 Hello, good afternoon. Welcome to the YouTube live session. Those who have joined in the session, I would request you to type in your names in the chat box so that I know who all are attending the session. You all can see the screen which I have shared and I'm clearly audible to all of you. Hi, man. Good afternoon. Hello, Vidyotha. I can see one more person in the chat and live chat. All right, so today let me, yeah. So let me discuss the agenda today. Today we'll be doing two topics. One would be, we'll be doing problem practice, problem practice on trigonometry because that's going to be coming for your tomorrow's UT till the compound angle identities, till the compound angle identities. And the second agenda for today is we'll be quickly doing sets with you till the practical problems, till the practical problems, or we can say the problems based on cardinal number of sets by the use of Venn diagrams. Is there anything else that you want me to do because I think these are the two concepts that are going to be tested tomorrow for you. Is that fine? So we'll do one thing. We'll keep the first one and a half hours for problem practice on trigonometry and again the later one and a half hours for sets. I'll tell you the important parts where the questions can be framed. Okay, so let us begin with the problem practice. So I'll touch upon all the concepts and sub concepts which we have already covered in trigonometry. So let me begin with this question. So there is a sector of a circle. So let me just draw it out. So here's the first question for you. There's a sector of a circle. So this is a sector of radius R and this arc is AB arc. The angle suspended by this AB arc at the center is beta radians. Okay, it's given that the perimeter of this sector, the perimeter of this sector is 18 meters. Area of the sector, area of this sector is 8 meters square. Okay. Find number one, the radius of the circle and number two, the angle beta. Please give your angle beta in terms of radians. Please type in your response in the chat box once you're done. So we have been given two pieces of information. One being the perimeter of this. Okay. And the other being the area of the sector. I have to use these two information to get the value of the R, that's the radius of the circle and the angle suspended at the center, which is beta radians. Yeah, anyone? Guys, let's not try to take more than three to four minutes to solve each question because roughly that's the time limit that will be provided to you in your UTs as well. Not sure when that is trying. First thing is what is the perimeter? Perimeter is basically this plus this plus this, correct? So this entire thing will constitute a perimeter, right? We know from the formula of the length of the arc and the radians that length of the arc AB, okay, would be nothing but it will be R times beta. Basically, OA and OB both are equal to R. Which implies the perimeter is going to be R plus R plus R beta and that's going to be 18, right? So the first equation that you get is going to be this equation. So let me call this as equation number one. Equation number two is that will be obtained from the area of the sector information. So half R squared beta is the area of the sector and that is given to you as 8. That means R squared beta is given to you as 16, right? Now, from here I can say R times 2 plus beta is equal to 18 and from here we have R squared beta is equal to 16, correct? Now what I can do is I can square both the sides. For example, of this I can square both the sides and write it as R squared 2 plus beta whole squared is equal to 18 squared, okay? Let me call this as 1, let me call this as 2, okay? Now let us divide 1 by 2. Let us divide 1 by 2. When I do that, R squared 2 plus beta whole squared by R squared is equal to 18 into 18 by 16, okay? And we have a beta over here as well. So R squared, R squared gets cancelled off, okay? This you can get cancelled by a factor of 2 as 8, again 9. This is by a factor of 4, okay? So we get beta plus 2 whole squared is equal to 9 beta and you can just multiply, cross multiply with a 4, okay? Let us try to simplify this. So we get 4 times beta squared plus 4 beta plus 4 is equal to 9 beta that is 4 beta squared plus 7 beta plus 16 is equal to 0. Now can we solve for, can we try to solve for beta from here? So 64, 64 is factorizable. Can we factorize this? Try to solve this from here. How we get the second equation, okay? Let's do this once again. See, we know that in any sector which subtends an angle theta at the center, the area of the sector is half R squared theta. Are you aware of this, Mehak? Yeah, this part is clear. Half R squared theta, how do you get this? Very simple. See, we know that 2 pi radians is equivalent to pi R squared. So how much is theta radians equivalent to? Just cross multiply. So theta into pi R squared divided by 2 pi. So it's just a unitary method which I'm using here. So half theta R squared, or you can say capital R squared will give you the area of the sector, okay? And according to my question, that is given to you as 8, okay? So what I did, I wrote these two equations. I squared this up. I got this and I divided by this equation. So that give me beta plus 2 whole square by beta is equal to, by the way, you'll have 9 into 9 over here. That's a small mistake here. That's going to be 9 square over here. Yeah, this is going to be 81. Yeah, so this will change slightly. So this will become 4 times beta square plus 4 beta plus 4 is equal to 81 beta. So 4 beta square plus 16 beta plus 16 minus 81 beta is equal to 0. So if you bring it on the other side, it will become minus 64 beta plus 16 equal to 0. Drop a factor of 4 from everywhere. So it's beta square minus 8 beta plus 4 equal to 0, okay? Sorry, 16, 16 beta. I'll just change this out. Yeah, 4 beta square plus 16 beta plus 16 is equal to 81 beta. So that's going to be 4 beta square minus 64 beta plus 16 equal to 0. So just try to solve for beta over here. Once you solve for beta, you can put it in any one of the equation and get your value of R. 65 in radiance. Give me the answer of beta in radiance. You can use your quadratic equation formula minus B plus minus B square minus 4 AC by 2A, okay? So beta will come out to be 16 plus minus under root. 256 minus 16 is 240 by 2, okay? So that's going to be 16. It cannot be a minus sign, okay? Let's take a plus sign under root 40 by 2. 240 I can write it as