Number nine, we study angles in plane geometry. So first we're gonna work mainly with triangles and we add circles. And lastly, we'll discuss some of the advanced techniques in geometry. So let's review some of the basic things. For parallel lines, if A and B are parallel to each other, then this L angles are the same, right? And this is vertical with it, so these are the same as well. And then the basic theorem, the sum of the angles in a circle, in a triangle is 180 degrees. So if you have a N gone, then you just divide into a N minus one triangles, and then you just sum up, and then you would have the total angle is the N minus two, 180 degrees. This theorem, the exterior angle theorem is a very simple theorem, but it really helps you speed up calculations. So ABC, alpha and beta are the interior angles at A and B, and gamma is exterior angle at C, then gamma is just equal to alpha plus beta, right? Because both the left-hand side or right-hand side are supplementary to this inner angle at C. Then the next thing is the review of all the steps of triangles. For an isosceles triangles, as we know that the bisectors is the same as the median, is the same as the altitude. And this one, we saw it last class, a 30, 60, 90 degree triangle is a half of a equilateral triangles. And we want to remember this by heart, right? MC is one, then AM is root two, three, right? And AC is two. And with E is the midpoint of AC, then MEC is an equilateral triangle. And in general, if we have a right triangle, then BD, the median here is half of the hypotenuse, right? And vice versa, if BD, if it's as a triangle, such that BD is a half of AC, then ABC has to be 90 degrees, right? So some of the basic theorems that I think we're gonna use extensively in today's class. And the next step is using congruent and similar triangles to define the angles. So just summary of what we covered last class. For congruent triangles, we have the three tests, For similar triangles, we also have a three test, and the last one is just angle, angle. So when we face with the problems, let's go through first the steps of triangles, and the next step is consider either congruent and similar triangles. Okay, so let's get started. Nia, can you please do this one? All right. So for this problem, we have a square ABCD and an equilateral triangle DCE on top of it. So it's asking, what is the measure of the angle AEB? So that would be the top angle, but from in the smaller triangle. I'll give a few more moments all right so since we know that this is a square and then an equilateral triangle on top of it we know that all of these sides basically all the outer sides are all equal to each other um and that's very helpful to know because from there there's a lot of different things we can do the idea off the top of my head though is that we know that this side triangle here this side triangle is an isosceles triangle and we know that this fat angle in the middle we know that that is just 90 degrees plus 60 degrees because the angle of the side angle of an equilateral triangle is 60 so that gives us 150 degrees and if we know this is 150 degrees we know that these two side angles these like tiny ones need to add up to 30 degrees and since there's two of them that means that each of them is 15 degrees most importantly this one is 15 degrees and since this is symmetrical we know that if this one is 15 degrees this one is also 15 degrees and we know that the total angle here is 60 degrees so it's just 60 minus 15 minus 15 which is equal to 30 degrees and that's our angle aeb so it would be answer choice b yes thank you yeah it's a good idea so to mark all the you know equal segment you have right on the pictures so let's go to the next one let m be the point where all the attitudes of the acute triangle abc meet if ab is equal to cm then what is the measure of the angle at vertex c of triangle abc right so we know that yeah I think this one I should highlight it here we just draw a and you know what for this problem actually what you want to do is that you don't have the handout because I think I didn't send and I think you don't need the handout it's just you know there's a one one thing in geometry so one type of problem is construction problem so you start with ab and then you choose any point here and you draw the attitude up right okay so I think yeah maybe not quite but anyway I think we have to go to the other way around but anyway try to draw such that cm is equal to ab like in geometry you want to draw as accurately as possible because a lot of time from that you you can see the result and then you can you know work like what to prove that And I also like to mark all the, you know, right triangle or indicate any angles that you know right on the picture. So, I mean, usually when you see a problem, the first thing you see, oh, you can look for if any special triangles, like what we had in problem one, and then if not, then you look for if any triangles are congruent, if not, then if any of them are similar, and you know, if none of this works, then we probably need to draw some actual lines. So over here, you don't see any special triangles. We do have some right triangles, right, and that, so apart from the fact that these two segments are the same, once you see the angles that are the same, you also want to mark it on the picture, right, right on the drawing. For example, over here, if I call this angle alpha, you just mark it alpha, because here's the right angle, so this alpha and this beta, and we know that alpha and beta add up to 90 degrees, right, but from that, do you see any other angles in the drawing that are also alpha and beta? For example, this one is also beta, because it's vertical with this angle here, and this one is complementary to beta, right, so that is also alpha. So you mark everything, and if you look at A prime B, then again, you have a beta as well. So from that, do you see any angles that are similar or congruent, any triangles that are congruent or similar? You see, right, so we have a bunch of angles that are equal to each other, but then we have the key thing here, we have the side CB is equal to A, CM is equal to AB, right, so we need at least one side for the two triangles to be congruent. So in this case, you want CMA is actually congruent to A, CMA prime is actually congruent to A, A prime B, okay, and when you write it, you want to write it so that the corresponding angles are the same, C is equal to A, and M is equal to B, okay, so should be A, B, A prime, and the test here is angle, side angle. So once that we have these two are congruent, then we see that the corresponding side have to be equal, so that means CA prime is equal to AA prime, right, so the drawing is not to scale, but if you're drawing to scale, then you would see that these are the same, and this is 90 degrees triangle, that means CA prime A is right isosceles triangle, and that follows that this angle here is 45 degrees, right, which is answer choice C. Go to the next one. So relationship between angles and sides in a triangle, so you probably, I don't know if you have a trick in school or not, but there's a famous law, it's called the law of psi, that says that A divided by psi alpha is equal to B divided by psi beta is equal to C divided by psi gamma, and actually, it sounds like, you know, complicated, but you actually, you can prove that, because if we draw the altitude from A here, it's AH, then you see that B times psi gamma is actually equal to AH, and that's equal to C psi beta, and then it's divided, you know, you can cross multiply it, and you see that this relationship holds, and you can do the similar thing for the vertex A. And what does it say, I mean, what is the consequence of this law, right, because we know that psi is a monotonic increasing function from zero to pi, so the bigger the angle, the bigger the psi, and from this equality, we know that the bigger the psi, the bigger the psi, the opposite side to that angle, so that means that the, you know, triangle, the smallest angle is opposite the smallest side, and the biggest angle is opposite the biggest side, right. But there's another way you can come up with that conclusion, is what we are looking for, is that suppose that in this triangle ABC, we have the psi C is smaller than psi B, right, C smaller than B, then, or you can say that maybe we can start by saying that the angle gamma is smaller than beta, right, this is gamma, and this is beta, then if we draw from B an angle beta or gamma, here is gamma, then it's going to intersect the AC at point M, which is inside the segment AC, and then this BMBC in isosceles triangle, right, you have MC, MB is equal to MC, and then you look at the triangle AMB, we say, we see that MA plus MB is bigger than AB, which is the triangle inequality, but MA plus MB is exactly AC, right, bigger than AB, then we also, you know, come up with the same conclusion here, so either you can use trig, or you can use this elementary geometry, but that is what we are interested in. Let's move to the next question. Okay, so this question, which side of the polygon shown in the picture is the shortest? So you can use the result from, you know, the theorems that we just show, or actually when I think about it, you don't even need to do that, but let's see. Give you a couple of minutes to use this question. So from the picture, we kind of obviously can eliminate A, B, and B, C, right? They look, I mean, the picture is pretty much drawn to scale. So A, B, and B, C are definitely not the shortest, and we are thinking between E, D, and D, C. Which one is that? This is 90 degrees, this is 120 degrees, so it's an obtuse angles and definitely this AB is way, way bigger than MB. And we also want to mark, right? You see, CMB is an isosceles triangle, so MB is equal to MC. So we want to just indicate this by marking random drawing. And then DMC is also isosceles, so MC equal to MD. And now in this triangle EDM, this is 49 plus 96, that is 138, 138. And then we take 180 minus 138, then again, we have another 42 degrees here, right? Another thing that whatever the missing angles that you have, you also want to calculate as many missing numbers as you can. And then that means EDM is an isosceles triangle, so DM is equal to DE over here, right? So ED is this. This CB is root two of this, so AB and CBs are question because they're so much bigger than this. And for this one, either we know that 61, 61 is 58, so either we invoke the theorem, which shows us now that DC is opposite the smallest angle, so it's the smallest in this triangle, or if not, I think you can, if you don't know that theorem, I think you can just make it into a equilateral triangle over here, and the 61, so this point is inside the triangle DMC, so that means DC is smaller than DM. I think that probably would be the best if you don't know the theorem. And in the end, that means DC is the shortest side, right? Okay so now we add circles to the legs. So let's review a little about the circle. So first theorem that an inscriber angle subtended by the same arc are equal, right? So this one actually has a very nice interpretation. So suppose that this is this one like let me mark it ABC. So suppose that BC is the screen in the theater, right? This is BC and then you take a seat at A and you say, oh this is the view, this is the angle subtended by the screen from point A. And like we ask ourselves what other seat has the same angle, right? Subtended by the screen from this point. And then the collection of those points is called the lobe, the locus, right? And how to find it? You actually from ABC, we know that there's a circle that pass unique, there's a unique circle that pass through any three points that are non-linear, right? Non-collinear. So you draw a big circle here at O that passing through B and C. So all the points that lie on the arc, the major arc BC would have the same view to the screen. This, yeah, there's the same subtended angle by the screen. And we can also show that each of this angle is equal to half of the center angle BOC. So I think that's a very nice theorem and that's also used extensively in geometry for circles. And then when we consider BC, the special case when BC is diameter, in this case AB of the triangle. So any angle subtended by a diameter is going to be half of the center angle 180 degrees. So that angle has to be 90 degrees, right? So that's another way to define a circle. This is a locus of points such that the angle subtended by this AB is equal to 90 degrees. And another thing that, another result that we usually use, the problem that we see is that tension lie to a circle. Either the two circles are tensioned to each other. In that case, whenever you see two circles that are tensioned to each other, you want to draw the lines that connect the two centers. And this line is going to have the length equal to the sum of the two radii. And if you have a line that's tensioned to a circle, then you really want to draw the radius from O to B that, you know, touches that point, that line's point B. Okay, so let's get started with inscriber angles. Okay, so what is the measure of the angle phi shown in the picture? Right, so we just draw a circle and pick any two-point cut approximately. We draw a circle and I put this is A and B. So we say that B is an inscriber angle, you know, subtended by arc AB and from point C. And we know, so these are intersecting at O. And we call O, maybe I don't call O because it's not necessarily the center of the circle, right? Okay, so we call M and this is D. Okay, so this is the first question, so let me just do it as an example to the theorem that we just described just now. So in the triangle ADM, we see that it's 30 degrees and this is exterior angle at M, right? So 70. So you can of course calculate that this angle is 110, but you can bypass that step, right? So this is 110. So of course you can calculate this 110, but you can bypass that step by invoking the exterior angle theorem, so this 70 and subtract away 30, then right away you have this one equal to 40 degrees, right? And now if we look at the angle, the circle over here, 40 is the angle subtended by the arc AB. That means all the angles subtended by arc AB must have the same measure, which is 40 degrees. So pi here is actually equal to 40 degrees. This could be new to some of you, but you know, we do a few problems, you're going to get the hang of it. So next one, kind of a little bit similar. In the figure here, you see points A, B, C, D on a circle, quad AB, this one is a diameter of this circle, and the measure of angle A, B, C is 35 degrees, and the question asked what is the measure of angles B, D, C? So you draw a circle and take point C and D, you know, 35 degrees approximately, you don't have to draw exactly, just take one point, C on one side of the arc AB and D on the other side. And the question says that quad AB is diameter, so you want to indicate it in the picture by putting, you know, putting O here, just let's reserve O to be the center of the circle, you want to put O here to indicate that AB is a diameter. You may need to draw an extra line, right, because, you know, just to make use of the fact that AB is the diameter. So I draw actual lines and see if we can figure out from there. So A, D, AB is diameter, so that means if I connect A and C, what do I have? Okay, C, put it here, and C, and similarly if I connect A and D, do you recognize any special angles in this picture here? So although we, I mean, we can actually prove that you have a A and B, and then you take M as a midpoint of AB, and then, you know, choose any point N such that A, N, M, N equal to, A, N equal to M, B, then A, N, B is 90 degrees, right, because these, both of these are isosceles triangles. So you can easily prove that without using the, you know, the theorem about inscriber angle that we just talked about just now, but if, you know, we use that, we know that if AB, quad AB is diameter, then ACB is 90 degrees, right, 90 degrees. And that just, just exactly same as what we show here, because CO here is a median, and OC equal to OA equal to OB. Yes, so, but we don't have to use that, we can right away say that ACB is 90 degrees, and that means this is 35 degrees, so the first step is just, you know, do angle chasing, so you know, find out all the angles, as many as you can find, 35, so it means this one is going to be 65 degrees, right. And 65, now you set up a standard point A and you look at OCB, so that means 65 is going to be the angles that, you know, subtended by the OCB over here, that is 65 degrees. And what is another angle, it's also subtended by OCB, right, so if I change our point of view, we move from A to D, and so D is exactly, it also, you know, exactly subtended by OCB, and that means this angle here is also equal to 65 degrees. And then if we look at AB, and also let me look at this point from this side, then ADB is also equal to 90 degrees, right, and that means this 65, so the left over here is 35 degrees, and now you look at the arc AC, the arc AC here, then we have this B, this angle, subtended by arc AC from B, that's 35 degrees, and it must be that the angle from D, right, subtended by arc AC from D is also must be 35 degrees, so sort of everything comes in nicely together. Okay, good, okay, let's go to the next one, yep, yeah, can you please do this one? Alright, okay, so in the picture, PT is the tangent of the circle C with the center O, and PS bisects the angle TPR, so we need to calculate the measure of the angle TSP, so that would be this angle right here, so I'll give a couple minutes to go at this one. Okay, I'll give a few more moments, and this is a question 27, so this is a significantly harder one But anyways the way You can go about this is trying to find Similar angles so obviously we have the two equal ones here, so we have we can call this alpha and We can call this one alpha 2 Let me draw that so we can call both of those alpha and then I don't know why okay my Cursor is being a little weird, but anyways what we can then do is we can draw a line here and kind of make this T O and R its own triangle and from there Let me draw this line a little bit better You can see This is a right angle right here and Since that's a right angle and these two these are both the radius right T. O and R. O. They're both the radius Which means it's a nice oscillates triangle, which means that these two Angles here, let me use a different color, but those two angles in red here are The same so we can call that Okay, we can call both of those beta And So that being said now we can just look at the angles and try and sum them up So the first thing I see is if we take the big angle or the big triangle TPR We see there's beta plus beta So we take these two outer betas right there Here and here, and then there's the 90-degree angle And then there's the two alphas there on the right side And that's all equal to 180 so if we do the algebra here we'll end up getting is that We have two alpha This is we get two alpha plus two beta Equal to 90 because 100 minus 180 minus 90 is 90 and if you divide both sides by 2 you just get alpha plus beta Equals 45 and we can't exactly do anything with that So we're just gonna leave it as is and now we're gonna look at the smaller triangle So we have our question mark angle there and in the triangle TSP. So we have the Here, I'll use a different color. We have the unknown angle Plus Alpha on the right side And then we also have that same beta plus 90 kind of at the top here this So beta plus 90 And we know that out and it's all equal to 180 and Then we know that alpha plus beta is Equal to 45 Which Means the entire thing so we have alpha plus beta So 45 plus 90 is 135 and then 180 180 minus 135 is 45. So that means that our question mark is Equal to 45 Which is answer choice B So any questions about that? Yes, any questions is fee free to send a check is it to me or leave Okay, if not, let's see Let's go to the next one. So last week remember there's a one question I was solving it using trick and then one student was able to do it with that trick, right? So that's what we're trying to do here. We try to solve I mean a few problems in different ways so Recall what we did now is though you look for special triangles you look for Similar to congruence and if everything else fails and a lot of time you need to draw that actualize and that comes with practice Like how to draw the actualized and then of course They have a choice of using advanced theorems like the law of cosine size in terms of other Beautiful theorems in geometry, but my experience that I mean in the in the exam They just don't make questions have that, you know Some just because you know more than you can you can actually do that So this is good to know, but I don't think it's a huge advantage in math can cool Exam and then of course if you know trigonometry you feel free to use it But as you can see that a lot of time you can you know get away without knowing trick but it's a good thing to touch on so we're gonna do all of this and Remember whenever picture or whenever problem you try to draw as accurately as you can So let's go for this one. Yes a drawing cut simple here. You have a B C D E F G H, right? It's basically you have a three Equal congruent square just like put together and then you draw H C the vertex here is the second point here and draw the diagonal again in here and These two lines intersect P and the question asked how to find what is the measure of this angle at P? So I didn't send a hand up, but I think it's a good skill to to have just draw as accurate and accurately as possible on the draft paper Three squares and connect this in this So this is an example of problems that use many different ways to do that If you know you trick you have to you can you trick but at least you have to draw some actual line So I give a couple of minutes it's a it's a good question If you have a protractor and then after you draw the picture accurately you can actually measure this angle right and you know exactly what it is and a lot of the time actually you know you know the answer and you work backward and try to prove it but in the exam when you don't have it and these angles are pretty close to each other, there are some choices. There are, you know, several different ways to do that, but let me just start it halfway and see if we can finish it. So one way I see this problem is that we look, we are asked to calculate this alpha, this angle over here, and if I look at the triangle APC, then I know that this is an exterior angle at P, so it's gonna just be equal to alpha plus beta, right? But these two angles are, you know, in the middle of nowhere, so I want to bring them together, and one way to bring them together is you draw a line here, if I, you know, extend this, you know, to the point that's equal to the sine length of each small square here, and I connect over here, right, then because, then because I call this point M, then because CM is, you know, equal and parallel to HA, then AHCM is a parallelogram, that means HC is parallel to AM, and that means angle beta here is gonna be equal to this angle MAC, right? So I successfully bring the two angles together, and also you can see it right from here because HC is parallel to AM, so right away you have this angle here equal to, you know, PAM, okay, so that is just one way to do that, and from here, you see, you can calculate what this angle is. So, if we look at, let me draw another line here, if we look at, if we connect E and M together here, then EFM is the right triangle, and the ratio of the two sides, EF and FM, is 1 over 2, and if we look at the triangle MCA, it's also a right triangle with the same ratio length, CM over CA is equal to 1 over 2, so that means MCF, we have these two triangles, MFE, is congruent to ACM. Right, do you see that? Yeah, so back to the first step, looking for any triangles that are congruent after we draw the actual lines, so that means this angle here is going to be equal to beta, and beta is complementary to this angle, AMF, and this is 90 degrees, so that means AME is also 90 degrees, 90 degrees, and then also because these two triangles are congruent, the hypotenuse are the same, we have ME is equal to MA, right? So, that means the triangle AME is a right isosceles triangle, and the angles we are looking for EAM is 45 degrees. Yeah, so really the hardest step in this problem is to realize this is a sum of alpha, beta, and try to bring them together, but if you do in a problem, it's kind of, you know, it's a systematic and not like you just have a random guessing, right? So, that's one way to think about this problem. Another way to think about it, oh, by the way, it's a long question, so I type up everything here, if you need to review, you can, you know, pause the video and watch the recording and stop at this page. And then another way to look at this problem is in trigonometry, so again, similarly what we did before, we noticed that this angle, square here is equal to alpha plus beta. But what is alpha? Alpha is actually a very nice angle because the tangent of it is a third, right? EA over ED, so alpha, the tangent is a third, and we look at beta, the tangent of it is AH over AC, which is a half. And from that, you notice that if you know trig, it's just a matter of calculation because we know the tangent of the sum, how it's related to the tangent of each two angles, so we just have to calculate these things out and do the inverse, right? And everything works out perfectly well. We have a 1, and the inverse tangent of 1 is 45 degrees, yeah. And in general, that's why it's very powerful because in this case, for geometry, we kind of work with angles that are pretty special. We use elementary geometry, but for trig, you can use arbitrary angles. So if you know that, definitely use it, but as we've seen several times, we actually do not need to use it in the exam, but I put it here. And then actually when I was doing it for the first time, I actually wanted to show this problem just to see how versatile our geometry tools are. So another way to look at this problem is like this. This is angle alpha, this is angle beta, right? But if we draw the diagonal from HD over here, then this angle is also equal to alpha. So in the end, I just redraw it. I, you know, just, you know, delete whatever the top part, and I only have HABC, AB equal to BC equal to CD, and I need to sum up these two angles, beta here and alpha here. And forget about anything else. I just have this very simple picture and what the sum of alpha plus beta. Okay, so let me see. From here, what we can do, yeah, another thing we can do is that we can extend H to B here, and then from C, I also extend all the way to M here. So M here, and this is 90 degrees. All of these sides, they are equal to each other, right? So we have a very nice, like a lot of right isosceles triangle, and same thing over here. This is also a right isosceles triangle, MCD, and then right away we have this angle is 90 degrees, right? But then we look at the triangle HMD, because DM is equal to MB equal to BH. So that triangle HMD is a right triangle with the ratio of psi length DM over MH is a half. But that, if you look at the triangle HAC, we also have the same relationship, this 90 degrees, and HA is a half of HC. So actually HMD, that triangle, is similar to the triangle HC, yeah, I like to write, so CAH. Do you see that? It's the psi angle, well, okay, psi angle, psi test, right? So if you have the two triangles the same, so it turns out that this angle here, beta, is actually equal to beta, right? And now, if you need to sum up alpha and beta, you look at the triangle HBD, and that's equal to the exterior triangle angle at this point, and that one is 45 degrees. So really, there's so many different ways to do geometry problems, and the shorted one might not be the best, because, you know, the more complicated one actually allows you to do more practice and learn more. So let's go to the next one to see if we can also do the same thing. Let AD be the median of triangle ABC, so let me mark from here, BD is equal to DC. If ACB, this triangle is 30 degrees, and ADB is 45 degrees, it's only 45 degrees. Yes, mark everything you can on the picture, so draw it and mark it, and, okay, I think that's it. And the question asks, what is the measure of triangle BAD? What is this triangle? So very simple pictures of, you know, two angles are given, these are median. Again, for this question, obviously, we need to draw an actual line, and as we did on the other one, we draw the actual line so that we have a parallel line, we end up with some special triangles, and from there, we can create similar or congruent, right? So as you try to draw actual lines, try to make use of these special angles, 30 degrees, 45 degrees, that rings a bell, right? What kind of special triangle you can create with these angles. i also mark another angle that i know from this 45 degrees is exterior triangle angle so that means this one is 15. i don't know if it's useful or not but you know just mark as many things you can okay if you have a protractor and you know a ruler you can draw exactly to scale and you know exactly what this angle is right and go backward but in this case um so let me get started and then maybe you can finish it um 45 degrees right I mean one way of making use of 45 maybe I draw something like this right 90 90 and then 30 degrees I maybe draw something like this because I have a 30 60 90 degrees right you know you can try different things that we just don't sit and you know look at the problem but we try different things it may or may not work out the only thing is that yes I probably can make use of the angles but what about the side here right these two bd d go to dc I mean these two segments are the same and how can I make use of that information and over here it kind of doesn't really allow me to make use that information so um over here maybe if I can draw like this one this line over here you know perpendicular to this I mean parallel to this one then I would have a 30 I still have a 30 60 90 and again now I have something else um this I have a right triangle and remember what we talked about a right triangle before the median that connects the right angle to the midpoint of the hypotenuse what's the property of that median right so maybe that is something that we allow so this line looks like promising because it allows me to make use of the fact that d is a midpoint of b and c which I know this actually doesn't you know allow me to use that information right so um if we come back to that we may be this is probably the most promising lines that we have among the ones that we've drawn okay so this one over here and usually is I mean once you got the right line things you know all come together very quickly right because once we have a 90 degrees and over here we have a 60 degrees so that's already promising and then we know that whenever you have a right angle and you have the midpoint you want to connect them right because we know that hd equal to db dc this three point you will draw a circle here bc is this diameter and db dc dh um this radii of that circle right so again because of that I also mark this one equal to um you know bd equal to bc and this 30 so that means it's also 30 degrees and that 60 then we have a equilateral triangle here right so that means hb is also equal to hd so I have all of this together and then this is 30 degrees again you know you can calculate this big angle but I really like the exterior angle theorem because it allows me to right away find out that the angle adh is also equal to 15 degrees here because you take 30 and subtract away 15 so again I have a hd is a isosceles triangle so that means ah is equal to hd and now everything is so beautiful because you have a tons of special triangle and bha is an isosceles right a right isosceles triangle so this angle is 45 degrees and that means the angle here is 30 degrees right so I really think that the person who you make this problem he draw a bunch of you know special triangle first and then deleted this actualize and left us with the question that we got and then our job is had to recreate it so that is what you can do with actualize and then again if you want to do if you want to do okay if you want to do trick you know trick is I think it's also a very good solution so the way we look at that is that we draw an actual line here again it involves drawing actual line eae is perpendicular to bc so what you see here you have a 30 degrees here so that means you know that b is going to be equal to this one's going to be root 3 of b right bc and that we know by heart right now this is a and a this is 45 degrees right this is 45 degrees so that means ed is equal to ea and that means ed is equal to b so from that I mean without going to the like all the details we know that we can calculate eb in terms of a and b and actually can find ratio of a over b so if you can find the ratio over b that means we can find the ratio of eb over ea and that means we know the tension of this angle right and if we know the tensions of this angle so in principle we know that and what that angle is because we can do the inverse tension and we know ead is a 45 degrees so with a subtract away that angle to get this angle with 30 degrees so that is kind of principle you don't want to go to the detail but you know that it can be done that way and then the way it's done that there are many ways to do that you actually don't need to use that but if you know you can use the tension but otherwise you can just calculate all bunch of like small thing maybe draw another line here do angle tracing length tracing and you can find out the 50 degrees but if you know the formula for the difference of tension you subtract it in and then you actually find out the tension of bad turns out to be one over root two of three which is 30 degrees so I really think that geometry is very exciting subject later on you probably do a lot of calculus linear algebra maybe analytic geometry but euclidean geometry is a very good subject and you know allows you to prove a lot of things from some basic axioms so I hope you have time to review everything if you need extra questions let me know I can definitely recommend some sources okay so anything please feel free to ask me and give and if not I will see you for next class for 3d geometry next week

plane geometry

angles

triangles

circles

exterior angle theorem

isosceles triangles

law of sines

inscribed angles

tangent properties

problem-solving strategies