 So in properties of triangles, the last concept that we stopped at was your cyclic quadrilateral. If I'm not mistaken, so properties of triangles continued. So we stopped at cyclic quadrilateral. Okay, so in cyclic quadrilateral, I think we had discussed about the cosine rule. So in today's session, I'm going to start with one of the very interesting theorems which is applied to cyclic quadrilateral is Ptolemy's theorem. Now those who have done geometry of PRMO, RMO level in your class 10 would have definitely come across this theorem, Ptolemy's theorem. So what is this Ptolemy's theorem? Let's discuss about it and we will also prove this. Okay, so we'll not only write the theorem, but we'll also prove it. So let's talk about Ptolemy's theorem. So let's say this is a cyclic quadrilateral. Okay, this is a cyclic quadrilateral. Let me just make the diagram. Okay, so any cyclic quadrilateral A, B, C, D will satisfy the fact that the product of its diagonals. So AC into BD will be equal to sum of the product of the opposite sides. So AB and opposite side to it is CD plus BC opposite side to it is AD. Okay, so what does it say? It says that the sum of the product of the diagonals is equal to the sum of the product of the opposite sides. Okay, so AC into BD is AB into CD plus BC into AD. So this is applied to a cyclic quadrilateral. So this particular formula, it works for a cyclic quadrilateral. Let me write it down here. Cyclic quadrilateral. Now, an actual question will arise in your mind, what if the quadrilateral is not cyclic? Then what will happen to this theorem? Okay, so a small change will happen if the quadrilateral is not cyclic, then AC into BD will be lesser than, in fact lesser than the sum of the product of the opposite sides. Okay, so this will hold true if the quadrilateral is a non-cyclic one. Non-cyclic quadrilateral. Is it fine? Any questions? Now, for a non-cyclic quadrilateral, we are not going to talk about it, but it is good to know that for a non-cyclic quadrilateral, the product of the diagonals is actually lesser than the sum of the product of the opposite side. So we are only going to focus on this game. So this fellow we are going to prove now. Okay, so let's prove this fellow as of now. Now for proving this, for proving this, let me recall my cosine law. In fact, let's talk about, okay, so when we talked about the cosine law in the last session. Okay, what was your cos of B? Let me name it also. So let's say this is your A square, not A, B, C and E. Right? So if you apply a cosine law, I mean the normal cosine law for the triangle, okay, let's say I call AC as X. Okay, so in triangle A, B, C, if you apply cosine law, so cosine law is nothing but A square plus B square minus X square by two AB. Okay, which means X square, which is nothing but square of AC becomes A square plus B square minus two AB cos B. And now what I'm planning to do is I'm going to replace my cos B. If you recall cos B formula we had learned was A square plus B square minus C square minus D square, where A, B, C, D are the sites of this quadrilateral upon two AB plus CD. Just turn the pages back. I think this was the last thing that we had done in our previous session. So what I did first, I'll again reiterate it for the people who joined in a little late. So we are doing the proof for the Ptolemy's theorem. And what I did was first I called this AC length as X. And for the triangle A, B, C, I wrote the cosine law, which is A square plus B square minus X square by two AB that we have already seen in our triangles. So we got X square in terms of A square plus B square minus two AB cos B. Now this cos B we had already seen in our last class for a cyclic quadrilateral cos B is given by this expression. Okay, so so far we have only done so much. Now we are going to do a simplification of this. Okay, don't worry. I'll prove it very, you know, in a very easy way. So I'm just doing a simplification of AC squared. Let me take the LCM so it'll become by the way this to this to you can cancel off. So you have a square plus B square times AB plus CD minus AB times a square plus B square minus C square minus D square. Okay, so let's try to cancel off few terms I can clearly see will cancel off and I expand it. This AB times this and this AB times this, they will definitely cancel each other out. So I'm not writing all the unwanted stuff. I'm just writing the stuff which is required for us. So I will have a square CD. Okay, I will have a square CD. I will have B square CD. Correct. And from here I will get plus AB C square and plus AB D square. Am I right? Is there anything which I'm missing out? Please do let me know. Okay, all good so far. Any problems, any concerns so far in whatever I haven't done. Okay, now what I'm going to do, I'm going to club these two terms together. The ones which have ticked mark with yellow and I'm going to turn these two terms together which I have ticked mark with white. So if I take club these two, please note that I'll be taking AC common and I will end up getting AD plus BC. And from these two I'm going to take BD common and I'm going to get AD plus BC. Okay, in short, this is going to get factorized. The numerator is going to get factorized as AC plus BD times AD plus BC. Okay, now this is the expression for, this is the expression for AC square. Now look at the figure, AC square. I'll just copy the figure once again so that we can see it clearly down also. So let me just copy this. Okay, so let me take this figure and copy over here. Yeah. So look at this figure, this is your AC square. Now I would like everybody. Okay. Shouldn't there be a two in the denominator right, you cancelled it before. Yes, yes, yes, yes, yes, yes, yes, yes, yes. Ah, thank you, thank you, thank you. Thank you, the Yasini. The Yasini you're the only person who, no Prisnim also pointed it out. Yes, thank you so much. I cancelled that two in the denominator so that will not appear. Yes, exactly. Thank you so much. Okay, now I would like you to tell me looking at the same expression, had I done the same thing for BD. That means if I wanted to you to tell me what is BD square. Can you take a clue from this and tell me what would be BD square. This is your bd let's say and I want you to tell me what is bd square take a clue. See we have ac plus bd Okay, let's say while we're finding out this we had ac plus bd. That means we are basically taking I mean, how would you like to read this? Product of the opposite side so that will equally be applicable to bd square also So we'll have ac plus bd there. Okay. Now see the second term ad plus bc a D Plus bc. Now. What are you doing here? You are basically taking the two sides which are on either side of that particular Diagonal and you're multiplying them and adding them so ad bc So if I have to do the same for bd square, I will do how ad Plus cd. Am I right? Ab plus cd. So that will that is what is going to come Yes or no Again, I'll tell you see you had ad plus bc, right? So let's say focus on ac diagonal and are Other two opposite sides of that particular diagonal and b and c are also two opposite sides So here while you wrote this expression, you can perceive this as if you multiplied this and this And you multiplied this and this and added them right in the by the same logic See, I don't want to derive bd square again just like I did for ac So I'm just trying to you know reach to that result in a faster way So for bd what I will do I'll multiply this this And I will multiply this this and add them. So that will give me ab plus cd In a like manner Is it fine? Okay, and what happens to the denominator? What happens to the denominator? Yes. Sorry. Yeah So what happened to the denominator? So now see here earlier It was abcd. That means you multiplied the two sides, whichever on the same side of that particular angle, right? Right. So here you'll do ad plus bc, right? So you'll do ad plus bc. Absolutely, right? Okay. Now, let us multiply these two Let us multiply these two So from these two if you multiply it, what are you going to get? You are going to get ac square into bd square. Let me write it down Let me write it down So ac square was ac plus bd times ad plus bc upon ab plus cd times And this was in yellow. I'll keep writing it in yellow so that Yeah, this is going to be a a lot of terms is when to get cancelled off This term gets cancelled off with this this term gets cancelled off with this ultimately you're left with Ultimately you're left with ac plus bd the whole square now Just remove the squares from everybody so from both the sides Okay, so both the sides remove the squares So you'll see something like ac plus bd on the other side. Now. What is it? ac let's check the figure A is your ab and c is your cd. So this guy is nothing but ab into cd And what is your small b? Let's check the figure again small b is bc Okay, and this is your ad Correct. This is what we wanted to prove This is what we wanted to prove. Is it fine any questions any concerns? Okay, so I think with this we have done a fair bit of our You know properties of triangles part Now we are going to start with solution of triangles Okay, so let's take solution of triangles part from here on which is again an important topic Few questions have been asked based on the same in your competitive exams So this was your Ptolemy's theorem and we are now going to take a look at some of the questions Ptolemy's theorem and we are now going towards towards the next segment of our chapter which is going to be the last segment, which is called sort solution of triangles Okay Now what is this meaning of this topic solution of triangles properties of triangles many of you make sense out of it So you learn some properties which are related to triangles So what is the solution of triangles many people call ask me sir? I don't I didn't understand the name of this chapter solution of triangles So solution is normally associated with you know solving equations, right? So we find solution of equations So is there any equation that is involved? So are we solving some kind of an equation? Actually, they're not solving an equation, but you're solving for a triangle. Now. What is the meaning of solving for a triangle? Now, let me explain this When you talk about any triangle, okay, let me just make one quickly a Triangle has got six parameters. What are the six parameters? The three angles ABC and the three sides small a small b small c So in all together a triangle has got six you can say parameters associated with it The three angles those are three parameters and the three sides. Those are also three parameters three plus three six Now it is said that In fact, logically that if you know any of the three you will be able to find the remaining three Okay, except when all the three given to you are angles So if somebody gives you any three of the six parameters You can actually find the remaining three also other than those conditions where all the three are angles So if somebody gives you all angles, okay, let's say I give this is 30 degree This is let's say, you know 70 degree and this is 80 degree. Can you find a triangle? You'll say there there are so many triangles with that particular angle, okay? Or with that particular set of angles so other than the three information being all angles If somebody gives you any three parameters, let's say two sides one angle or one angle two sides or three sides You can definitely find the remaining three as well So finding the remaining three from the given three is basically called solving for a time Clear is that here? What is the meaning of this chapter solution of triangles? So I'll write it down. So this chapter deals with finding finding triangles with Three given parameters, okay three given parameters except when except When the three given are all angles is basically called solution of triangles. Okay, so this chapter deals with this concept Okay, plain and simple So, let me see let's let's see how many situations can be given to us and what all properties that we have already learned in our First part of the chapter is going to help us to find the unknowns which are there in the triangle, okay? So let us start with the condition when you have been given All three angle, sorry, all three sides Okay, please note that all three angles if somebody gives you You I wrote all three angles all three sides All three angles if somebody gives you there will be infinitely many triangle with those those angles, okay? So let me start with the first case where all three sites have been given to you So let's say in a triangle you have been provided I'll just make a triangle once again You have been provided with small a Okay, so let me write the ones which have been given to me in yellow Okay, and let's write down the ones which have not been given to me in white Okay, so small a small b small c are given to you. How do you find capital a capital b capital c? How do you find the angles of a triangle? And can you tell me how to find it out anybody any suggestions Which law? Which concept will you like to use here to get the unknowns? Anybody? It's not a rocket science Cosine law absolutely, right? They're just sitting absolutely, right? So by using your cosine law, you can find your all the angles isn't it? So nothing to worry about it. This is something which is already known to us. See I'm not teaching anything new here It's like solving for a triangle using the properties already known to us, right? So there's I mean, of course you'll be learning few new things, but idea is you have to implement You have to apply the concepts that you have already learned in the first part of the chapter Okay, so using this you can find your a Using this you can find your b Okay, and using this you can find your c Anglesy, okay, when I say c means angle c Yes, I know yes, I know Okay, is it fine any questions any questions any concerns all right Let's take the second case Let's take the second case. So from here. I'll just write it down from here. You can find your Find your a b and c. All right Let's take the second case Second case is where you have been given two angles two angles and any one side and Any one side so as I told you three things must be given to you. So in this case, I'm assuming that you have been provided with two angles Two angles and one side. So how do you find the remaining? Let me just make a dummy triangle over here Okay, a B c. Okay, let's say the question setter has provided you with angle a okay I'm writing I'm writing down whatever has been given Okay, and let's say he has given me one of the sides. Which side you want to take? Let's say this side Okay, small a how do you find out? As I told you, I would write the known things in yellow and unknown things in white So this is not known to us. This is not known to us By the way, this is small c not small a small c. Okay, and this is small b. So, how do you find? how do you find small a small b and Capital C when these three things are known to you any idea So you'll say sir for small c Sorry for capital C. You just have to do 180 degree minus a plus b There's nothing much to do about it Yes or no Correct So nothing to do for finding Capital C right, how would you find small a? Any idea? How do you find small a? Okay? Shalini shardhuli, okay, so I think Somebody suggesting me sign law. Okay. Now, let's say I use sign law now that you found your C You can use C by sign C Is equal to let's say I want to find small a right small a by sign a Okay, can I use this very much? I can easily find it out by using this Okay, so small a can be obtained from here Let me write it in white as I already told you. This is our unknown. So I'll write it in white Yes or no, can we now find out from the same expression? Small b as well Correct. So this is going to be b divided by sine b So b also can be found out So b will be c sine b upon sine c easy Yes or no, so sine law is going to help us out Nothing more to do here Now coming to the third case Any doubt related to the second case? These all cases are very easy. I mean the the trickier ones are going to come little while down the line So now we have let's say given two sides and an included angle And the included angle I should say because there will be only one included angle between two sides So let's say the question center has given you Two sides of a triangle and the included angle. So again, let me make a diagram. Okay, so let's say Let's say a b c is our triangle Okay, and the question center has provided you with Let's say small b and small c And he has provided you with this angle a Okay Then how do you find out the other things again? Let me make this in yellow because it is already given to me Yeah, how do you find the other things like small a Angle b angle c Okay, so here b c and a are given to you How would you find small a small b? Sorry small a capital b capital c What tactics will you use? Okay, so Sharduli has an idea So she's saying that you can use You can use cosine law to get your small a Okay, so we know a so we know cos a cos a is b square per c square minus a square by 2 b c So from here you find out your small a Okay, so this now will be known to you Now small a is known to you Then how would you find out capital b now here is a call that all of you, you know, please take it So if you know a you are going to do a by sine a because you know a You know small a now also because you are found out by using your cosine law Now let's say you wanted to find angle b and you are going to do small b by sine b So the problem here that would arise is let's say you find out sine b by using this formula b Sine a by capital a Now let's say let's say if this comes out to be half Let's say hypothetically Would you take your b as 150 degree or will you take it as? 30 degree Right, so there may be a confusion especially when your a is A is not an angle big enough to rule out 150 degree Okay, so if let's say a 60 degree then you can say automatically sir a is already 60 degree Then b cannot be 150 because you don't want an angle of a triangle to become more than 180 Right, but let's say a was a deceiving angle and you ended up getting sine b as a half The problem with sine is It gives you the same value for two angles which are supplementary And both could be possibilities in a triangle So this might create a confusion. I'm not saying always you'll you know get into a problem. So this might Might be an issue this might create confusion Okay, so I normally refrain from using sine law in such case Okay, what I would do is see I've already found out all my three sides so why not use cosine law and Cos b and cos c to get our remaining side. So cos b is what a square plus c square minus b square by two ac And cos c is what a square plus b square minus c square by two a b Why not use this? Okay, so better to use this And refrain from using this. I mean because you know, it may cause a confusion But it depends on case to case. I'm not generalizing it. It's not going to happen always There's actually one more way to get it and that is by the use of tangent law Most of you would remember tangent law napier's analogy That also can help you to get the assault for the triangle. So this is one way Okay, this is one way. This is your second way to do it and third way is your use of of Of tangent law now see here b and c are given so we can use the formula tan b minus c by two Is b minus c by b plus c cot a by two, correct? So, please note the question center has already provided you with small b small c and angle eight, right? So from here, you can find b minus c. This can be found out correct And since a is known to you b plus c is also known to you Right because a is known to me. So from here, can we not solve for b and c? Yes, I know We will take some questions. Don't worry later on later on during this session and take some questions So that is another way to get the you know unknowns from the triangle. Is it fine any questions any concerns? Any questions any concerns? Okay Now the last case which is the fourth case that is very important Okay, so we'll talk about that but before I move on if you have any questions, please get it addressed Okay, all fine So let's move on to the fourth case fourth case is where you have been provided with two sides two sides and a non-included angle a non-included angle Okay, this is called as the ambiguous case Okay, it's has been given a name as an ambiguous case, but don't get scared Sir ambiguous case means nothing is known. No, nothing like that. It's just a fancy name which has been given to it So when you are given two sides And an angle which is not included Between the two sides, so there will be two non-included angles, right? So any one of the two is given to you Then how do you find out the complete triangle? So let me just show the diagram here So let's say the question setter are you The question setter has provided with a b Okay, and let's say a c And a non-included angle. Let's say this angle Okay, fine. So what is given to you small b small c And capital b they are given to you Okay, how do you find out the other unknowns? In a triangle that is what we are going to discuss under this ambiguous case So listen to this very very carefully Yes, you may have multiple triangles satisfying the condition So that's what we are trying to figure out So why multiple triangles come out and what are the possibilities that can come out? We are going to analyze it in a very great depth So let us Utilize our cosine law first of all. So when we apply cosine of b, we end up getting something like a square Plus c square minus b square by 2 ac Now, please note that bc are known a is not known So in short, if you see here, you basically give you a quadratic expression in a A is an unknown mind you Up till now, we have not found out a so this gives me a quadratic expression like this Okay So this becomes a quadratic in a so when I say a quadratic in a obviously We all know that a quadratic equation can have two roots Okay, so let me write it down quadratic Equation in small a right So that means two values of a can come out correct So from this situation I mean it is not necessary that all quadratics will give you, you know, uh, to equate two roots Maybe a quadratic can have a single root also. It may have no roots at all also, right? So please understand here very very important If this quadratic equation If this quadratic equation gives you Two positive values of a Okay Then only you have two triangles possible Okay But if it gives you Only one positive value Another value is negative Okay, then Then there's only one triangle possible Right So basically the number of a's that we get the number of positive values of a that we get from this equation that will decide how many triangle will be there Okay, so if I'm getting two positive values of a That means two triangles are possible having the same b c and capital b Okay But if I'm getting two answers to real solutions out of which only one is positive and the other is negative Or zero for that matter then only one triangle is possible Okay And if I'm getting let me write it down here also and if I'm getting Two negative values of a Two negative values of a or non real values of a value of a or or Non real values of a Okay, then in that case there is no triangle possible Okay, so in this case there will be no triangle possible So all depends upon All depends upon what are the roots that come out from this equation So if two positive roots come out well in good two two triangles would be there which will satisfy the same condition If one positive one negative is there That means only one triangle is there with that given but particular b c l's capital b And if both the roots are negative or both the roots are non real means No such triangle is possible with that given b and c so everything depends upon the Nature of the roots that you get from this quadratic equation Don't worry this quadratic equation You can easily solve it by using any of your methods known to you to solve quadratic equation Normally we use our sridhar acharya formula factorization method whatever is known to you So once you solve it see what is your a values coming out? So depending upon these three situations you can take a call So normally a question will come that you know, how many triangles are possible Which will have this given small b small c and capital b So like that for that type of questions you can basically form a quadratic in a And check what are the nature of the roots that you are getting? Okay But we are going to do a more deeper analysis of this So we are going to see what type of situations will arise and why those type of situations are actually Arising we'll talk about that in much more depth. So first note this down Let's do a small analysis Since we were talking about the root of this equation playing a very vital role. Let's see Let's see what all situations can arise. By the way, I'm just assuming that I'm copying things correctly I'm a very short memory. So you have Yeah c square plus c square minus b square. Yeah plus c square minus b square So all of you, please pay attention Now if I talk about the Discriminant of this quadratic equation, let's let's let's talk about the entire value of a. Okay Let's not only talk about the script and let's talk about the entire value of a itself So as per our Shridhar Acharya formula A value will be Minus b. Please note that here b is negative 2c cos b. So minus b will be 2c cos b Okay, plus minus under root b square minus 4ac So what is b square here? It will be 4c square cos square b minus 4ac Please note that a is actually I mean the a which I'm talking about is that quadratics a coefficient of a x square So in that case, it is a one whole divided by Whole divided by 2a which in this case is 2 Okay, so let's do one thing. Let's simplify this I think 2 2 factors can go off from everywhere because I can see a 4 right over here and there is a 2 also sitting over here So if I get rid of that extra 2, I'll end up getting something like this Okay, minus c square plus b square. By the way, this can be clubbed as This can be clubbed as c square times cos square b minus 1 which is minus sine square. So I can write it like this Any doubt any concerns so far related to this So now we are going to see what all situations We will have two roots. What two triangles possible? What all situations will have only one triangle possible? What are the situations when we will have no triangle possible also? Okay, so everything will be hidden in this particular a expression. So it all depends on a Right, so if you want your a to be, you know, if you want your triangle to exist It should give you a as positives Right, no negative values of a no real values of a so let's let's analyze that into more depth So I'm now going to shift my screen to the right side Hope you have copied this down. I'll copy it once again. Don't win. So now the analysis starts analysis So your a is c cos b plus minus under root b square minus c square sine square b. Yes, this is what we had written Yeah, b square minus c square sine square. Okay. Now three situations will arise over here. Number one situation If your b c remember these three are given to you. Okay Don't forget this these three things are given to you. Okay If you realize that your b is such a value which is c times sine b Okay If you realize that your b is c sine c times sine v means you realize that b square is c square sine square b Which means you realize that b square minus c sine square b is equal to zero Then what will happen to a value? What will happen to a value? It will just reduce to c cos b correct Agreed? Yes or no? Yes or no? Yes or no? Okay. Now tell me How many triangles will be possible? How many triangles will be possible? Looking at this, what conclusion do you draw? Okay. Shardini says one triangle is possible. Anybody else? See this is where I want you to think So as per her, only one value of a will come out. Right? For a given c For a given capital V, you'll have only one value of a But what Shardini Will happen if that b was obtuse angle Then no triangle will be possible. Right? So here is a catch Your answer was actually half. Half correct. Okay Here if b is acute See if b is acute then you're definitely your a will be positive Okay, so a will become positive So only one positive a value will come out from there and hence only one triangle will be possible Okay, and that triangle will actually be a right angle triangle at c Okay, so one triangle is possible Okay, so let's check. Sorry, not right angle. Let's see. It will be Let me just show you. So one triangle is possible And that triangle Everybody please note this down a Is equal to c sign b right? I'm sorry a is equal to c cos b. Yes or no Now if b is acute Only one triangle is possible And that triangle, please understand here Is going to have I'm just going to write it over here So a by let me make a diagram out of it. So let's say you have been provided with You have been provided with b You have been provided with a c Okay, and you have been provided with an angle b. Okay now see everybody please pay attention So this is your let's say I'm just making a Complete triangle Let me make a complete triangle a b c So question center has provided you with a small b the question center is provided with a small c And the question center has provided you with this angle b Okay, now you are trying to find out your a location. Sorry, uh c location, correct Yes or no That means you're you're basically trying to find out your small a If your b happens to be if your b side happens to be C sin b now c sin b is actually a perpendicular correct Yes or no, so what are you doing here? You are basically having this length Equal to this length actually So this becomes your b Are you getting my point So when you're when your small b is equal to c sin b look here You're you're you know perpendicular from a on to bc becomes your small b That means this is the only location that you can have for your small c Are you getting my point? Okay, so that means the triangle is going to be a right angle at c Make sense. So that triangle is going to be a right angle at c Right angle at c getting this point very very important Now this can be proved very easily by using your sin law as well Okay, so we already know in sin law We already know in sin law that b by sin b Is equal to c by sin c correct Is equal to a by sin a Yes or no Ha will come to that obtuse part will come to that obtuse part. That's why I made one branch out of it Okay, I've only made created one branch out of it Okay Now see here if you put your b as c sin b over here in this part of the equation you put your b as c sin b Okay So it becomes something like this Sin b sin b goes off Small c small c also goes off. So that gives us sin c as one value And if sin c is one there can be only one possible angle for that and that is 90 degree thereby Reinforcing this fact that you will get a right angle at c Yes or no but what if b becomes obtuse If b becomes obtuse You will end up getting a as a negative value because cos b is negative if b is obtuse and in such case No triangle is possible Are you getting my point? So no triangle is going to be possible in such a case In other words, you're trying to say that This length b is actually a perpendicular drop from a to Let's say bc Okay, and at the same time b is also obtuse. It cannot happen right because there's already a right angle triangle at c And you're saying b is obtuse also. How is it possible? You cannot have a triangle with 190 degree and one obtuse angle inside it Overall the total angle will exceed 180 degree Right So that thing is taken care by the fact that you automatically get a as a negative value And because of that, I mean we know that in a triangle, we cannot have a negative side, right? So because of that we claim that there will be no triangle possible in such case Is it fine Okay, so Of course everything will be conveyed to you by that quadratic equation So you don't have to worry, but this is something that if you want to do a faster analysis And you realize that oh my my b is equal to c sin b and b is obtuse Immediately you write no triangle possible immediately you mark the option which says no triangle possible and move on Okay, so you don't have to even formulate that quadratic equation Are you getting my point? But again, this comes with a catch that you need to remember these conditions And by the time you reach class 12 That time I will ask you how many of you remember these conditions Okay, so many things we don't you know keep in our mind. We are not able to keep in our mind for long long time Okay, and one of those things is basically these conditions. So we are going to see much many more conditions here So the safe bet here is to form a quadratic Solve for it and see how many roots are coming from there simple as that So this is those situations where you will end up getting one a value as positives Right Right and other other value will actually come out to be negatives. Okay. So in those cases you will have you know this kind of a Conclusion draw. So this is was your first situation when your b is equal to c sin b. Let's talk about second situation I Will rewrite this again in the next page. Don't worry about it So our a was c cos b plus minus under root b square minus c square sin square b cut Okay, so second situation is a situation where My b is greater than c sin b That means b square minus c square sin square b is positive Okay, that means this whole thing is This is actually a real quantity. Okay. So and this is I mean, basically you're trying to say that the discriminant here is discriminant here is Positive, I mean discriminant is greater than zero. Okay So what happens in such case? What happens in such case? Now all of you, please pay attention Everybody, please pay attention In fact, let's take one Before this we take this guy less than zero That will be easier for us to deal Okay So let's complete this because third case will take some time. I mean this case which I was discussing will take some time So I'll take this case first. Okay If this is negative that would that means what that means your a is going to be non-real in nature Right and if a is non-real in nature, there will be no triangle possible Right. So the moment you realize that your b happens to be lesser than c sin b Immediately your conclusion is what's no such triangle is possible with that given b with that given small c and with that given angle b And even sin law will justify it. So even sin law justifies it like see if you're saying b by sin b Is equal to c by sin c. Okay. Now you're trying to see from here from here. You're trying to say that b by sin b is lesser than c, right So that means you're trying to say c by sin c is lesser than c That means you're trying to say sin c is greater than one How is this possible? This is not possible In any in any angle sin c cannot be more than one, right So this is basically coming from our sin law as well, but I'm getting the same from that quadratic expression also So I'm not saying that quadratic is a you know, must think to have to solve for that particular triangle You can make your judgments. You can take your calls based on the other laws that you have learned so far So what is the conclusion here that if b happens to be c sin b Immediately don't waste even in a quarter of a second and mark the fact that there is no triangle possible Because of course the discriminant is negative a will be non real and second thing is that you will not get a You will not get a c which satisfies this condition. Okay. Now. Why does this happen? You know people want to know from the geometrical point of view See geometrical point of view Again, I'll just make the okay. So this is known to you b Okay, this is known to you c and this angle is known to you. Okay. Now this perpendicular length is actually b sin c All right, sorry c sin Sorry, I just interchange. Yeah now Let's say this was not known to us Okay, this position was not known to us. Okay. So what will you do? Let's say From a you are trying to you take a compass Okay, you take a compass you keep the you know needle of the compass over here and you fixed b length on that compass Correct And you're trying to see where is that line? Where is this line? I mean the line which you have drawn here Okay, so let's say this line Okay, so if you want to locate your Uh, if you want to find your small a you need to locate your Small capital c over here So for that you are keeping your compass over here and trying to make a cut on this Right, so you're basically keeping your compass at a fixing a length b on it Okay, and you're trying to see where it cuts But if this b happens to be lesser than c sin b, what will happen? You will not be even able to touch it This line that is let's say I call it as bx this line bx will not be even touched by that particular length So if you take a length which is even smaller than the perpendicular drawn from a on to bx You will not be able to touch this also forget about cutting it Right, it will not even touch it If it is not even touching it, you will not be able to locate your capital c If you're not able to locate your capital c, you will not get your small a over This is what has been mathematically explained by that particular quadratic expression So if somebody asked you that if this is happening, then what exactly is geometrically happening that you are not getting any triangle So you will say that geometrically if I take If I put my compass at a take a length b on that compass, it will not even it will not even touch this bx extended or b line extended This line extended Okay, so no triangle is possible Is it fine? So see I'm showing you from various angles From quadratic perspective. I'm showing from sign law perspective. I'm showing from geometrical perspective. Also. I'm showing why is triangle is not possible for the point Okay, so we'll go on to the last case. The last case is which By mistake I started with but I immediately erase it off. So the third case is where you're I mean again, I will write my a expression I think it was C I think it was c cos b plus minus under root b square minus c square sine square b, right? Okay. Now your third case is where your b is greater than c sine b That means b square minus c square sine square b is positive. That means of course discriminant is positive. Nothing to worry. Okay By the way, I'll just call this quantity as a p Okay, and this quantity as a q for the time being So here you have a certain that you're This quantity which is your q Okay, of course, it'll be a positive quantity for sure. Okay, so this guy will be a positive quantity So now all of you, please tell me If q is positive Okay, I don't know about the nature of p right now How many values of a will be possible? How many positive values of a will be possible? Now here In order to answer this We'll have to account for two more conditions When your b is acute Okay, and when your b is obtuse. So let me write it over here when your b is obtuse Okay, all of you please pay attention. So to further, you know Analysis we need to perform Yes, I'll be very similar to what you are saying. We'll discuss it about So if b is acute it means Your q p value Which happens to be c cos b This will also be positive Okay, now both are positives p is also positive q is also positive. Okay So you'll say sir if both are positive Then definitely with a plus sign I will have a triangle or I will have a value Correct So this will also be positive So this is one triangle possible Correct one triangle we got from it But what if I take a negative sign in between C cos b with a negative sign What will happen to a in this case? If I take a negative sign, what will happen to a in this case? Now, please note This is also positive. This is also positive, but you're subtracting it Does it mean that a always will be negative or a will be positive or it could be either of the two It could be either of the two right shantarik now see here very very important if your if your magnitude of c cos b Exceeds The magnitude of this fellow Then what will happen? Then in that case a will come out to be positive and another triangle will be possible over here Are you getting my point? See here are two positive quantities. This is your p and this is your q You are subtracting it correct So when you're subtracting two positive quantities, you can get a positive answer also right But for that to happen this p should have a higher magnitude no as compared to q Isn't it so for that to happen this guy should be Having a magnitude more than the other guy Correct, which means let me further simplify this condition. That means C square cos square b should be greater than b square minus c square sin square b Let's take this on the other side That means c square cos square b plus sin square b should be greater than b square Which means c square is greater than b square Correct, which means c should be more than b Now here is the final conclusion So what is the conclusion that I'm drawing from here that if b is acute If b is acute and c is more than b Then you will have two triangles possible Isn't it? But if your c happens to be lesser than b Only one triangle will be possible Now why two triangle because you take a plus or a minus between Let's say here there was a plus minus no so both plus and minus is going to give you a positive value of it So two triangles can exist Right, so in such a case the quadratic that you will solve Will give you two positive a values Plain and simple is just that I'm doing a further deeper analysis on it Right, so if a is acute, sorry b is acute and c is greater than b two triangles will be possible But if c is lesser than b then remember only with a plus sign you can get that means only this triangle will be possible With a minus sign you will not get a triangle Are you getting my point? Let me not put right or wrong next to it Okay Is this clear or not Now geometrically speaking why it is happening? Why am I getting two triangles here? Why I'm getting only one triangle here can geometrically me understand it Okay, now see the idea is the same the idea is the same This was let's say This was your side length c okay c was given to you this angle was given to you now what you did on a compass On a compass you took You took b length, okay And now see here if your b is more than c If your b is more than c then what will happen when you run that arc When you run that arc that arc will cut something like this it will cut on this side also Yes or no See from here you put a compass. Let's say look at my hand My little finger you are putting on that a and my thumb is where you are going to cut your b extended Let's say I call it as b extended okay line So if your small b is more than c It is going to cut here. Let's say this position. Let's say c1 and this position. Let's say c2 Correct. Yes or no Yes or no Now if you are cutting it at c2 Can I say this can be my triangle? You will say no sir because in that case b becomes obtuse This will become your b. No, not this one, right? That is why when b is more than c Only one triangle is possible and that triangle will be the one which you will obtain by connecting a with c1 This will be your required triangle Not the other one. This one will not be your triangle Are you getting my point or not? Understood, but if your b is lesser than c Please note my dear students if your b is lesser than c. This is your situation number this guy Okay, now for the situation where your b is lesser than c Let me again make the diagram This was your a this was your b. Okay and now You're taking an arc and that arc is going to only cut on one side of this that means it is going to cut like this So again put a compass at a Okay, and it's going to cut like this. It is not going to exceed here See, please understand b is less than c, right? So it cannot come on this side of c So if it comes on this side then this length will become more than b No, but my b is lesser than c. So it cannot come on the left side of b Are you getting my point? So it is only going to make an arc on the very same side of b. Don't know c1 c2 will be on the right side of b Are you getting my point? So with this you can frame two triangles one with this and another is this that we make a different color here So one is like what you see on your screen and one other is with the blue line So this triangle is possible abc or this triangle is possible abc Are you getting my point here? So this could be your small b or this could be your small b Okay, so this could be your a this could be your a That depends upon the Okay Is this fine? So this is your second situation when your b is lesser than lesser than c Is it fine Any questions? Let me do one thing. Let me put Figure number to this. Let's say this is figure number two. This is figure number one So that when you're referring your note, you can see those figures So this is basically c figure c figure number two. Okay, and this is your c figure number one Right, so when you're reading your notes, you can understand why one triangle was possible and here why do triangle is possible Right now. This is not the end of the story. We have further more analysis to make Okay Now let us talk about b is obtuse. This is something which we have not covered yet Okay, so let us talk about that also. Let me write it Yeah Let me make a fresh diagram on the next page. Maybe that would be more fruitful So this is part a No, not this is part. Sorry. This is part a part a is where your b is acute Okay, part b where b is obtuse. Let's see what happened in that case So for that, I'll go to the next slide So if you have any questions in this slide, please do let me know Any questions any concerns So any dates that you have received for kvpy? I mean, I did not check the website lately, but have they announced any dates The new dates No, not yet Okay, so now here we'll take a situation where b is obtuse Okay, by the way, I'll just rewrite the a again for you all to refer to so a was c cos b Plus minus under root b square minus c square sin square b. Okay, so here we have already assumed that b is more than c sin b Okay So here please understand If b is obtuse then this guy which is p and this guy which is q Okay, so p is by the way, this is what I named it in the previous slide also or did I named it the other way around I want to be consistent in naming actually What did I name it? Ha PNQ PNQ. Yeah I keep forgetting what I gave us the name of the two expressions. Yeah, so this is p. This is q So q is definitely positive. So because of this q is definitely positive Right, it exists and is positive. Okay, so q is positive Now what will happen your p which is your c cos b will become negative now Because of the fact that b is obtuse Okay So now when you're talking about A triangle you cannot get a triangle if there is a negative sign taken over here So if you take the negative sign Okay, like this They cannot be any triangle form because this is anyways negative And you're subtracting a positive quantity from there. So overall this will be negative So no triangle will be possible in such scenario So no triangle possible Okay, so if c is greater than Sorry, if b is greater than c sign b and you realize that b is obtuse No triangle possible straight away you drop the matter there right but But with a positive sign What can happen with a positive sign what can happen? Right, let us check If I take a positive sign that is c cos b Plus under root b square minus c square sin square b now understand here This guy is in inherently negative quantity Okay, and this guy is positive quantity and you're adding it So you can get positive answer also from there. You can get negative answer also from there depending upon whose magnitude is more right, so if If the magnitude of c cos b is lesser than the magnitude of b square minus c square cos Sine square b under root Then what will happen the positive quantity will dominate the negative quantity means overall it will make things positive Isn't it because this is more powerful than this guy. So even though this is negative Overall on adding it will become a positive quantity Right, by the way, this is something that we have already solved I mean in our previous slide also, but still I will do it again That means if you're I'm just rewriting it completely That means if your c square is less than b square that means if your c is less than b Then what will happen in this case one triangle will be possible Okay But on the other hand if your c is more than b then again no triangle is possible No triangle is possible So let us try to see this condition Let us try to see this condition see here I mean, I'm just summarizing it So the conclusion here that we are drawing is that if b is obtuse If b is obtuse and your c happens to be lesser than b then only one triangle is possible Okay, and that triangle is possible when you're taking a positive sign, right? but if your c is More than b then no triangle will exist Are you getting my point here? Okay, so now this one triangle that is possible is the one which we had ignored in our previous diagram Okay, so let's again make the same scenario So let's say this is my Let me make it in white color Yeah, so let's say this is a Okay, this is b And let's say I extend this. Okay. I don't know where is my c because you know that information is to be figured out from here So this angle is known to you. This is known to you And you have you have basically given that c is less than b or b is more than c So b is more than c Please realize that if you keep your compass at a And try to locate your c it will be cutting this line at two points One is at this position c1 other is at this position c2 Right. So which of the two will have which which of the two will give you the case where b is obtuse And b small b is more than c this guy is going to meet that criteria. So this is the only triangle possible Are you getting my point? This is not going to work out Okay, so this will be your b. Isn't it? Yes or no? Is it fine? Are you getting my point? So this is the triangle and this triangle is not going to be possible clear Right, that's another valid point. That's another valid point Okay So if if c is greater than b, please note that in those cases what will happen Your entire triangle will be formed with the side on the right side That means it will be something like this I'm just making a quick diagram It's just give me a second So what is going to happen in this case? When you're taking a compass and you're trying to cut And if your if your b is smaller than c it is going to cut This side only right c1 c2 And none of them is going to work out. You can't have a triangle like this Why because if you have a triangle like this your b is not obtuse it is acute in this case This is not obtuse in that case Right, so this triangle is not possible If you try to make a triangle with this c1 this also is not possible. Is it fine? Any questions? So that is why no triangle is possible for such a scenario Right And Shardul is right If b is obtuse it is supposed to be the greatest angle. That means Sorry, if b is obtuse it is supposed to be the greatest angle. That means b is greatest than any other side So how can b be smaller than c? So this is not possible That means the entire scenario is faulty. No triangle is it going to exist like this? Okay So these are the analysis that you need to keep in mind while you are solving questions So now let's see what type of questions are going to be asked Okay, let's take that Where is my Question bank Okay, let's start with this question Find the number of triangles that can be constructed with the given information that your small b is 3 small c is 4 and angle b is 5 by 3 Please solve it and let me know your response on the chat box Yes, done See, which is this case? You have been provided with This length This length And this angle Okay, so two sides and a non-included angle is there. So this is actually an ambiguous case, right? So we have we are basically looking into an ambiguous case, right? So again, don't panic Don't panic if you've forgotten those conditions. Those conditions have all come from the fact that you are supposed to get a quadratic equation for me Let's say this is your a right, so you end up getting Cost 60 degree as a square plus four square minus three square by two into eight to four In other words, half is equal to a square. This is going to be seven by eight a Okay, so let's solve it. So a square plus seven is equal to four a Correct. So a square minus four a plus seven is equal to zero. Okay. Now If you look at this particular case, you would realize that you're discriminant. That is your b square minus four ac is negative Correct, that means your a is going to be non real Right So whether you take plus minus doesn't matter a itself is going to be a non real So how many triangles are possible? No triangle is possible. Zero triangle is possible. No triangle is possible with such given scenario Right and you would realize that Okay, you would realize that even if you use those conditions Right that condition is going to be basically fulfilled Is it fine any questions here? Anything that I missed out Please feel free to highlight Good enough Okay, let's take another one Yes, any success Now here also what has been given to you is These two and a non-included angle. So this is again a case of ambiguous case So two sides in a non-included angle. Okay again start from basics. Don't need to see your small c is your unknown, right? So first write cost of a Okay, so what is cost of a? Aja Okay, let's let's solve for it. So cost of 60 degree cost of 60 degree will be what sorry cost of 30 degree will what c square plus 8 square minus 7 square by 2 into c into 8 Yes or no Yes, let's solve for it. So this is going to be root 3 by 2 and I think this is going to be uh 64 minus 49 Which is 15 And this will be 2 into 8c. So 2 2 goes off. So you have c square minus 8 root 3c plus 15 equal to 0 right Now you can solve for If you solve for your c values, let's check what happens Now see the presence of a real root is not important the presence of real and positive root is important You can't have real and negative values of a side, right? A side cannot be negative in a triangle, right? So when you solve by using your sheatharacharya formula, you get B minus B plus minus B square B square only 192 minus 4ac will be 60 divided by 2a So that's going to give you 8 root 3 plus minus by the way, this is 112 by 2 Not 112 132 132 by 2 Now please understand here that even if you take a plus you'll get a You'll get a positive value of c even if you take a negative also you'll get a positive value of c Right that means both plus and minus are acceptable So this is also possible. This is also possible So two values of c are possible and hence two triangles are possible So off so here the answer is two triangles Is it fine any questions? So what I wanted to convey here is that it is not necessary that you remember all those conditions that we discussed Okay, you can just make use of your quadratic itself to get your required result Let's take this one So this is a case where a is given to you B is given to you And the included angle is given to you find the other two angles All right any success? Or sure take your time 30 seconds is okay. Okay, Prisham Okay, so many of you would have actually a very good show the many of you would have actually used your cosine law How many of you use cosine lawyer? Just say me on the chat box Okay, some of you some of you used for a change. I will not solve this by cosine law I will solve it by using my tangent law. Okay Why because ab is given to me and a included angle is given to me. So I can use this formula tan a minus b by 2 Is a minus b by a plus b cot c by 2 Okay, so let's write This as it is and a minus b will give you correct me if i'm wrong root 3 minus 1 and root 3 plus 3 Which is actually root 3 plus 1 correct Cot c by 2 cot c by 2 is cot 30 degree cot 30 degree is root 3 This will be one my my bad. Yeah So if you cancel this off, you get something like root 3 minus 1 by root 3 plus 1 now all of you please pay attention If you express it like this This actually becomes tan 45 minus tan 30 by 1 plus tan 45 tan 30 Which is clearly a compound angle identity for tan of 45 minus 30 Okay, and this is clearly 15 degree correct So basically you're trying to say Basically, you're trying to say a minus b by 2 is 15 degree Which means a minus b is 30 degrees And a plus b is already 120 degree. The reason being c is known to you as 60 So a plus b is 120 So if you add them to a is 150 that means a will be 75 And b will be 45 There you go. So a and b are found out straight away Okay Now what I'll do in order to find my sign c. I will use my sorry in order to find my small c not sign c small c I will use my sign law Correct. So can I say b by sign b will be small c by sign c so b b is given to you as 2 Okay, sign b is 1 by root 2 Small c is not known to you and sign c sign c will be sign of 60 degrees sign of 60 degrees root 3 by 2 Okay, let's try to solve it So this is going to be uh, let's let's take it on the other side So this is going to be 2 into root 3 by 2 into root 2 correct Correct me if I'm wrong 2 2 goes off So your answer is going to be root 6. So small c is going to be root 6 So I think I think uh prisham your c value was not right Sharduli Yes, I think You didn't find your small a capital a by the way But yes, your b is correct from that equation and c is also correct Is it fine any questions any concerns anybody has? So we are going to stop this topic over here itself and we are now going to start with a new topic