 Hey everybody, what's going on? Tutor terrific here. Thanks for stopping by my live stream Whether you're watching it live or in post good to see you guys So second live stream in the try this again. I Was happy to get a few people on here last time checking it out. Let a few more people know So tonight I have a couple things planned, but as for usual If you guys want to interrupt the stream for some questions, please put them right in the chat if we get enough people Trying to get questions answered then I'll prioritize those who do a super chat for me And I'll thank you in advance for that supporting me and my family and my work as a tutor and an educator so feel free to do that, but also I'll be I have some stuff planned For tonight in case I don't get questions and that's totally fine I'll put what that is in the video description after I post it that does live goes post and But feel free to interrupt that at any time with your Requests alright, so without further ado, let's shrink me and let's get my screen up there Alright, and we're gonna share that right now Here we go So What we have here is our first topic for tonight, and that's balancing chemical reactions All right It's a topic I got from one of my private students a couple days ago And I think it's really really really really important to Go over this topic Because a lot of chemistry students really seem to struggle with this and I'm gonna go through multiple ways as you could see over here to do This balancing reactions System that we need to do but let's get started here With understanding what we've been looking at for those who aren't really clear on what this is Hi for those of you who are joining now. Thanks for stopping by Feel free to put any questions you have in the chat to interrupt my scheduled programming Okay, so what we're looking at here is a bunch of reactions, so let's just zoom in on number 11 for the moment so we have P for phosphorus and it's gonna combine with oxygen to make diphosphorus pentoxide and What the the the goal for this exercise is is to balance the number of atoms on each side so that we obey the law of conservation of mass if You have a reaction that is not balanced And doesn't have the same number of atoms of each type of element on both sides. You are completely and utterly Disobeying the law of conservation of mass and that is a fundamental law in physics and chemistry and all the sciences So we cannot do that and so we balance our reactions to avoid that now our first one here phosphorus Oxygen and P2o5. I'm gonna explain before I start doing this the two methods by which We balance reactions. So first we have here. I know it's wordy. I'll go through this one step at a time the traditional method Okay, the traditional method It has many different forms a lot of different people discuss the traditional method in different ways and that's fine and you won't find any any Symbiosis with anyone everybody has their own version. So this is mine I like and I learned this when I was in high school To start on the very left with the very first element and you take inventory of the first element and countered on both sides of the reaction What I mean by taking inventory is a mental inventory. We go back over here. How many of The first element do we have? Well, we have One phosphorus Adam on this side and on the other side. We look for that P and we have two Okay, great So we've taken a mental inventory And now if we go back to step two if there's an imbalance in that very first element that you encounter You need to correct with coefficients, which are multiplicative. Okay, I'll explain what I mean by Multiplicative in a second. Let's talk about what I mean by correct with coefficients. What is a coefficient? a coefficient is a number that goes in front of the compound or the element and what it does is it multiplies all the numbers of that element the number of atoms of that element that Succeed it so they've come after it before any pluses or arrows and by the way you read an arrow as the word yield Okay, when I have phosphorus and oxygen combined like this it yields Die phosphorus pentoxide, so that's how you have to understand that but that's not the focus for today The focus for right now is balancing it So we look and we see our inventory for phosphorus and we see there's one phosphorus on the reactant side which is before the arrow and there's two phosphorus atoms on The product side so that is an imbalance. We do not have the same number So we can't have that we've got to deal with this now what we're gonna do is we're gonna Fix that by putting a two right here What this does is it multiplies the number of phosphorus atoms that I already see on the reactant side So now there's two times one or two phosphorus atoms on The reactant side so now we have a match we have a match of two phosphorus And two phosphorus perfect. So phosphorus itself is balanced But we're not done. So let's go back to Our traditional method rules here what we're gonna do now Is we're gonna move all through all the elements left to right We leave oxygen and hydrogen for last because in some of the compounds you've seen these early chemistry classes Oxygen and hydrogen is all over the place And so it's best to leave them for last because if you start with them You really really mess up any balancing you try and do and you might have to go back and fix things So not a good plan. So we leave those elements for last but Like I said, we're going to move through all elements left and right left to right. So now we've got oxygen Let's take our inventory unbalanced inventory. We've got two oxygens here And we've got five oxygens here We've got an imbalance and when we have Particularly an odd number. That's when things can get a little dicey It's really hard with multiple multipliers to balance odds like this Sometimes what we end up doing is we double anything odd so that it becomes even And that really helps with this process. So let's fix it. Okay What would I do to turn two into five? Well, immediately if I want to balance this I have to remember I can only do multipliers The coefficient that I would add here is a multiplier. Okay, it's not I'm going to add A certain number of oxygens and then I'll have five. So you can't just put like a three here And then I have five oxygens. It's a multiplier. It's three times two. That's six So that's no good. Okay, we can't have that But what we are going to have to do in this case if we're not allowed to use fractions, which I'm saying we're not We have to come back to the other side The product side and we have to double The number of diphosphorus pentoxide compounds. Okay molecules And so that's going to be a two right here Okay, let's fix our inventory now. We have four phosphorus over here and 10 Oxygens over here because that's how these coefficients work. They multiply everything in the compounds that succeed them Okay, so there's a p2 here. Well now there's four because it's two times two There are uh 10 oxygens because it's two times five. That's how this works Okay, now What we're going to do next is come back over here to this oxygen. Okay. I want 10 oxygens on this side. Now I can do that which is the beauty of this Doubling like this now I can do that. I can make this number of oxygens 10 easily by putting a five here five times two is 10 change my inventory now. It's 10 So now oxygen is beautifully balanced But if you as you noticed something happened to my phosphorus when I doubled the diphosphorus pentoxide I Caused it to be in balance and that's when we're going to go to the uh the traditional method steps I'm going to look at what happens each time you correct an imbalance and another Balance for a different element. It's thrown off. That's what we just saw You have to go back and correct that balance and any other subsequent imbalances that are caused Then we only have one that was caused and that is phosphorus was imbalanced So what we need to do now is we need to come back over here and Do what we need to do this coefficient to make it four four phosphorus on the reactant side and What do we do? We go like this we put a four here instead. So it's four times one phosphorus on the Reactant side. So that is a four All right now we look with these coefficients. Do we have four phosphorus on each side? Yes Do we have 10 oxygen atoms on each side? Yes So with this set of coefficients The reaction is balanced and we obey the law of conservation of mass Okay, so the final answer. Let's get rid of all this work. This is confusing to look at would be this Okay Now if you do it this way you are ensuring that you're getting the smallest possible ratio of atoms We don't want anything larger. I wouldn't want three times these numbers 12 phosphorus 15 oxygen and six died a died phosphorus pent oxides I wouldn't want that because that's not the smallest ratio when you do these. Do you want the smallest ratio? That's what chemists like to see I know in real life We've got millions and maybe even billions of these atoms reacting and maybe the tip of tip of a match Or something like that. That's all we care about. We care about the smallest ratio to make the reaction possible Okay, so all of that to say we can go faster with this step But let's look at what I said was the last the last last last step in this process Okay If you find that you have to write a fraction for a coefficient to balance the reaction Multiply the entire reaction by the denominator Of that fractional coefficient you're trying to use to remove the possibility of any fractional answers, right? Like I said, most teachers don't allow fractional coefficients. I've seen some that do I just think a little with a little bit of math this math right here in step five You can fix that and you don't have to deal with that. Okay So what we're when did we come up with that? Right here when I had before I doubled this and I had 0 5 I was thinking. Oh, maybe I'll just uh go ahead and put a What do I have to put here to make two five I have to put five over two Okay, five over two if I put a five over two in this spot And I had a two here and I had a one here What rule five says is you have to multiply all of these numbers by the denominator of this fraction To get rid of the fraction obviously when you do that you're going to get the numbers below So that would have happened anyway. So if honestly it's up to you What do you think is faster? I think that's not faster I don't like working with fractions if I don't have to especially improper ones like this So I avoid that and avoid fractions by doubling. That's just my preference You're welcome to your preference, but your teachers requirements are what matter most Okay, let's try and do another one faster. Okay Same rule though, and then we're going to go back and learn the algebraic method We're going to see how awesome it is, but it takes a little getting used to Okay, sodium Combined with water makes a violent reaction as you may know Let's off a little heat for sure Create sodium hydroxide, which is a solid and hydrogen gas which you actually see bubble up in the water sodium suspension And it's very nice if you get your hands some pure sodium. It's one to try but do it safely That's what alkali metals do in water. They react violently Especially as you go down the period it gets more and more violent So let's look here. Let's start our inventory. I'm not going to ride all these numbers out this time I'm just going to say we're looking left to right at sodium first. I have one On the product the reactant side and I have one on the product side So I'm good. I don't have to do anything So I move on to hydrogen I don't really have anything but hydrogen and oxygen left and I said those are the ones you want to say for last And so I'm doing that I'm saving them for last because there's nothing else to do. Let's look here. I've got two hydrogens on the products the reactant side and I have one two three Two three. I've got three On the product side. That's an odd number. That's no good. Okay My little shortcut is to double what makes it odd because you're gonna have to double it anyway to get rid of fractions and so we're going to do that right now I have a one hydrogen here and so that's going to make Adding one to an even number. What does it do? It makes it odd. Okay, so we're we're going to not have I'm not going to be able to deal with that. Okay So we're going to put a two here To get rid of that fact that that one H makes an odd number of H's on the product side Okay Now I haven't balanced H yet. How many do I have now on the product side? I have two Two times one and then plus two more. So four Okay, I have to come back over here to balance hydrogen And I have to put a two here so that there's four on the reactant side Notice how I messed up my perfect balance with sodium So I have to go back and fix that. I had two sodiums here That's why I'm going to have two sodiums here. That's going to make two on each side It's nice when you have elements by themselves because you can adjust how many there are And fix any imbalances that result from trying to correct other ones Okay, so our hydrogens are balanced at four each our sodiums are balanced at two each now Let's go to oxygen. So we got two oxygens on this side The reactant side and we've got two oxygens On the product side. So guess what? We're good to go. We're all done We've got two sodiums two oxygens And four hydrogens. So we didn't need a number here and when there's no number there It's assumed that you have a one. Okay We don't write a one ever. We just write anything larger than one Okay, all right, I'm gonna just pause for a second and look at my email and see if anybody's emailed me any questions they have Okay All right, not yet. And that's totally fine. I'm enjoying myself Seeing that people are watching Okay, let's look at this Other method, okay Algebraic methods beautiful methods. It's not super complicated or some crazy method or anything but Using algebra a lot of the time Once you understand what you're doing and can get past the abstraction of it can really Really help you it can really really help you Algebraic method for sure is preferred when you have really complex reactions to balance Okay I've worked with students who learn the algebraic method in class And I tried to get them to go back to the traditional method and they said, uh, uh, no, please don't I understand the algebraic method. Please work with me through that and I'm gonna I just said, okay I'll do it and then I started to work with it more and I realized how good it is So first step in the algebraic method is to assign variables for each coefficient location So let's go back here We already know the answer for number 11 four five two make a mental note of that. I'll put it up here So the algebraic message should provide the same exact Uh Coefficients as its answer. So that's what we hope And what we're going to do is we're going to erase these and we're going to do what it says. We're going to put Variables I can start with the alphabet lowercase letters a b and c in the places where the coefficients should go At step one. Okay, let's go read what step two is step two for each element Determined relationship between variables required for balance Emphasis on each element here. We're doing each Element at a time. We're going to make a separate equation for each element Here's an example of those equations for this other reaction. Just so you can see an example There's one for aluminum And there's one for oxygen. Okay, they're separate and so we're gonna that's how we're going to do this But let's do it for our example, okay So for phosphorus, I'm going to have a equation and for oxygen. I'm going to have an equation, okay for phosphorus What do I see? I see that there's one a on the reactant side and I see that there's two A phosphorus on the product side. So how does that factor into the equation? Let's look at this example to see if I have one aluminum on the A reactant side I put a single a there And if I see next to the c over here a two I put a two next to that c Okay, it's like you're multiplying the c by the number of atoms of each element you have So when I apply that over here a very similar reaction with different elements I have a a single a because there's one p next to it. That's going to equal two C's The reason this works so well is because I need two times c For a in order for there to be a balance between the phosphorus atoms I need a to be double the size of c. Again, we don't know what they are We just know that a needs to be double whatever c is Then we go to oxygen, okay Oxygen I have two on the reactant side so two b And that has to equal on the other side Five c Okay Five c So Now that we set up these equations, how do we solve them? Normally what's done is you assign a value to one of the variables to get started Now you have to be careful when you do that. Let's explain why The smallest positive integer possible is A little funky. We already know what the answer is going to be And so I'm going to assume we didn't and we set one of the variables to equal one And what we're going to do after we do that is we're going to solve for the other ones using algebraic substitution Okay, this is not super complicated substitution Like you might have learned in your algebra 2 or pre calculus class if you're currently in it It's actually pretty simple but You must You must use that to solve if you get fractional answers You must deal with those. So let's see if we get fractional answers after we do step three Just assign a value to one of the variables Usually what's done is you assign one to the variable with the smallest coefficient In your two equations my two equations have a with a coefficient of one And so what I do is I assign a equal one and I see what happens Okay If a equals one then the p equation we can solve for c because now I know what a is One equals two c Okay So If we have that we could solve for c by dividing by two Hopefully if you're watching this you have at least a small At least a first full year of algebra under your belt And we're going to divide by two and get c equals one half. Okay. We got a fraction We're gonna have to deal with that, but let's keep going And see what happens if we continue to use these Now we need to find b. That's the last of the three variables What do we have? We have five c equals two b. Okay So we're gonna plug one half in for c and then find a b two b Oops forgot to be Two b equals five Times one half for c. I plugged in c. Uh, so two b equals Five halves five times one half is five halves if I divide by two That's the same as multiplying by one half I'm gonna cancel I'm gonna get b equals five fourths. Oh dear. I've got lots of fractions now. This is no good. Okay When I look at I'm gonna put these up here next to each other all of my answers in fraction form What I need to do is multiply by the Largest denominator I need to multiply all my answers by the largest denominator. I have the largest denominator. I have is Five over it's four. Okay. And so each one of these answers needs to be multiplied by four Okay Times four. What do you get? You get a equals four b equals five it's because it cancels and c What is one half times four everybody? It's two c equals two And when you get these nice beautiful coefficients like this you are assured You don't have to check. They are the smallest possible ratio Okay, so let's take these numbers and put them in here four five two look at that They are a match. Okay. This is the algebraic method Once you get your hands and you get the hang of it Uh You might prefer it. I'm starting to it's it's grown on me over the last couple uh years So I would say it's pretty nice Let's try it with the next one, you know Let's do a couple. Uh the next one number 12. We know the answer two two two and one Okay, those are small nice small ratio of the numbers. Let's do the algebraic method a little faster Okay, so moving over here All right, we'll do the work in this area I've got na a h2o b naohc h2d so we've got Four variables that we're assigning to each of these. Okay, great. Oh my gosh four variables is gonna be hard You're probably gonna have three equations, but that's okay. Okay, you're gonna have the number of equations that you do elements So let's start making our equations sodium I've got one sodium over here. So that's one a And what is it equal on the other side? It equals c because I have one sodium on the other side So it's coefficient should be c it should be one To one ratio with a Okay, great Let's go ahead and do h next Okay, h was that one that was a little complicated where to see how that complication shows up in this here Situation this equation. So I've got two hydrogens next to b. So that's 2b And that's gonna equal I've got a c next to a single h so c But I got more h is over here the plus sign in the middle. So it's gonna be plus 2d Okay, so this equation has three variables in it. I'm not the funnest Um, I see you on the chat. Hey, welcome. Uh, oh, sorry not in the chat I see your viewers out there. Uh, if you're new and you're just joining you can interrupt my Uh, regularly scheduled programming here with any questions you have just ask them in the chat All right, um back to what we're doing here my goodness, I have two equations and Four very possessed not going to be enough even with substitution. Thankfully I've got another Nice Element to deal with Okay, and that's oxygen. So oxygen. I see one oxygen next to h2o next to b. So b I see another oxygen over here And uh, that's a single one next to c. So b equals c is what we're dealing with right now And look at that. I got a equal c and b equal c a lot of things kind of equal each other And by transitive property a and b and c all if to equal the same thing I don't know if you remember that three of those coefficients were actually the same Okay, and what did I teach you choose step three for one of the variables to be Uh, this, you know a nice small whole number. So I'm not going to cheat We don't know what the other answer was we're pretending we're doing it only this way And we're going to choose a equals one again, okay Again, this is the difficult arbitrary step you get to take it will fix itself if there are fractions. I'll show you Okay, so if a equals one you automatically know that c equals one because a equals c in the first equation Down here if c equals one then you know b has to equal one because b equals c in the third equation So I have a one one one for a b and c. Okay, that was easy, right? Let's try and find d in the second equation plugging b and c in So I've got a two times one for this part To be equals and the c is one Plus a two d then d is unknown Let's see what happens when I solve for d We're going to subtract this one Two times one is two minus one is one equals a two d. Yes, okay I think I see where this is going. Maybe you do two divide by two And you get d equals one half Oh d equals one half that's uh not good But we know what to do final step Just refresh the steps solve for others using algebraic substitution any fractional answers required you To multiply all answers by the least common denominator to remove all fractions That's what we're going to do now The two right here is the only denominator here. So we're going to multiply by two. Okay So everything gets multiplied by two that means we're going to have a equals two b equals two c equals two And d will equal one because it was one half. Ah getting ahead of myself here d equals one Okay, so there you have it We come over here a b c d. So a is two b is two c is two and d is one A is two b is two c is two and d is one. We don't write ones. Look at that. It's exactly What we had the other way So you decide guys you decide if you like the traditional method better Or the algebraic method better I'm telling you you will once you get to college and you have to do this at a college level chemistry course You're probably going to prefer and you're going to learn the algebraic method and your teacher's probably going to tell you It's better and for larger equations. It's definitely better. It's just not as intuitive. So that's the that's the tradeoff Okay, so that's balancing chemical reactions If you'd like me to do more just write in the chat that you'd like me to do more. I'm happy to do it All right now second topic for tonight In our hour together is geometry practice Circle properties. Okay circle properties So if you were given a bunch of problems like this and you were asked to do that What would you do like how would you tackle these things? So we're moving and we're totally switching gears here I mean you got to think geometry. Well, there's lots of interesting Shapes and there's lots of interesting figures in Circles when it comes to geometry. Okay, let's look at what those are Okay, so some of these you probably like this is easy The center of a circle is defined as a point that's equidistant from all points on the circle. Okay That distance between the points in the circle and the center Is called the radius. Okay, so all of these points on the circle Are one radius away from the center. Okay by definition Then we have a cord, okay a cord is not a diameter which is Two radii that's one way to think of it a cord is just a line segment that connects two points on the circle A line segment that connects two points on the circle It turns out the diameter is the largest possible cord and it's the only cord that goes through the center Okay, so that's a cord Then you have this thing here called an arc. Okay An arc is a section of the circle now. There's two measurements for arcs. There's a degree measurement And there's a length measurement We're not going to get into arc length today, but there's a formula to find it However in geometry when you're more reasonably Interested in is the arc degree. Okay arcs are measured by the angle they make With the center Okay, what I mean by that is how much do I have to rotate at the center to go from one end of the arc to the other? That's called This angle here the central angle The central angle that's my symbol for angles. Okay So those are the same the degree of the arc and the central angle that subtends it's called Uh is the um That is the same. Okay the same measurement So it's getting a little busy. Let's erase that Okay, then we have these lines that go through the circle They're they're full lines. They don't end on either side. Those are called secant lines. Okay, secant lines Now one section One line sorry that uh connects at only one spot because secant lines connect at two spots It's called a tangent line tangent lines intersect a circle at one point Okay, that point is called guess the point of tangency. Okay Uh something to note is that the point of tangency in the tangent line And that radius that connects that point of tangency to the center of the circle It looks you know looks and geometry you never go by looks you go by proofs But it looks like that's a perpendicular situation. Is that a 90 degree angle? Well, let's find out. We're not going to do proofs We're just going to look at the theorems. Okay Uh one thing I didn't discuss are what are called inscribed angles. So let's come back up here And let's draw an inscribed angle. Okay An inscribed angle would look like this It's an angle whose vertex is on the circle. Okay the very important angles Okay, we talked about central angles inscribed angles do not have their vertex at the center They have their vertex on the edge of the circle basically just on the circle Okay, uh a very important truth about inscribed angles versus central angles Here's a central angle looking at the same arc here The measure of this angle will be half the measure of this angle every time A central angle and an inscribed angle that subtend the same arc Will be at a two to one ratio central angle will be double the size of the inscribed angle. Okay So that's a very important truth. I'm going to leave that there Now if we have a an inscribed angle that has an arc That's literally a semi circle meaning half the circle That arc length is 180 degrees And so this the inscribed angle here has to be half of 180 which is 90. Okay So if you have a triangle inside a circle and one side is the diameter Then the opposite angle is a right angle. It's a right triangle. Okay But only if one of the sides is literally the diameter of the circle has to go through the center Okay over here. We have a really cool And very interesting Um It's quadrilateral all of its vertices touch the circle This type of quadrilateral has a lot of interesting properties that has a lot of constraints on its shape and it's called a cyclic Quadrilateral cyclic quadrilaterals We can do a lot of interesting proofs on what has to be true about them But one in particular is that opposite angles are supplementary. Do you guys remember what supplementary is? It means that they add up to 180 degrees So the two green angles here these add up to 180 degrees the two yellow angles those add up to 180 degrees Every time Not consecutive angles next to each other. No relationship, but opposite angles are supplementary. Okay, so that's one to know Then we have This guy we already talked about this inscribed angle versus in a central angle Okay, half as big as the angle at the center. They don't even use Angle at their circumference. That's stupid. They're called inscribed angles. Everybody calls them that silly book all right next Angles in the same segment. What does this means is angles that subtend the same arc Inscribed angles specifically. So it's like I took this vertex and moved it over here Is it the same angle measurement? Yes, I could move it again and put it way over here Moving it down the circle like this It would still be the same measurement. It's pretty interesting feature But it has to be based on the other things and it's happening with these two angles here looking at the other direction it's happening there too so That is a very important thing about inscribed angles very important conjecture Let's look At this one. This one isn't used very much okay But it is rather interesting if we have a tangent Okay And we have a triangle inside that's inscribed If one of the vertices of the triangle is at the point of tangency We've got some congruent angles here based on these colors This exterior angle will be congruent to the opposite interior angle. This exterior angle will be congruent to the opposite interior angle Really really really interesting, but it is true If this is a tangent line only if it's a tangent line does this happen? Okay, we're at that point of tangency earlier here. We have another one It's a tangent line And this is the radius at the point of tangency Notice how we were correct that that is a right angle Okay and then lastly If we have two tangent lines which start at the same point Basically tangent rays and they keep going on Tangents which meet at the same point are of equal lengths Okay, so this tangent The section of the tangent to the point of tangency is the same length as this one Now the really interesting thing that happens here is if you draw the points of tangency and the radii that connect them to the center This one and this one Well, these are have to be congruent because they're both radii. You have this quadrilateral here, which is called a kite a kite has Two consecutive pairs of congruent sides meaning for those of you are really familiar congruent means Same measure, okay So those two radii obviously have the same measure because that's the same circle and those the conjecture says that these two um these two line segments from this Concurrent point that they both emanate from And to the points of tangency those are also congruent. Okay There are many more but that's how we can get started. So if you see something like this And you're asked, uh, what do I do I have to find a? Well, look I have a central angle. That's 130 And it subtends this arc And then I have this inscribed angle a that subtends the same arc, but it's inscribed So what does that mean? That means a will be half 130. It'll be 130 divided by two Which is 65 degrees Okay So that's an example of how you use these conjectures to solve problems that you may have Next be How do we find a b? Hey, look at this. This is that's a certain type of triangle, isn't it? It's a triangle with one of its sides going through the center. So this side here is a diameter I taught you that the angle opposite the diameter must be 90 okay So if that's 90 and this angle is 60 You have to use another a priori truth One that you already know and that is that the sum of all angles In euclidean space meaning a space that's a plane. That's flat. No curvature that Uh, all those angles in well, not just this triangle any triangle all the interior angles always add up to You guessed it 180 degrees So if I want to find b, I'll just take 180 degrees And subtract every other angle And that gets you of course 180 minus 90, which is 90 minus 60, which is 30. So b is 30 degrees Alternatively if this is a right triangle, you know That uh, the two non-right angles are complementary meaning they add up to 90 And so you could have just done 90 minus 60 Which of course is the same thing. It gets you 30 degrees also voila So we're using these conjectures to solve Problems. Ah, look at this number three cyclic quadrilateral. Oh, yes A nice cyclic quadrilateral What do we got here? We've got a 95 degree angle right here And We've got this 120 degree arc right here Sometimes this uh, depending on what book you use this book this page is from a book called discovering geometry Which is a interesting book not a whole lot of lessons more like a whole lot of discovering, you know, I'm not a huge fan of that in that textbook, but You know without teachers. Yes, you should be discovering knowledge too. So that's a arc like of 120 degrees Okay, it wants me to find this arc over here. See this blue one. Totally different arc So when you see something like this, we're going to use what we know And if it's not enough, we might have to construct something or at least sketch or something If this is 95 degrees, I actually know this angle because they're opposites okay They have to be Supplementary if one of angle in a supplemental pair is 95 the other one is 180 minus 95, which is 85 So this is 85 degrees Okay, now let me explain why this is useful Because 85 and 95 they're part of a cyclic quadrilateral, but they're also inscribed angles Okay, what does this mean? This means that this 85 degree angle Subtends this arc So that arc has to be double. Let me try and do that better. This arc has to be double The degree measurement of this 85 degree angle. What's double 85? 170 So that entire arc is 170. Now stop What Is missing From knowing the whole circle. We already know a whole circle is 360 degrees The only part of this circle missing is C So if I want to find C, I can just do a 360 Degrees minus 170 degrees minus 120 degrees And I'll have C right off the bat 360 and then we've got so 170 and 120 that's 290 360 minus 290 is what we're doing right now. That's 70 Okay 70 degrees, that's what C equals So we used a different conjecture set for that And the a priori truth one that we assumed For this one was that The total arc length the arc degree of an entire circle arc is 360 All right, let's look at this one. Oh interesting This one will involve what I've said might happen last time Use what you know and if it's not enough you might have to construct or sketch something Well, that's okay. We've got h What is h equal h is an arc. It's this little blue arc. I'll try and highlight that over Just there to that dot Anybody remember what this Dot is called It's called a point of tangency. We have a tangent line here. This guy is a tangent line. Okay Great We know what to do with that Even though there's nothing drawn. So this is a 40 degree angle here That's outside the circle so we can't relate it to h by itself And so I have nothing else to go off of Okay, that's already drawn or constructed here. So what I'm going to do I saw that point of tangency. I know that the radius that connects that to the center Is involved in a conjecture and that conjecture is That the tangent and this radius Are 90 degrees or perpendicular. Okay So perfect that gives me a triangle number one And an angle in that triangle number two So now I've got a right triangle that I've made with a 90 degree angle and a 40 degree angle So You can find this angle now Let's use the complementary angles conjecture in the right triangle the two Uh acute angles in a right triangle add up to 90 So one of them is 40 So 90 minus 40 That's going to give me the other one. It's going to give me this big ol 50 So this angle right here is 50. I don't think you're done. I mean you did a lot But that's central angle. Is it the same as h? Which is the degree of that arc h? Yes, we talked about that before central angle conjecture The central angle subtends an arc and the arc will be the same degree measure as the angle So this right here H equals 50 degrees Oh, yes, so cool So cool so cool All right As you can see I'm using a stream yard. I'm using a free version And so I haven't seen anything come to the chat. I just want to make sure that there's nothing actually in the chat I'm going to do is I'm going to go on to my hoon live stream And I'm going to make sure That I'm going to do is I'm going to go on to my hoon live stream And I'm going to make sure that I'm going to do is I'm going to go on to my hoon live stream Okay There was quite a delay there as you saw that was going to cause a feedback loop But uh, I'm not seeing actually any posts in the chat at all including mine So what I'm going to do right now is I'm going to go ahead and test Because that was kind of weird Nothing was coming through And so I'm going to go back on my live stream and check and make sure that that Chat was there. Otherwise the stream yard you're gone man. If this chat doesn't work You're out Okay, let's see Let's see Hmm nothing there Nothing was there including my test. I've written three chats and none of them are there So I'm going to try the I'll look at the playback and see if that's what's causing it But um, I got to have my chats Okay so All right, so that about concludes what I wanted to do today. I'm going to leave these up there. We'll do some more next time but um Hope if you're watching this in post that you find it helpful I'm glad there were a couple of you hanging out with me on the chat itself So, um, let me know if the chat is showing up in the comments if you watch this My posts there are not I didn't see them when I logged on it could just be a glitch with stream yard Which is not a good one So, uh, let me know in the comments if you saw any Chat Chats from me or anybody else. Thank you all for watching for all you made it to this point. I hope you learned something I hope you learned something And I'm going to stop screen sharing so you can see my beautiful face again So next time I'm going to pop on here is monday at eight And eight p.m. Pacific time And if you guys are really like, oh man, I missed it. Let's do some more. I can do a weekend one I'm willing to help you guys if you're willing to help me and this works out I want to be someone who can really give you guys Assistance on the fly when you need it get regular times up here that you guys have come on and get your questions answered So I'll check the email one more time before I leave make sure no one's giving me something over email ha academic solutions at gmail.com feel free to go on there And check the spam folder just to be sure nothing. Okay Yeah, ha academic solutions at gmail.com if you want some help If you want something and it's a little bit longer harder to explain just write that just write me an email And we'll get to it. Okay When I said before is when a lot of people show up start asking in the check. It's pretty big We're going to do a super chat Um preferences, so I'll go for those who want to support my work that way Uh a priority prioritize them. All right guys 50 minutes long. I think we're going to stop here Thanks you guys so much for watching. This is falconator signing out