How far is your h from a b c? How far is your h from the sides? And we have already done it for the case of circum circle Center also how far it is from the vertex r r r and how far it is from the sides, which was r cos a Etc etc r cos a r cos b r cos c etc So we are done with the critical distances. Now using this we'll be talking about Few more distances. So let's move on to the next slide. All of you have copied this Everybody has copied this. Can I move on to the next page? Yes, sir Yes All right, so next thing that we are going to talk about is the distance between Distance between Yeah distance between The circum center and the in center number one Circum center and in center now This length is given to us. Let me call this as s. Let me call this as i symbolically So the distance between s and i is given by The simpler formula to remember is r square minus 2 r r And a complicated version of it is both are both the same thing only We will prove one of we'll derive one of these distances because I'll be also talking about distance between Ortho center and the circum center etc etc So let us first derive one of them. You'll get an idea about how we can use it How we can use this uh to find out the other Proofs also. So let me draw a diagram for the same Let me draw a diagram. Let's say I draw a diagram now. We know that s is obtained by perpendicularly bisecting the Sites it is a very very obtuse case. So I'll take another This is your s Okay and uh I also let me draw Okay, I'm just showing two of the internal angle bisectors This is your eye point Okay, so how do you find out this length? Let's say si How do you find si length? Do they lie along the same line passing through the opposites side? Are you saying a i and s lie on the same line? Need and be Need not be Yes Any idea how to deal with this? So no coordinates are not given no sir. Nothing Okay, sir I'll give you a hint If you connect if you look at the triangle as i As is r Okay, and what is a i? Correct Okay, can I somehow figure this small thin angle? Can I figure that out? Can I figure that out thin angle in between? What is this angle? See Very simple This angle would have been angle c right So if you drop a perpendicular from the circum center on here Actually, this whole thing is 2c right? So can I say this is c? This is also c correct. So this is c and this is 90 degree. What is this angle going to be? 90 minus c Right, this angle is already a by 2 Correct. So what is this thin angle? So thin angle will say minus 20 minus c a by 2 minus 90 minus c correct now a by 2 Minus 90 Plus c now 90. I will write it as minus a by 2 minus b by 2 minus c by 2 Can I say it becomes just c minus or c by 2 minus b by 2? Correct. No Correct. Yes. Can I write cosine law for this thin triangle a c i? Or sorry a s i a s i not c i a s i This looked like a c to me So let's write the cosine law for the triangle a s i So cos of c by 2 minus b by 2 will be what? Can I say this will be a i square? Okay a i square My plus r square minus x square by 2 into by 2 into By the way, I can use my uh direct formula also here r is small r is known to be 4 r sine a by 2 sine b by 2 sine c by 2. So let me use that only So that will be 4 r sine b by 2 sine c by 2 square Okay So this will become 2 into 4 r sine b by 2 sine c by 2 into r. So r square Is that fine Any questions any concerns here Just a second sir So could you scroll up a bit here? Yes, sir. Yes, sir Done sir. Thank you, sir. Yeah So from here, can I say x square is going to be just correct me if I'm wrong It will be 16 r square sine square b by 2 sine square c by 2 plus r square Minus 8 r square sine b by 2 sine c by 2 into cos of c minus b by 2 For the time being let us keep our r square aside That's just a bottleneck in our calculation And let's write it as 1 plus 16 sine square b by 2 Sine square c by 2 minus 8 Sine b by 2 sine c by 2 and let's expand this which is cos b by 2 cos c by 2 plus sine b by 2 sine c by 2 Uh, don't worry about the switching of the order in cos in cos. You can always write it as b minus c by 2 as well Not a problem Okay Okay, some things will get adjusted as you can see 16 sine square b by 2 16 16 sine square b by 2 sine square c by 2 will get adjusted when these two multiply Right, so if I'm not mistaken, it will give you 8 sine square b by 2 sine square c by 2 minus 8 sine b by 2 sine c by 2 cos b by 2 cos c by 2 Take your 1 plus 8. In fact, you can take minus minus 8 sine b by 2 sine c by 2 common So it will be left with cos b by 2 cos c by 2 minus sine b by 2 sine c by 2 Now this is conveniently cos b plus c by 2 And cos b plus c by 2 is conveniently sine a by 2 Don't ask how you're dealing with a triangle Right So overall I can say let me just write it overall I can say this will reduce to 1 minus 8 sine a by 2 sine b by 2 sine c by 2 And hence your x may be written as r under root of 1 minus 8 sine a by 2 sine b by 2 sine c Okay, and we already seen that your small r is 4 r sine a by 2 sine b by 2 sine c by 2 So this hefty looking term over here, can I write it as r by 4 r? So if you place it over here, it becomes r square under root 1 minus 8 r by 4 r Correct, which is actually 2 r by capital R 2 r by capital R And if you introduce this r inside it will become r square minus 2 r r Okay, that is the formula for the distance between s and i Distance between the ortho center and the in-center Now a very interesting fact I would like to share with you s i is under root r minus 2 r r right Okay And you'll be happy to know that s i 1 Now what is i 1? S5 circle Right, so that would be r square plus 2 r r 1 So plus will come over here and r will become r 1 s i 2 will become under root r square plus 2 r r 2 And s i 3 will become under root r square plus 2 r r 3 Try proving one of them at your home. The approach is exactly the same. So I will not Discuss this because of the paucity of time right now Okay So Yes, that's it So in the April's analogy somehow related dramatically because we're getting b minus c by 2 there? No, not because you have to have a tan right tan b minus c by 2 I don't think so it is helpful much Of course indirectly somehow you can relate it but not directly I don't think so it is directly related over it Okay Next is the distance between Circum center and ortho center. Let's say I call this as s and I call this as h So the distance between the ortho center and the circum center is given by r under root of 1 minus 8 cos a cos b cos c Again, I'm leaving this proof as a homework. So please try doing it. The process is exactly the same You make that thin triangle and figure out the one of the angles there and apply cosine rule The approach is exactly the same. No difference between them. Okay Another interesting fact that I would like to share Let's say this is your This is your ortho center. I mean, I'm just Assuming that this triangle happens to be a this thing and let's say this is your Circum center s. This is your h Okay Centroid Let me make it a different line Centroid This is your centroid now remember centroid Ortho center and s they lie on the same line Okay, and this line is called as the Oilers line and g divides h and s in the ratio of two is two one Okay Now this this becomes a limiting case if your triangle becomes an equilateral triangle in your equilateral triangle h G and s are coincident. In fact, all the critical points are coincident Circum center in center ortho center Centroid they're all at the same point. So this two is two one then becomes a limiting case of that ratio Okay, so note down. This is the oilers line and The centroid divides the joint of ortho center and the circum center in the ratio of two is two one Okay So if somebody asks you this question, hey, how far is centroid from the ortho center? You already know the distance of sh. This is a distance of s. You just do two third of it You'll get the distance of centroid from the ortho center If you want to know how far is centroid from the circum center, you just do one third of this distance That will be a distance of the centroid from the Uh circum center Okay, so all those critical distances can be figured out from this particular fact That h g s lie on the same line, which is called the oiler line G divides the joint of h and s internally in the ratio of two is two one Is this fine any questions? Okay So a few things related to quadrilateral also we'll discuss because we have Very less time now. Uh, and we have to also discuss some solution of triangle cases Quadrilateral Okay, so first thing that we need to know is what is the area of any quadrilateral Area of any quadrilateral Sir, is it that root s minus a s minus b s minus a s minus two? Yeah, we'll talk about it Okay, so the very basic formula that you should all keep in mind is area of a quadrilateral is the product of its diagonal Into sign of the angle included take any one of the angle. Let's say this is angle theta Okay, so it is half Let me write it area of a b c d Is half product of the diagonals a c b d Into sign of the angle between them Please note. It is not going to matter. Even if you had taken this angle Because this angle will be supplementary of this. So sign of that will again give you the same thing So it doesn't matter. So half Uh, the product of the diagonals into sign of the angle between them So in case of special figures like a square Okay, uh, you can find the area of the triangle area of the square by just doing half into product of the diagonals Because the angle between the diagonals is 90 degree there. So sign 90 degree will become a one Okay, so make a note of this another formula that is very important and very less known Is you can also write it in the same way as you wrote the heron's formula Okay. Now in heron's formula, there was an s in the case of Uh, uh, quadrilateral also you will have an s which is your sum of these sites. Let's say I call these sites as Uh, a b c d Okay So there is an s for the Quadrilateral also so it is a semi perimeter So area of the cyclic quadrilateral. Sorry, not cyclic quadrilateral my my mistake area of a quadrilateral Is given by Note down, uh, I'll write it slightly shifted here Yeah under root of under root of s minus a s minus b s minus c s minus d minus minus a b c d cos square theta Is that fine? Excuse me Yes, sir. Tell me Uh, sir, could you repeat the relation relationship in the Euler's line? In the Euler's line, it says Euler's line is basically a line which connects orthocenter Centroid And the circumcenter so they all lie on the same line. That's what this line is called the Euler's line And this ratio is two is two one Yes, sir. Got it, sir. Okay Now this theta that you see over here is half of Any of the two opposite sides Angle by two I should use a different. I think notation here because you'll get confused with the formula then So if it becomes a case of a cyclic quadrilateral In case of a cyclic quadrilateral, you know a plus c is 180 degree so in case of a cyclic quadrilateral A plus c will become 180 so a plus c or for that matter b plus d will also become 180 So a plus c by 2 is 90 degrees which is also b plus d by 2 So what will happen? Uh, this term which is cos square 90 degrees that term will vanish So in case of a cyclic quadrilateral, so this is in this is in general any quadrilateral But if you have a cyclic quadrilateral cyclic quadrilateral Your area will become area of a cyclic quadrilateral a b c d Will become under root of s minus a s minus b s minus c s minus d because this last term a b c d cos square alpha will vanish That will become a zero. So why did you take opposite sides there? Where for alpha This is how it the formula comes out. So when you derive this formula, which again, I'm giving to you as a homework Okay, I'll also give you a hint. How to derive it So area of a quad uh cyclic quadrilateral area of any quadrilateral can be obtained by just adding up the areas of these Uh, you know triangles Okay, so when you do that you'll automatically get alpha as A half of the angle between the opposites half of the sum of the angle of the opposite vertex While derivation, you'll come to know Richard. Okay. Yes, sir Okay, so please derive it on your own and we'll discuss it Sir, could you go right? Sure Yes, sir, just Sir, this cyclic quadrilateral thing we can extend to any polygon or sorry Uh, I'm not sure about any polygon Depends upon the case I will not generalize this actually Okay, sir All of you have noted this down any questions One interesting thing I would like to discuss here is the tolmi theorem All of you know tolmi theorem What is tolmi theorem anybody can explain me? Those who have done preparations for prmo you have done so much of geometry So tolmi theorem says It's something like when you multiply opposites Yeah, tolmi theorem says if you have a cyclic quadrilateral If you have a cyclic quadrilateral Then the product of the diagonals That is ac into bd is ab into cd that is product of the opposite sides sum So ab into cd and bc into ad So this is the case. This is a Formula for tolmi theorem again. I would leave up to you to prove it This can be easily be proved by taking cosine law On this angle and this angle and we are knowing that they are supplementary. So if this is Theta, this is 180 minus theta So please prove this tolmi theorem So good. It's a good theorem This is a good theorem, which will help you to solve a lot of different types of Geometry based problems on quad cyclic quadrilateral One more interesting thing I would like to share with you If this quadrilateral is not a cyclic quadrilateral Then what will happen to this? What will happen to this inequality? What will happen to this equation? This will actually become an in equation, right? And now I'll leave up to you to decide whether this will be less than equal to symbol or this will become a greater than equal to symbol Okay So a lot of homeworks I'm giving you because of the paucity of time. I cannot do everything If it is not a cyclic quadrilateral If abcd is not cyclic If it is cyclic, then it will be equal to That is fine that we can easily prove If abcd is not cyclic, what symbol will come over here? Greater than equal to or less than equal to that's what I would like greater than equal Figure out figure out figure out. So right now is will not waste time on that Actually another thing I would like to discuss with you for a cyclic quadrilateral the circum circle that you have drawn over here Can be linked up to the sides and that is and linked up to the area also So it is given by four one by four delta under root of All possible divisions of group two you can make between The sides Oh, sorry Okay, so this is like making all possible divisions of four people in groups of In two groups of equal sizes a cbd adbc abcd okay, so this is the Radius of the circum circle which circumscribes a cyclic quadrilateral Okay, please make a note of this very very interesting. You may include delta inside and you can say it is under root of ac bd adbc abdc or abcd by area of a cyclic quadrilateral we already know Okay s is a plus b plus c by A plus c plus d by 2 s is always the semi perimeter Okay, now last half an hour of our session we'll talk about solution of triangles It has been uh Very very fast So are you gonna take some problems? I'm sorry So are you gonna take some problems? We'll see we'll see let me first can uh Finish off solution of triangle that is important None sir, thank you None Yes, sir Yes Solution of triangles Solution of triangle is basically a concept where See every triangle has got Six Parameters angles That we know abc And side lengths correct Now solution of triangle is basically a concept where it says that if you know any three of the six unknowns Except all angles So if any Three If any three Of a b c capital a capital b capital c is known Except All Angles Then It's possible It's possible to find Triangles Why are all triangles is because they can be more than one triangle also having Or it is possible to find possible to find the unknown unknowns of the triangle And they can be see it With a given three information You may have multiple triangles also possible. I'm not saying that you'll always get a unique triangle and you'll get a unique value of the unknowns I'm not trying to say so that's why I normally include this word s because see many a times what happens when I give you three of the uh given uh information You may end up getting two triangles Which has got the same set of unknowns uh same set of uh parameters Okay, so there need not be always a unique triangle that will come out from this particular exercise That's why I have written triangles over here Okay, now why not Why not we can figure out uh with this uh all angles given because as you know that If angles are infinitely many triangles And it's very difficult to find out uh for infinitely many triangles. What are the unknown parameters? So if all are not angles, then only we can easily figure out what are the unknowns So now what are what are we going to take here? We are going to take up some cases Where you know, uh, what are the possibilities that can be given to us? The first case is where all sites are known Okay, that means if you know abc of a triangle What are the unknowns here? Unknowns will be angles Can you figure out the angles if you know the sites which rule is the best for this? Use Sign rule cosine rule Correct cosine rule is the best way forward And cosine rule why we normally recommend is because when you know the cosine rule cosine rules are very sensitive to quadrants Right, so you know exactly when is the angle acute and when is the angle obtuse But sign creates the issue because sign 30 degrees sign 150 degree both will be half and both can be possible Angles in a triangle Okay, so cosine rule is the best put forward when you want to find out the When you want to find out the uh, you know unknown, uh, angles of this triangle Now i'm not saying this is the only way there are a lot of other means by which you can figure it out If you find out the area of the triangle which you can easily find out by knowing your semi-perimeter from here Okay And you can use the fact that sign of a is equal to two delta by bc But the problem here is that if you get an answer like half You won't we won't able to figure out whether a was 150 degree or a was 30 degree So that creates an issue. So i'm not i'm not ruling out the possibilities of sign rule for it But sign rule make create confusion Okay Tan is also a very good formula. For example, you can figure out tan a by two by using the formula delta by ss minus a remember we had uh Done this when i was trying to find out the Uh in circle radius. Okay. This also is a good way to find out the angle Okay, similarly tan b by two tan c by two etc. But the best is your cosine rule You need not find out uh, delta and all those things you can easily deal with abc directly. That's the best way to get Okay Case number two Case number two One more thing I would like to ask If three sides are known Will I get a unique triangle for the side or will I have various different triangles for the side? We'll get a unique triangle. Yes We'll have only one angle opposite to one. Only that combination is good. Correct Now next is when two sides and included angle are given When two sides and included angle are given Included angle is is known. For example, if let's say somebody gives you Somebody gives you uh, let's say side length a side length b and this angle c Right, will you be able to figure out the other angles and the unknown side? If yes, how? In this case, the best formula to use is your Napier's analogy So I can directly use here and get my And get my unknowns very easily Okay, so from here, I know abc so I can use it on the right hand side So a minus b will be known from here A minus b will be known a plus b is already known to me because I know c So from these two I can figure out my a and b values Now, I'm not saying they're not other ways. You can also go for you can also use your uh concept of uh sign rule So you can find your opposite c by using the formula a sin c by sin a correct That is also possible Okay, so a lot of things can be done The third case which I am going to do That is slightly irritating And that is called the ambiguous case case number three So case number one and case number two. I hope nobody has any issue Okay case number three This is called the ambiguous case Here Two two sides will be known Let's say I just make a diagram let's say, uh Let's say I give you this side bc And I give you this side b And one of the angles which is not sandwiched between the two sides is given. Let's say angle a is given. So let's say You have been given a b and angle a then how do you find out The other unknowns from here. Okay, now you can have many Possibilities. Yeah, there are many possibilities that will come out from here. We'll take up uh one by one in the interest of time So first write cosine formula So now c is unknown for you. Remember c is unknown for you. So this is unknown for you So let's write down cosine formula cosine formula will give you something Like this So this will actually result in a quadratic in c Oh, let me write step by step So this will give you c square minus two b c And you'll have one Plus cos a right? Am I right? Oh, yeah, sorry, sorry My brain has stopped stopped working. I think So yeah, so b square, uh One second one thing I think I was thinking ahead of time So b square plus c square, uh minus a square is two bc cos a So you'll get c square minus two bc cos a Plus b square minus a square. This is a constant term like this. Yeah Okay, this is what we get now. This is clear cut quadratic Quadratic in c now when there's a quadratic in c you already know that you can have various possibilities You can have distinct roots. You can have equal roots. You can have No real roots also. Correct. So let us try to figure out What happens and when it happens see If you take The quadratic equation formula So what do you end up getting is? b cos a plus minus Uh four will be cancelled out from here. So it'll be b cos square a plus a square So it'll be a square minus b square sine square a let me know if this is clear. This is your c Okay, so two I cancel out From this four four and two here b square cos square minus one is minus b square sine square and minus of minus a square is a square Now a lot of things depend on this guy A lot of things depend on this guy if if a square If a square is less than b square sine square a Right in other words a is less than b sine a Okay, that means if sine of Let me write it like this That means uh A by sine a is less than b Okay Now what does it mean when a by sine a is less than b Remember Normally a by sine a is actually b by sine b Correct. So you're trying to say b by sine b is less than b Right indirectly you're trying to say something which is not possible Okay, in other words if this becomes negative Then there cannot be any triangle possible So let's say if you're a b and angle a were such That a became less than b sine a Then a triangle will not be possible. So there'll be no such triangle. Are you getting my point? There will be no such triangle possible So if you draw a diagram out of it Let us say What is given to us we have been provided with this angle a And we have been provided with this Sides a b and this angle a Okay, now remember What is b sine a What is b sine a Oh, sorry b b sine a would be this this length This length will be b sine a So if this this length is smaller than this That means you're trying to say your perpendicular is You know longer than your hypotenuse Okay, so such a triangle cannot be existing at all such a triangle cannot exist at all Okay, so please note this now This is first of your case three Let me call it as a similarly if if Uh, I should do it sidewise here if your a becomes equal to b sine a Then definitely Your uh, the term here would be zero And you will only get You will only get one such triangle And that triangle will be a right angle triangle adds At b Am I right? Yes, yes or no Right, so if this length is If this length is your this length that means your triangle is cut short to here Correct, that means your b will automatically become a right angle triangle So for such a case you will get a unique answer So you will end up getting only one triangle possible one triangle possible Sorry one triangle possible Okay Now your c part Where your a is greater than b sine a let's talk about it If you a is greater than b sine a Now let's go back to the roots here So the root was b cos a C was b cos a plus minus under root of Under root of a square minus B square sine square a So if we get it greater than we have to check the root itself whether it's negative or not Sorry, I didn't get you So if we get it greater than b sine and then we'll have to check the sine Right now this is possible. This is positive. Correct. Now if if you take a plus over here If you take a plus over here Okay, then what will happen? Can I say Can I say let me write it like this. Let me write it like this If b cos a is greater than Under root of a square minus b square sine square a Okay Then in fact, let me simplify this b square cos square a is greater than a square minus b square sine square a That means b square is greater than a square. That means b is greater than a Then in that case Then in that case in fact, I should write mod over here. Anyway, it doesn't make a difference Then in that case you will have two possible values of c because even with plus or minus you will be able to find a positive c Am I right? Am I right? So this will result into two possible c Should we check that it's less than a plus b? Two triangles are possible. Sorry Should we check that this will be less than a plus b? Which one uh c? Yes, sir But that basic inequalities you have to always check So won't that come automatically or again we'll have to check on top It should come out automatically It should come out automatically Okay Okay, but if c is but if a is less than b what will happen If this condition is not rectified that means if a is less than b what will happen if a is less than b then Acha here one thing we have to keep in mind that our angle b is acute I'm sorry a is acute else. What will happen? Uh, there will be a problem with this guy because if I take a minus here And this also happens to be negative term more in magnitude then there will be a problem So I'm assuming that a is a cute angle over it Okay So if a is acute and b is greater than a then there are two possible values of c and hence there are two possible triangles But if a is acute and a is less than b Then what will happen? We can only have the plus case working out Not the minus case because minus case Will make the whole thing negative because magnitude wise this guy will be smaller than this, isn't it? See if a is less than b means what are you trying to say? You're trying to say magnitude of this is less than under root of a square minus b square sine squared Am I right? And if this happens to be lesser than this Then c will become negative If I take a negative sign over here So only plus sign is a possibility that can work here. So only one triangle is possible in this case Are you getting my point here? Okay What if a is obtuse If a is possibly become negative Now if a is obtuse the only one case I will get when This guy supersedes this in its negativity Yes or no So I will always have when your a is less than b because then only This fellow will supersede this guy in the negativity, right? Correct and I can only have when this is a plus I cannot have when there's a minus because minus means this will also be minus this will also be minus because a is obtuse So only with a plus and only when a is less than b I can have a triangle So one triangle possible Okay, and if a is greater than b no triangle possible Why because this guy has been beaten in magnitude by this And this is already negative So no matter you keep it positive and you and this guy is already negative Anytime the answer of c will come out to be negative So you cannot have a negative side of a triangle your basic inequalities must be fulfilled side of a triangle must be positive So no triangle possible. Are you getting my point here? So the third case is a complicated case Where you have got three possibilities coming in itself because of the nature of the roots And the third one itself has two more possibilities The third one itself has two more possibilities where a is acute and a is obtuse A is acute we'll have possibilities A is obtuse we have possibilities Are you getting my point here? Yes, sir It is Last part is something which you need to do it on your own also and try to figure out this theory once again So in the interest of time one question can be taken here. I cannot take more than one So can you go to the slide before this slide take 34 slide 34 yes, sir So nps hsr next sunday you will not have a class i'm repeating once again Next sunday you will not have a class you will probably have a test or something like that and uh Rnr has to have uh rnr and ypr will have a class on sunday So tuesday which is coming and next sunday you'll have mass class And you any tests if it if at all it is given to you you will be writing it in the evening time because nine to one they'll be class So that would be your last class. I believe uh for the session Thank you. Thank you, sir. My question is If a is 30 degree Small a is seven b is eight Right Then how many triangles Can be constructed with this dimension find the number of triangles possible with this dimension so just recall we have b cos a plus minus under root of a square minus b square sine square a right a is already acute no issues with it and This guy that is 49 minus 64 sine square a sine square is one fourth correct This is positive Correct. So this is positive This is also positive Okay, and Let's also check What is the relationship between a and b sine a Oh that we have already figured out what is the relationship between a and b b is greater than a Am I right b is greater than a that means even if I take a negative sign over here Or a positive sign over here There is a possible value of c coming out from there. So there'll be two triangles possible fine So, uh, I think this is all we have uh for our properties of triangle. I've done the chapter to the best of My time constraints, but yes, it is subjected to uh, how much you practice out of it I will send you some, uh A worksheet on it So at least try to complete all the questions in that worksheet It will be not much. I think 40 45 questions will be there So just try to solve them that will be more than enough for you Don't spend too much time on this because It's just a part of your big chapter called trigonometry. So probably even not even one question may come from this Thank you Sir Yes