Grade 12 Math Live Class | Euclidean Geometry | Members Only

Añadido: 2026-05-04

[00:00:08]You'll tell me when you are ready.

[00:00:18]Good evening. Z.

[00:00:23]>> Good evening, sir.

[00:00:27]>> Are you well?

[00:00:28]Yes, I'm good.

[00:00:43]Okay, just give me a second and Okay. All right.

[00:01:05]Okay, ready when you are.

[00:01:12]>> See, we're going to practice again.

[00:01:18]>> Okay, so 10 nine.

[00:01:38]Okay, let me behave myself before I get overly excited. Okay.

[00:01:46]All right, let's get right into it. Good evening. Good evening. Good evening, everybody.

[00:01:54]most particularly to you that are joining us and who is doing the background work over there. And good evening to those of you that are joining us on our YouTube channel. We did say that we wanted to find a way to continue going live on our channel particularly for those that are members. Please, if you want to know how to become a member, you just click on that join button and you will find the top tier there which is the pro membership and there you can be a member too.

[00:02:34]>> You can be a member. You can be a member. Everybody can be a member.

[00:02:38]>> All right. So today we're going to start with ukidian geometry.

[00:02:44]>> Now I had already started in my in my classes. Um, we started this on Saturday and what I want to do today is just a quick recap on our grade 11 uklidian geometry and the reason for that is that please remember >> that we are still going to encounter it quite a lot and you know you can't have proportionality theorem which is the grade 12 part without necessarily mastering every other thing that you've done both in grade 11, grade 10 and all the other grades before that. So, so here's what I want us to do. Just quick reminder of what we have.

[00:03:28]>> So, when we look at geometry in its totality, so firstly, we looked at triangles, right?

[00:03:37]The kind of triangles that you've got, what are triangles? Scaling triangle, all sides are different.

[00:03:44]>> Okay? So we know that once we've got the only thing that we know about triangle is that the sum of all angles is equal to 180.

[00:03:57]>> And then we said we've got an isosles triangle. Please keep this in mind because once we've got radi we usually come back to this guy. So when we've got an isosles triangle it means that two sides are equal. And so what that simply does is that it makes even the base angles of an isoclesles triangle equal.

[00:04:20]So we said if we given P R is equal to PQ then we know that we can conclude that then angle Q is equal to angle R. And what we say as a reason there is that these are angles opposite equal sides. Okay. Right. And then we've got a a uh right angled triangle. What we know about a right angle triangle is that the square of the hypotenuse is equal to the sum of the squares of the other two sides. That's Pythagoras. Okay, that's the Pythagorean theorem. And uh we pythorus that's how we write it.

[00:05:05]Okay. So what we do know with a an a right angle triangle is that once you know one angle then the other angle becomes 90 minus theta. Of course we know already that the one angle would be 90° and we also have I forgot to mention this other triangle what we call a uh an equilateral triangle. Okay.

[00:05:31]Right. So Zanda, I didn't mention this with you on Saturday. We've got an equilateral triangle. And please remember, once you've got an equilateral triangle, it means all three sides are equal, but it also means all three angles are equal as well. And not only do we know uh the h the the the angles that they are equal, but we also know the size of the those angles will be equal to 60°. Right? Now what we said is that remember in a triangle we say that the exterior angle of a triangle meaning that when I'm looking at the angle that is on the outside it is equal to the sum of the interior opposite angles. So it means angle P plus angle Q will be equal to angle P RS which is this guy over here right and if this is 60 that's 60 so which means this guy would be 120 please don't forget all of these because they're going to feature quite prominently when we are doing ukidian geometry okay right and then we know that the sum of angles on a on a straight line is equal to 180. Now, very quickly, guys, I'm going to just recap on this. Uh, we said when we've got parallel lines, the moment that we know that we've got lines that are parallel, we've got corresponding angles that are equal. So, if you look at this angle and that angle, they are equal. If uh we also have alternating angles that are equal, okay, there we've got alternating angles um that are equal, right? Of course given the fact that they are we've got parallel lines. Now if we've got uh parallel lines again we also have co-interior angles. Now please note with co-terior angles we do not say that they are equal but we rather say that they are supplementary meaning that the sum of the co-terior angles will equal to 180°. Now guys here's what I want us to do very quickly. go through the grade 11 circle geometry.

[00:07:48]Right? Now, we said it is subdivided into three parts.

[00:07:54]The first three theorems. Now, remember you've got nine theorems. The first three have to do with centers.

[00:08:01]The second three have to do with cyclic quads. And then the last three have to do with um uh uh Ga what do they call this? um uh tangents right so the last three have to do with tangents so the first one let's look at the theorem number one so theorem number one says that if a line is drawn right if a line is drawn from the center and it is perpendicular to a uh to a chord then it automatically bicts that chord it cuts that chord into two equal parts So we said if we draw a line from the center then we know that O right. So if we draw a a line from the center that is O no no no this was supposed to be O R is perpendicular to PQ.

[00:08:59]So then we know that P R this side here will be equal to QR or you can say RQ right but the converse of that theorem is that if we draw a line from the center and it bets the chord this time around we're saying all right if we knew that P R is equal to RQ therefore our conclusion is that O R is therefore perpendicular to PQ. Now guys, I want us to really really pay attention to this because we are going to encounter it quite a bit when we now do proportionality theorem.

[00:09:44]Not only that, remember once you go to analytical geometry h there's a whole lot of this and the thing is in analytical geometry they don't even say that uh you know you have to apply uh principles of ukidian geometry but they just give you questions and they tell you specific things. So please remember that. Right. Now the second theorem it says h the angle that is subended by a chord that is at the center is twice the angle at the circumference. So we say simply put angle at center is twice the angle at circumference. So if I knew that the angle at the circumference is theta, then I automatically the angle of the center becomes 2 * theta. Okay? So whatever that angle becomes, the angle at the circumference is twice that. And please remember it's not the other way around. So if they tell me that's a 30° angle, then automatically the angle at the center will now be equal to 60°.

[00:10:56]Right, let's go to the next one. The next one says, well, in this case, we say angle at semicircle.

[00:11:04]Right, so if we've got an angle that is subended by a diameter, all right, an angle at the circumference that is subended by a diameter will automatically be 90° equal to 90°. So in this case we say this is the angle at semicircle right and then let's go to the next set.

[00:11:29]So we said the next set of of uh uh theorems have to do with cycling quads right am I going too fast.

[00:11:38]Are you guys still okay?

[00:11:42]>> Yes sir.

[00:11:43]>> Okay. You still you still okay?

[00:11:47]>> Yeah.

[00:11:48]All right.

[00:11:50]So, um, okay. Someone has asked us to go back to the exterior angle of a triangle. So, what I was saying about the exterior angle of a triangle there, if I just go back. So, please remember I said in this case, if we've got an exterior angle, right? So if I look at this angle over here, this is equal to so if let's say angle P was equal to X and angle Q was equal to Y, right? Then it means that the exterior angle here would be equal to X + Y. All right? So we say the exterior angle of the triangle is equal to the sum of the opposite interior angles. Okay? Right? that may come in handy at some point. Right? Now let's go to the next set of theorems. We said we are now looking at cyclic quads. So firstly remember that a cyclic quad is a four-sided figure whose vertices meaning that whose corners right touch the circumference of a circle. So even if sometimes they don't tell us that a figure is a cyclic quad, if all its vertices touch the circumference of the circle, then we know it has to be a cyclic quad. Okay?

[00:13:21]Right? Now please note the first theorem of those or theorem number four said says the opposite angles of a cyclic quad are supplementary meaning that they equal to 180. So meaning that when we add angle A plus angle C they are not equal but we know that the sum of them will give us 180°.

[00:13:47]But what it also means is that if we take angle D plus angle B. So angle D plus angle B is equal to 180. And how we say that is that they are opposite angles of a cyclic quad.

[00:14:07]Okay. Right. So the next one we've got a cyclic quad and we know that the exterior angle of a cyclic quad. So once they extend the line right usually they would say to you line CD is produced till E. So they are trying to tell you that's a straight line. Okay.

[00:14:29]So once they extend a line and so we know the exterior angle h of this cyclic quad will be equal to the opposite interior angle. And so when we put the reason we just simply say this is the exterior angle of a cyclic quad. And then the next one says we call it the butterfly theorem or you can call it the bow tie theorem. It's really up to you.

[00:14:59]Right? So this says the angles that are subended by the same chord right would be equal. All right. the angles at the circumference that are subended by the same chord are equal. So in this case when we look at angle A it is equal to angle B. Right? And the reason is that these are angles on the same segment.

[00:15:25]Okay? Right? Uh but also what it means is that angle D over here would be equal to angle C.

[00:15:36]Apologies there.

[00:15:39]Right? So that would be equal to angle C and the reason is exactly the same.

[00:15:45]These are angles on the same segment.

[00:15:48]Now please I want you to note the moment they tell us to prove that a a figure is a cyclic quad we will have to use up all of those uh theorems. I I mean we not all of them but we will have to use any of the three to prove that we've got a cyclic quad. Okay, I've got one such a question today and I really want to go as quickly as possible so that we can get to those questions. Right. But what we also do note is that if in the same circle we have got uh chords that are equal in length right then it means that the angles that they subend will be equal. So what I said is that if we say BC line BC is equal to line EC note this is not a cyclic quad. We've got two different triangles there. So what it then means is that angle A would be equal to angle F. But what do we say? We say these are angles that are subended by equal chords. Okay. Right now let's go through the last set.

[00:17:07]The last set has to do with tangents. So we say when we're looking at tangents, right? So the first one, remember what is a tangent? It's a line right that touches the circle only once. So if we draw a line from the center and in this case it um or rather it intersects with the tangent right it will be perpendicular to the tangent at the point of tangency. So meaning if we've got a line from the center here's our radius here. Okay. And in this case it will be perpendicular to the tangent at the point of tangency. So please do note that this makes a 90° tri I mean angle. So we say 10 is perpendicular to the radius. And then in this case if we've got tangents from the same point so if it is given to us that we've got the same point a common point B and we've got tangents from the same circle that meet at a common point that is B those tangents will be equal but please note in this case what it means is that if we were now to make this into a triangle then it means that angle A will automatically be equal to angle C.

[00:18:34]And why is that? These are angles opposite equal sides. Okay. Right.

[00:18:43]Sorry. Uh remember what we can also say in that regard is we can also state that uh these are angles um or or rather we can say base angles of an isocles triangle. We can also use that as a reason. Okay. Right. And then finally the 10 chord theorem. Right? So if we had A, we've got B, we've got C. Right?

[00:19:12]And let's say uh this is D over there. Okay?

[00:19:20]Let's call this E. So we know that the angle E A B right. So this is this angle here E A B will be equal to angle A C B right so we call this the 10 chord theorem.

[00:19:50]So that's the 10 chord theorem. Okay.

[00:19:53]Right. So automatically uh again it means that if we call this angle X right uh that will automatically become angle X. All right. So we call this the 10 chord theorem. Now please guys I'd like for us to take some time now. Um Zand you are you ready to to do some to practice today?

[00:20:22]60% >> 60%.

[00:20:26]Are you ready to practice?

[00:20:30]>> Yes, sir.

[00:20:33]>> Right. So, I'm going to take u I'm going to take a past exam question. All right.

[00:20:41]And please don't be scared. Don't be shy.

[00:20:45]Right. And we're going to try and dissect this guy as much as we possibly can to find all the things that are related to this uh uh to this diagram.

[00:20:59]So they say in the diagram below we've got O which is the center. Now please remember the moment that they give me O as the center I am going to look for firstly my radi.

[00:21:15]Right? So please look there's my radius there's another radius over there okay and we know what do we know about radi they're equal right so which means that o must be equal to o what I also would look for is is there any angle at the circumference that is subended by the uh um so that is subended by the diameter.

[00:21:48]So there is no diameter here because if I look at BF, it doesn't reach the other side of the circle. Right? So that is not a diameter. Right? Is there diameter? Absolutely not. So in this case, we're not going to um yeah, we're not going to have an angle at semicircle, right? And then they say to me uh with always the center of the tri of the circle a e b c d. So meaning all of those points touch the circle the circumference there they are right and so my question is is there a cyclic quad over here?

[00:22:40]Is there a cyclic quad?

[00:22:46]Is there a foursided figure that touches all the >> vertex touch the circumference?

[00:22:54]>> No.

[00:22:55]>> No, there isn't. So as a result, there is no cyclic quad. So in that case, we're not going to apply the principles of a cyclic quad. So they say to us, F.

[00:23:08]Yes.

[00:23:09]Is A B C D not a cyclic?

[00:23:12]>> A B O Yes, you are absolutely correct.

[00:23:17]You are absolutely correct. A B C.

[00:23:31]You've got this. You've got this. Right.

[00:23:34]You're absolutely correct. We do have a cyclic quad which is ABC D. Thank you for that.

[00:23:41]Okay, so we do have a cyclic quad over here. Right. So they say to us F lies on AD, right? And they say P O is 100°.

[00:23:56]So there's the angle that they're giving to us there. That's 100° and uh D1 is equal to 20° right they say the size of some of the angles are given in the table below in each case supply a valid reason I like this question because they've done the work for us they just want us to uh to give them a reason why that is the case Right.

[00:24:30]So um let's start with you. So they say to us A2 is 50°.

[00:24:40]Why would A2 be 50° >> angle at center?

[00:24:50]>> There we go. There we go. Angle at center. So which one is the angle at center? It's the angle B O D or you can say angle O2. Right? So angle O2 is the angle at center. And look at this. We can draw a chord there.

[00:25:12]And you can see the the angle at circumference is also subended by the same chord as well. Do you see? So which means that will be the angle 50° and so that means angle A2 is 50°. Why? Because angle at center equals to 2 * angle at circumference.

[00:25:47]Okay, great stuff.

[00:25:49]All right. Um >> sir 2011 is 80°.

[00:26:02]>> Do you know why?

[00:26:04]>> Yes sir.

[00:26:06]>> Mhm.

[00:26:07]>> Angles on a straight line.

[00:26:10]>> Right. Sum of angles on a straight line.

[00:26:15]Right. So that's angles on a straight line.

[00:26:25]Okay, that's equal to 180. So that would be 180 minus 100. And so that gives us this angle to be 80°. All right. So so far so good. It's going okay.

[00:26:40]Okay. Right. Um, all right. Uh, I see there's a question there. I'll I'll come back to that question about the cyclic quad. So, they said f_sub_1 is equal to 80°.

[00:27:01]There's f_sub_1 over there.

[00:27:07]Okay.

[00:27:09]Uh Z.

[00:27:13]>> Um so I have to say the interior angles of a triangle.

[00:27:18]>> Okay. Which triangle?

[00:27:20]>> Um F O D.

[00:27:22]>> F O D. I totally agree. So in this case, we would have taken the interior angles of a triangle, right?

[00:27:36]uh let's say triangle F O D. Okay. So if we added the 20 plus the 80 that gives us a 100. But I want you guys to also note that um you do you notice that we could have also used the interior angles of a the exterior angle of a triangle, right? We said the exterior angle of a triangle is equal to the sum of the opposite interior. So which means O2 would be equal to F1 plus uh D1. And so if this is 20 that's 100. So that makes this equal to 80. So another reason is that we would have we could have used the exterior angles of a triangle.

[00:28:32]All right.

[00:28:34]Okay. Um right we are back.

[00:28:44]So A1= to 30° A1.

[00:28:55]Mhm.

[00:29:12]All right. Anyone uh also on YouTube who wants to weigh in, why would A1 be equal to 30°?

[00:29:32]Uh if you see it uh Xand as well please you can you can weigh in All right, I want to uh I want to just put back all the angles that we've just found.

[00:30:05]So, we said this guy is 80, right?

[00:30:15]We said this guy is 80.

[00:30:19]That's 20°.

[00:30:22]We said that was 50.

[00:30:27]Uh, is there any other angle that we found?

[00:30:35]Okay. All right.

[00:30:38]Uh, Nabulo, have you found it? Not yet.

[00:30:44]>> Not yet, sir.

[00:30:45]>> Okay. All right. Um, let's see. Okay, have you have you gotten it?

[00:30:55]>> It's okay. Please don't forget what did they tell us here.

[00:31:05]Got our parallel lines, isn't it?

[00:31:09]Yes, sir.

[00:31:14]>> So, So there's our second parallel line. So tell me what would be the relationship guy F1 and angle A? What type of relationship would they have?

[00:31:37]>> So corresponding >> corresponding angles. What do we know about corresponding angles?

[00:31:45]they are equal >> corresponding angles are equal. So we know that f_sub_1 is equal to a angle a right but we had already found out that angle a2 is 50°. So if the entire angle is 30° then it means that A1 I mean if the entire angle is 80° rather then it means that A1 would be equal to 30°.

[00:32:15]Does that make sense?

[00:32:19]>> Yes.

[00:32:20]>> Yes sir.

[00:32:21]>> All right. Beautiful stuff. So uh let me just remove this for the sake of just creating space. So firstly we would say um firstly uh f_sub_1 is equal to angle a1 and 2.

[00:32:50]Okay. And these are corresponding angles.

[00:33:02]Um remember we need to state the the the sides that are parallel. So EA is parallel to BOF.

[00:33:21]So therefore a1 would be equal to 30°. Okay, we already had found the value of um we already found the value of uh of of A2, right? And then finally they say to us B is equal to 30°.

[00:33:47]Okay.

[00:33:52]alternate angles.

[00:33:54]>> There we go. So, we've already discovered that A1 is 30. So, which means this guy would also be 30° and so that will be alternate angles and in this case because EA or we can say BF is parallel to EA. All right. Uh you guys have done well.

[00:34:19]Uh, can I give you a huge round of applause?

[00:34:26]Right. I'm sure all of you uh do agree that my two champions here have done the right thing.

[00:34:36]Right. So, what I want us to do is let's go on to the next one. Right.

[00:34:43]But before I do that, can I just quickly uh just check? There was a question and the question was Okay, I see all the nice, nice, nice. I like it. I like it.

[00:34:58]I like it a lot.

[00:35:00]Okay. Um, someone was asking about the exterior angle of a cyclic quad. Now, let let me just uh quickly show this on here. Right. We had found that we've got a cyclic quad.

[00:35:16]Okay. So there's our cyclic quad.

[00:35:20]Now suppose I extended ah suppose I extended this line here. Okay I can't get it to be straight but okay you get the point. Yeah. So suppose I extended this line. So which means the exterior angle this angle here would be equal to the opposite interior angle of the cyclic quad. So which means um that angle over there E E D let's let's say Z there. So the angle E DZ would be equal to the angle A B C which is that angle over there. Why? Because they would be the exterior angle of a cyclic quad is equal to the opposite interior angle.

[00:36:14]Okay?

[00:36:16]Right? So I want us to do this quick, fast, and in a hurry. Right? So let's go on to the next question.

[00:36:28]So they say to us P, Q and R points on the circumference of a circle with center O. And I said to you ma the moment they tell me about the center I am going to look for my radi. Whether I'm going to use them or not is insignificant. But it's always good for you to note your radi. Okay. So which means O R is equal to O S is equal to OP. Right? Now uh they say to us P R which is where is PR?

[00:37:07]This guy here.

[00:37:09]Um what type of a a a a side is PR or what type of line is it in relation to a circle?

[00:37:23]Um the diameter >> the diameter. So what does that say to us?

[00:37:31]>> Is one of the theorems?

[00:37:34]>> Yeah.

[00:37:39]>> Is the angle that we would kind of know?

[00:37:43]>> Yes.

[00:37:44]>> Know that O is 90°.

[00:37:47]>> Say that again.

[00:37:49]>> Um is it O Q? Sorry, Q.

[00:37:52]>> Yes.

[00:37:52]>> 90°.

[00:37:54]>> Yes, Q is 90°. So, we already know. Uh, why would that be the case, Conj?

[00:38:02]>> Um, angle at semicircle.

[00:38:04]>> Angle at semicircle. So, we've indicated that we've got our diameter, our circumf not circumference, rather our radi. We know they are equal and we know that Q is a um is 90° right? So they tell us P R is the diameter of the circle.

[00:38:27]QSO is equal to X. So they're telling us this angle is equal to X. But they also give us ops which is equal to 3X.

[00:38:42]So before we continue, you know, I like working into the diagram, is there any other angle that we would be able to kind of determine whose value we know? Is there any other angle that we can we we know here?

[00:39:11]Um, can we say no? Q S P is 2X.

[00:39:17]>> Uh, angle >> Q SP, right? So, uh, you said Q. Okay, let me just try to Q S P is equal to >> 2X.

[00:39:40]>> Okay. Right. So in this case I'm going to ask a a a a just a a a question.

[00:39:51]When do we have angle at center twice angle circumference?

[00:40:00]One should be at the center, right? The other one should be at the circumference. But please note, they must be subended by the same code.

[00:40:13]>> Are you with me?

[00:40:17]>> Yes.

[00:40:18]>> But then like OS is a radius. So is OP.

[00:40:24]>> Uh, say that again.

[00:40:27]O S is a radius and OP is also a radius.

[00:40:31]>> Yes.

[00:40:33]O S is a radius.

[00:40:37]>> OP is also a radius.

[00:40:39]>> Yes. Yes.

[00:40:44]>> So one angle O P S and O uh P O S be equal.

[00:40:54]Um, say that again. Angle.

[00:40:59]>> OPS.

[00:41:01]>> OS.

[00:41:06]>> And L.

[00:41:07]>> Oh, yes. Okay. All right. I see what you're saying. So, you're saying this angle here would be equal to this angle here.

[00:41:16]>> Yes.

[00:41:17]>> Absolutely. I completely agree. I completely agree. Now that's great.

[00:41:24]That's great. So meaning tell me why you were telling us that.

[00:41:33]>> So as you know uh ops is equals to OSP.

[00:41:41]>> Therefore uh entire annual is 3x right?

[00:41:47]>> Yes. Mhm.

[00:41:50]>> So we already have that uh QSO is X.

[00:41:55]Therefore PSQ is 2X.

[00:42:01]>> Beautiful. Beautiful.

[00:42:03]Do you see what you went about this differently as opposed to using angle at center twice angle at circumference?

[00:42:11]And in this case I completely agree. So this would be equal to 2x. Why? Because 2x + x would give us 3x. All right. So now we know that little angle over there is 2x.

[00:42:33]All right. Um we can determine as many tri as many angles as possible. Yeah.

[00:42:39]But let's let's go for it one step at a time. So firstly they said find out what is angle SQR and I want us to take it step by step. Hey I don't like it when they give us that SQR PQS. It's better when they give us 01 O2 P1 P2. Okay.

[00:43:00]It's much easier. Okay. Right. So let's go for it. Um so we said the first one that we're looking for is SQR right.

[00:43:18]So we are looking for angle sorry S Q R.

[00:43:31]All right. Um what would be the size of angle S QR?

[00:43:40]Let's start with uh let's start with Xander.

[00:43:45]How would we go about that one?

[00:43:54]>> Um if I'm not mistaken. So we have a cyclic quad, right?

[00:43:59]>> Um which one is that cyclic quad?

[00:44:02]the cuz I was going to use the butterfly theorem cuz PQ I mean P S uh P S Q and R cutting the circumference of the circle.

[00:44:21]>> Okay, I agree.

[00:44:25]But would then need to make a construction line over there.

[00:44:29]>> Okay.

[00:44:31]>> Okay. for us to do that.

[00:44:34]Do you want us to do that?

[00:44:38]>> So, that's what I can think of at the top of my head.

[00:44:42]>> It's It's not illegal.

[00:44:44]>> It's not illegal. Um All right. Okay.

[00:44:48]So, you do you want to do that or do should we leave it without?

[00:44:55]>> Let's leave it. So, >> okay. So, in its current form, it is not a cyclic quad. Okay.

[00:45:02]>> Okay. So, we're looking for angle SQR.

[00:45:07]So, we're looking for this angle over here, right? Um >> any other way >> can't we say uh sqr is equals to 3x angles in the same segment >> 100%.

[00:45:30]So if we look at this, so you were right Z in saying it is a cyclic quad, right?

[00:45:42]But we couldn't use it in the traditional sense. There's our butterfly.

[00:45:48]So this says angle SQR would be equal to angle P I mean SPR, right? So that would be SPR and this is equal to 3X. What's the reason?

[00:46:08]Almost in the same segment.

[00:46:10]>> In the same segment.

[00:46:14]Okay. Right. So we got that guy to be equal to 3x. All right. And then they say they're looking for angle P QS.

[00:46:27]P QS.

[00:46:30]So we are looking for PQS. So we are looking for this angle over here, right? Are we able to find out what the value of that angle is?

[00:46:54]Don't forget.

[00:46:57]Don't forget >> what we know about this angle here.

[00:47:08]What do we know about angle PQR?

[00:47:14]>> PQR is 90°.

[00:47:16]>> 90°.

[00:47:18]So if the entire angle is 90° and this guy is 3x, what would be the size of this angle over here?

[00:47:41]Okay, please remember.

[00:47:44]So that will be 90 minus 3x.

[00:47:50]Okay?

[00:47:52]Because if we add those two. So if we take pqs plus sqr.

[00:48:09]So we took this one plus the one that we just got. They are equal to 90°.

[00:48:17]Why is that? We said angle angle on a semicircle.

[00:48:34]Do you guys agree?

[00:48:39]Right.

[00:48:39]>> Yes.

[00:48:43]>> So in this case, if these two gives us give us 90°, so we know PQS, that's the one we're looking for. So PQS + 3X will be equals to 90. So which means pqs will be equal to 90 - 3x. Does that make sense guys?

[00:49:13]>> Yes sir.

[00:49:21]>> Makes sense.

[00:49:23]>> Okay.

[00:49:24]Right. So we found this angle here.

[00:49:29]That's 90 - 3x. All right. Now, let's go to the next one.

[00:49:35]>> Sorry. So, would I be wrong if I were to to use the sum of angles in a triangle for PQS since well, we said um PSQ is 2x.

[00:49:51]>> Okay. So, we said this guy was 2x.

[00:49:55]All right. So for you to find this one here.

[00:49:59]>> Yes.

[00:50:01]>> Um you would be incorrect in that remember you don't have this entire angle.

[00:50:10]>> Oh okay.

[00:50:11]>> Yeah. So that's why it would not be it would be incorrect in that case. Right.

[00:50:18]The triangle we could work with is this small triangle here. Right. Because then you know the angle 2x you also know the angle 3x. So we can find out what is this angle over there. Make sense?

[00:50:38]>> Yes sir.

[00:50:39]>> All right. Beautiful stuff. Okay. Let's go to the next one. We are looking for angle PSQ.

[00:50:48]Right. So we're looking for angle PSQ.

[00:50:55]So, we are looking for this guy. Oh, I thought we already found that angle G.

[00:51:03]Okay. Oh, actually, we worked it out for ourselves, but we never really uh did it uh in in terms of the the angle. All right. So, how would we then justify finding that angle? Now please I want you guys to note you know when it comes to ukidian geometry remember that it's a deductive uh uh section so meaning that you need to make your points you you need to make them or construct them in such a way that they make sense this follows this follows that okay so I can't just start erratically anyway right so how did we get to the point Uh Z, if you remember, we said this guy was equal to 2x. But how did you get there?

[00:51:56]Um so we said that both the radi so we said P O uh or rather let's say OP is equal to O S they are radi that's the first thing right and so what do we note It means that angle O P S uh ops is equal to angle OSP.

[00:52:49]Why is that? These are angles opposite equal sides.

[00:52:57]Okay, are we still okay?

[00:53:02]Oh, by the way, which means they're equal to 3x.

[00:53:07]But we know that OS P if this is equal to 3x.

[00:53:14]So, OSP is equal to O S Q plus Q or PSQ.

[00:53:32]So, I'm simply saying that entire angle is equal to this plus this. Okay. If this guy is 3x, Osq is x, then it means psq would be 3x - x and that would be equal to 2x.

[00:54:02]Are we good, guys?

[00:54:06]>> Yes.

[00:54:08]>> All right. Beautiful stuff. Okay, now let's go on to the next one. Now they're looking for P R Q.

[00:54:20]P R Q. So we're looking for this guy here.

[00:54:30]So we found this guy to be 2x that one equal to P RQ. How would I find the angle P R Q?

[00:54:42]Uh, anyone wants >> Mhm.

[00:54:48]>> almost in the same segment.

[00:54:51]>> Okay. So, once again, we've got our butterfly.

[00:55:05]We found this guy to be equal to 2x and so this guy would also be equal to 2x. So we know p rq would be equal to p sq which is equal to 2x.

[00:55:24]We said these are angles on the same segment.

[00:55:31]Right? Iman, you guys are taking all the points. Eh, strong as a wild ox. All right. Uh, I know that the diagram looks really nasty right now, but we are almost there. So, the last one is QPR.

[00:55:53]Okay, let me just remove this QPR.

[00:55:57]So, we're looking for this angle here, right? How would we get angle QPR?

[00:56:07]Z, do you want to take this one?

[00:56:12]Yes, I can. So, um, since you know the angle of Q, the smaller one, can we not use that?

[00:56:24]And um, since= 180X Yes, sir.

[00:56:31]Um the only problem is if we do you want to take sum of angles in a triangle?

[00:56:38]>> That's what I was thinking.

[00:56:40]>> Okay. Okay. Actually, which triangle do you want to take?

[00:56:50]>> I was going to take Q QPS.

[00:56:57]>> Okay.

[00:56:58]and then work your way towards this entire triangle. I mean this entire angle.

[00:57:07]>> Yes sir.

[00:57:08]>> Okay. And thereafter then you can work out that smaller triangle.

[00:57:15]Okay. That would work. It would definitely work. Um I want us to use a a a an efficient more efficient method.

[00:57:25]Remember, remember who you are, Simba.

[00:57:34]What do we know about this guy?

[00:57:39]Do you remember that's a 90° uh angle?

[00:57:46]>> Yes.

[00:57:47]>> Yes. So, we can say uh sorry, I'm just going to have to remove one of them. Let's remove this guy here.

[00:57:58]Sorry.

[00:58:01]So in triangle.

[00:58:04]So I'm telling my examiner which triangle I'm working in. So in triangle PQR, right?

[00:58:17]I already know that angle Q plus angle P plus angle R is equal to 180 and this is angles in a triangle.

[00:58:39]Okay. So these are the sum of angles in a triangle. Right? But what do I note in this case? I am looking for that angle P over there.

[00:58:52]I know angle Q is 90.

[00:58:59]Angle P + 2X is = 180. So what does that mean?

[00:59:07]Means that that angle P is 90 - 2X.

[00:59:14]180 - 90 will be 90. If I take the 2x to the other side, it becomes - 2x. So that will be 90 - 2x. So angle uh QPR is 90 - 2x.

[00:59:32]And look at that guys. like it's uh it's amazing how many marks you would be able to pick up from just being able to observe these things. And I do understand uklidian geometry once you in fact I would say to you please make a point of doing at least one rider a day. Okay, just a one rider a day. All right, I'm I'm just going to take this last one. um they said to us uh in the figure we've got QPR and S which are points in the circumference of a circle with center M right so we know we've got a center there if it or rather it is given that M1 is 4x + 40 and S is 5X + 20 right and they want us to calculate the size of angle Q with reasons.

[01:00:39]They want us to find this angle here with the reasons.

[01:00:46]How would we go about that?

[01:00:55]Can we find angle Q using uh uh opposite angles of a cyclic quad since well QRSP as a cyclic quad >> right? So if I took angle S plus angle Q.

[01:01:16]So angle P plus angle Q.

[01:01:23]This would be equal to >> it's actually SL.

[01:01:28]Sorry about that. Angle S plus angle Q is equal to 180. So what does that mean?

[01:01:39]We know angle s is 5x + 20 plus q.

[01:01:49]Oh sorry uh we need to give a reason.

[01:01:52]These are opposite angles for cyclic quad right. So this is equal to 180.

[01:02:03]So what would be the size of angle Q?

[01:02:07]That would be 180 - 20. That gives me 160 - 5x.

[01:02:15]Correct.

[01:02:18]>> Yes.

[01:02:25]>> Right. Um well they would probably say to us in this uh in this question we need to find out let's say if they said calculate the value of x uh I'm surprised that they didn't go that far.

[01:02:42]Um then in that case what would I say?

[01:02:47]I would say well angle Q is equal to or rather angle M M1 is 2 * angle Q.

[01:03:02]Okay. So if I've got m1, m1 is 4x + 40, which is equal to 2 * 160 - 5x.

[01:03:17]All right, I'm sure you can see right now we can do a lot of uh gymnastics in terms of algebra and then we can now find out what our value for. Uh so that would be 320 - 10 x and if I take that to the other side that will be 14x is equal to 320 plus that would be 360.

[01:03:46]Okay. Oh no no no not 360. 320us 40. So that would be 280.

[01:03:55]And you can divide both sides by 40.

[01:03:58]that will give us the value of x. Are we together guys?

[01:04:04]>> So forgot to give the reason angle at center twice angle at circumference.

[01:04:12]All right. So I am going to leave it here. Um we are still going to look at our proofs and so on. It is very crucial ladies and gents. We've got an entire playlist on on utilian geometry. I would say to you guys now that you've started with uklidian geometry do not miss a day without at least doing one rider up because remember your instincts in answering questions comes from doing this thing on a regular basis. So I would say to you do this as often as you possibly can so that you can maximize and and in fact even cut the time short that you are going to answer these questions. Of course you'll be a little bit rusty at the beginning but the more you keep doing riders the better at them you become. Okay. Right. Uh I'm sure you are thinking I finally let the day end.

[01:05:16]Even so, were you beginning to enjoy it?

[01:05:22]>> Yeah, >> can be nice.

[01:05:28]The torture utility and children can be nice.

[01:05:31]>> Yeah, but um it isn't such a bad section. So, please do note. So, what we're going to be starting with um we'll practice this. I'll give you guys a a um some work uh that you can do on a daily basis.

[01:05:46]uh but what I want us to do on Saturday is that I'd like for us to start with the proportionality theorem right to start looking at how we relate uh this to proportionality theorem we'll still be coming back to this but of course we now are going to mix it with a bit of proportionality here and there all right to those of you that were joining us on our YouTube channel thank you so so much please don't forget we do have a um a workshop that is coming up. We are going to let you know about the dates. Uh this is before your June exams, right? So we will have a workshop. Um and the different workshops will have not only are we going to have workshops, but we're also going to have master classes, right? One of those master classes will be the master class on uklidian geometry h but we'll also have master classes on every other section that you'd require.

[01:06:44]Please guys continue to work hard and if you have not applied at university what have you been waiting for? Please apply.

[01:06:54]All right, have you applied?

[01:06:59]>> Yes sir.

[01:07:02]Have you applied?

[01:07:05]>> Yes, sir.

[01:07:06]>> Okay. I hope all of you on YouTube have also applied. All right, guys. Let's leave it here for today. H I'll see you guys again next time. Please enjoy the rest of your day from me for now.

[01:07:27]Shut up.