Circle Theorem Questions Lecture Notes

Oh Oh [Music] [Music] right in this video a selection of sirs was their own questions number one this episode slightly differently I'm gonna run through quite a lot of questions here so what I would suggest is obviously having a go make sure you've watched the previous video on all the circle theorem see if you can apply any of that knowledge see how if I can get with each question and then obviously watch me go through the go through like the explanation of all of these different questions now I'm gonna run through about five questions of just sort of what I would call slightly easier questions and then I'm gonna have a look at three specific or relevant exam questions towards the end as well so we're gonna get kick start over this question but obviously if you want to try and draw this out and make some notes and have a go Eunice obviously pause the video at the start of each question and then run through the example with me so this question here that and there's normally quite a lot of language I've tried to reduce the language as much as possible but it says be OD is the diameter of the circle so there's obviously this line just here beta o to D is the diameter I says BC and AC are tangents to the circle and we want to work out the size of angle D oh a Saudi okay let's have a look that is here so we're trying to work out this one if I label that with an X and it says you must give reasons now for these questions I'm not going to be writing down the reasons but we will sort of speak of all the reasons as we go along obviously the bits that you do need to make sure that you're writing down within an exam okay so have a look at this question now let's see where we can go with it now it says Bo D is a diameter and normally when there's a diameter given we're looking for a right angle somewhere but this is actually not the case in this question which is quite interesting that is actually more just telling us so that's a straight line I think because obviously there's no triangle being made via that diameter within that semicircle so I think that's kind of giving us a hint there that there's a straight line going on we're gonna have to use that at some point but if you ever look we have got tangents involved and they are meeting the radius so we've got some 90-degree angles here that we can draw on the straightaway and obviously see if that helps it's actually moving forward now just looking at that now we've got a triangle here I'm just gonna highlight that we've got a triangle here where we've got one miss so we can work out that other angle in the Triangle and that's what these sort of circle theorem questions are all about so just forget about what you're looking for for a minute and just see what you can find so actually looking at that question there if we do 180 degrees take away the 19 to 30 for obviously starting that off we could do 90 plus 34 which gives us 124 and take that away from hundred and eighty so hundred eighty take 124 that's 524 okay take 124 leaves us with 56 degrees so that's gonna be 56 just there there we go and actually what we've got is two tangents we can meet and equal in there as well so these are going to be at equal lengths which actually means that that length o to see splits that angle there in half at the end so this is also 34 and this is also 56 there we go so actually looking at right now with an and if I highlight that we've got three angles on a straight line we've got the X we've got the 56 and we've got the other 56 making a straight line there so if we have to give those 250 sixes so 56 plus 56 that gives us a total of 112 and if we take that away from 180 for the straight line so 180 minus hundred and twelve gives us our final answer here what is it 68 degrees so it goes to 68 degrees would be our final answer obviously there's a couple of reasons that we would have given along along the route there that I've sort of mentioned as we were going we have tangent to a radius at 90 degrees obviously giving us our first right angles we've got two tangents meeting equal lengths and obviously forms that little by so angle bisector there that would give us 34 and 34 and let me hand goes on a straight line to finish it off there we go right all those reasons down along with it but I said look at our next question okay so this question says work out the size of angle AEC and you must give your reasons now ADC is this angle just here and all right that will put Alex in there now what we've got if we sort of ignore that line CD and ad is we've got that sort of Arrowhead of all this these if we've got this angle at the centre here made via ANC and we've also got this angle at the circumference made over there so I've got a little Arrowhead sort of circle theorem going on and actually you can also see that there's a quadrilateral formed inside this Glantz we've got a cyclic quadrilateral going on as well now first things first we can just find this angle here cuz we've got 168 so we can have that for this angle and that would be 84 degrees okay and that is obviously you've got your angles at the center angles at the circumference let's get rid of that okay what was that 18 let's just get rid of all of that and rewrite it without that arrow this highlighter no egos that'd be 84 degrees there we go angles at the center being double angles that are circumference there now we can have a look at this in terms of a cyclic quadrilateral because these two are opposite angles within that cyclic quadrilateral so we can actually just obviously take that away 480 it's these opposite angles are supplementary they add up to 180 so takeaway 84 and that leaves us with 96 and there we go our reason for that would be angles in a cyclic quadrilateral add up 280 and there's our final answer there are different ways of doing this and to be honest with a lot of these questions there are more than one way of actually solving these problems but there's just one way and obviously you might find your own way of doing that that's absolutely fine but I'm not going to talk through all the different ways I'm just gonna go through the first way that sort of pops into my head is that as I'm doing them okay so there's one question let's have a look our next one okay so this question says AP and BP are tangents to the circle work up with sides of angle X and you must keep your reasons so straight away obviously it's mentioned that it's tangents so we know there's some right angle so we've got a right angle here which is made up of those two different angles and we've got a right angle down here which is made up of that angle X and the other little angle within this triangle also again got tangents meet and equal length so we've got this sort of idea here which means that obviously that triangle there APB is an isosceles triangle and if it is an isosceles triangle then we can work out the two base angles so for switch colors we can work out this angle here and this angle here just by using isosceles triangle rules and obviously and to be honest these isosceles triangles are going to be very very relevant in these circle theorems we're going to see a lots of isosceles triangles throughout different questions so I would actually always be looking for isosceles triangles and I think we're gonna have more than one in this question but we'll see where we get to so if we take away 86 180 180 take away the 86 that leaves us with 94 degrees Mele that's gonna be split between two base angles in the isosceles so if we divide that by two so 94 divided by 2 gives us 47 so we've got 47 degrees that's gonna go in both our base angles here 47 or 47 I always make sure you just label that all over the diagram and now we should actually be able to find the size of angle X straight away there because these two angles that are made up we've got the 47 and the X those two are gonna add up to 90 obviously because there's a tangent to the radius there so to finish that off friend of X here we're gonna Stu 90 takeaway 47 and that leaves us with 43 degrees there we go 43 all right there we go that's our final answer for that one we could actually be left under look at different angles I'm just I think if we could have done that in a slightly different way because obviously something that I noticed straight away was that we had another isosceles triangle here this one here because that length o2a and owed to be are both the radius so that was also an isosceles so there's probably another way of actually going about solving this berthing that's quite a nice easy way forming that isosceles in the first triangle and they're just taking that base angle there away from the 90s for our reasons were tangent to radius meeting at 90 and also we have two tangents meeting at equal length as well as well as isosceles base angles being equal there we go there's an answer to that one some look our next question okay so this question says ABC is a tangent to the circle B II is a diameter of the circle that's just highlight that so BT is a diameter and work up the size of angle abd and DB so we've got two questions within this one and it says you must give reasons again so let's start with abd where's that abd is just here and I should be nice and easy for us to do there because that's where that tangent meets the radius so if the tangent meets the radius there the total angle here is 90 so we can do 90 take away that 35 that's going to tell us what's left over so 90 take away 35 is 55 degrees and there we go there's our final answer for that first part so 55 degrees that's just how it like that there we go now the next one is asking us to find out the size of de Bie so de Bie is this angle over here so we can find out the size of that as well there's a couple of ways that we can go about this actually um let's have a look at this two ways we could get it in one step or we could get it into no what we have got is we've got this semicircle angle so we've got this angle here being 90 because angles in a semicircle equal 90 at the circumference so we can do that and we can add them together and take it away from hundred and eighty and it will tell us the size of this angle here but actually we've also got our alternate segment theorem that we could use in this one as well if you have a look as we walk through this angle and go up the line okay it was here so it's equal so that we've got our alternate angles I've got rid of my over there we've got alternate segment theorem saying that that um angle there is also 55 degrees as well okay so we could use two different rules here so we can use angles and a semicircle adding up to 90 or we can have a look at our alternate segment theorem we're all there to say that that one's 55 so a couple of different ways to approach that that one's a little bit quicker there and it requires one step rather than to the both ways absolutely fine so I've looked under the question okay so this one looks a little bit more complicated we've got a TB is a tangent to this circle calculate the size of angle otq and you must keep your reasons now otq is this angle just here so it's this little angle in between now we've got these lines on this triangle here something it's it's an isosceles that they must be the same length and we know we've got a tangent so we've got a 90-degree angle just here and if we've got anything else nothing else that stands out to me at the moment so let's just have a look what we can find so in between the 58 and the radius there we can obviously find that out so this angle here so we can do 90 takeaway 58 as obviously tangent metre radius at 90 so 90 minus 58 and what does that leave us with 32 degrees there we go so that angle there is 32 and we can draw that in there we go now there's nothing really that massively stands out that you can actually go for on this next movement now obviously there is a couple there are a couple of ways of doing this they're just spotted to do little things we could draw another line in here and perhaps making little isosceles triangle but I'm just thinking now that any of the circle theorems that we can look for we haven't got a diameter we haven't got any angles in the same segment we've not got a little our little angles at the centre or angles of circumference circle theorem but we have got like a cyclic triangle here for a meeting a tangent so we can actually apply also in a signal theorem again so actually if I go through this angle this 58 and I go up the length of the triangle it makes this angle here so that there is going to be equal to the 58 so this angle here is going to be 58 again there's our alternate signal theorem now looking at that obviously we've got an isosceles triangle and we know base angles and the most awfully sorry cool so if we work out the size of these two base angles here we can do 180 take away 58 and that would leave us with a hundred and twenty-two degrees there we go 122 now that is needing to be split in half obviously cuz it's two base angles so if we divide that by two that would give us 61 degrees so both the base angles in this isosceles this are 61 no I'm just about label that one there is 61 there we go and now looking at this value of x look we know that part of that angle is 32 and the following all there is 61 so a little value of X must be what makes 32 up to 61 so we can finish this off by doing 61 to take away taking away 32 and what does that leave us with that is 429 degrees there we go so 29 degrees would be our final answer that one okay a couple of different ways that we could have approached that actually I think it could have also done a little bit of an isosceles triangle here had two base angles of 32 then found the angle at the center here and half the angle at the circumference and worked your way through it that way but there's always a couple of ways to solve these questions I think that way obviously just involved a little bit less drawing and having to sort of think outside the box but there we go there's a way of doing that question as well and someone go another one right okay now after this question we're gonna have a look at three exams style questions as well but this is our last little question on some of these what I would say easier to type the questions but it says PT is a tangent to the circle and it says POS is a straight line again so thinking about using our straight line rules as well in this question so work out the size of the angle marked X and you must give a reason for now at the moment we've got a sir I'm gonna highlight it we've got this right angled triangle here and we know it's right and cooled because we've got a tangent meeting the radius sir so straight away I'm going to draw that right angle him and then we can obviously work out this missing angle in the Triangle here I'm just thinking about what I can work out without even looking at the X for the moment now that is just gonna be angles in angles no triangles we've got the 1932 at the moment or we could just do 90 take away 32 is this 90 degrees left once we get rid of that right angle so 90 take away 30 tea leaves with 58 degrees there we go so there's our first bit of working out we've got 58 degrees just there now we've got angles on a straight line so the angle next to that we can work out straight away so it can get this one here and that's gonna be a hundred and eighty takeaway 58 I'm Stan goes on a straight line equaling a hundred a/c writing down those reasons so 180 minus 58 leaves me with let's have a look 122 I believe double-check that I didn't I - yep absolutely fine now this is again one of these isosceles triangles slip coming in because we've got this triangle here made by these two loads that are the radius so that forms a little isosceles and again we know base angles and rice off so these are equal so we can find the size of X along with this one when we find out those base angles so if we do 180 take away the hundred and twenty-two obviously again leaves us with 58 it was just the opposite way around of what we did a second ago and then we can split that into so dividing that by to these is 429 degrees for that one there we go sweet 29 degrees for this question as well there we are so that's our answer there 29 degrees finishing that off and just obviously let's just have a look at what reasons we use there so the first reason that we had was angles made by the radius in the tangent being 90 then we had angles in a triangle the male angles on a straight line and then we have base angles in our sauce leaves being equal so obviously not a huge amount of circle theorem is actually going on in a lot of these questions it's more just two one circles there and possibly two and then a lot of triangles and straight lines and why sort of normal angles that were used to using all right let's look at some example questions right okay so a lot more language in these questions here's I'm not shortened anything down just here it is as it but as it would on exam so a B and D are points on the circumference of a circle Center zero e DC is a tangent B DC is 57 that's not given to us on the diagram so let's label that B to D to C there we go that's that one there and that's 57 degrees find the size of angle AOB so it's over like a OB is that on there now again let's not worry about sort of how we get to our angle of the members have a think about we can find I've spotted two things that we can find so one thing I'm thinking about is obviously where the radius meets the tangent is 90 so I'm straight away thinking we can find this one here so if we do 90 take away 57 that's going to give us that one so 90 minus 57 and again our reason here is angles where the tangent meets the radius is 90 so let's take that away and we are left with 33 degrees and there we go we can label that in the diagram let's get rid of that and label that in there 33 degrees there we go that's just in there now the next thing I've also spotted is that this angle here that we've started with given to us in the question is the angle where that tangent meets that triangle and that triangle there let's have a look does meet on all points on the circumference of the circle so we can do our alternate segment theorem we can go through the angle up the line and there we go it makes this angle here so if we get rid of that we know that this angle here has to be equal to about 57 due to our alternate segment theorem so angles in alternate segment theorems are equal sorry angles alternate segments are equal there we go so that must be 57 degrees so we've got a couple of angles there let's look at what we can find now moving from this so we've got I'm gonna highlight this we've got that this triangle here as two lengths are the radius so therefore they're equal so actually that has to be an isosceles triangle we do actually also have that on the other side as well so just high like this we've got the radius there the radius there so I'm actually on earth an isosceles triangle here so we could probably find our way through the triangle fire that route but actually us we've now got this 57 and we know the base angles and my sources are the same this here must also be 57 so actually we've just got wine left in an isosceles triangle and I can always draw that out to the side if you'd like you can always redraw some of these triangle 5757 and we're trying to find that top angle there so if we add them together that's 114 degrees and we want to take that away from hundred eighty so 180 take one hundred and fourteen leaves us with 66 degrees and that's our final answer for this question so that's there is going to be 66 degrees there we go 66 degrees final answer but again we could have taken some different routes here just have a quick look at this one so if it didn't take that route so if we get rid of a little bit of this let's get rid of that and forget we found those so we're trying to find this one come and get there with another another route now we go now I hope that this one was 33 and I said obviously this is another isosceles so we've got a base angle going on down there so that's also 33 so we could find the top angle in that triangle instead we could add together the two 33 so that would give us 66 and take that away from hundred eighty and that would leave us with that's if I think one one ball there we go and that means that if this one's one one four now we've got angles on a straight line so angles on a straight line you could do a hundred eighty take away the hundred and fourteen which obviously again leaves us without 66 there so you can actually do that with just the one circle theorem we didn't even have to use the alternate segment theorem you just do the tangent meets the radius at 90 and then base angles into isosceles and then angles on a straight line okay so that's prob maybe an eaglet may be a bit of an easier route for some people it may be struggle with using the alternate segment theorem you could have actually done that without it there we go let's have a look another question all right okay so this question says work out the size of angle C ad so C ad is this one here and we've got a few angles are given to us with what angle ad e equals 54 so ad equals 54 this one here there we go we've got angle ABC equals hundred 14 so A to B to C that's that one there one one four there we go and we need to try and find out this angle here now let's have a think about will become fine and we haven't got any radius meeting the tangent so we can't do any 90 degrees but we have got a triangle meeting that tangent so we can apply the alternate segment through we have a cyclic quadrilateral within there so I think we just thinking about those two circle theorems the alternate signal theorem and the cyclic quadrilateral obviously those opposite angles being equals 180 so that's one of the easiest ones find straight where and go opposite this hundred and fourteen I can do 180 take away the hundred and fourteen and that leaves us with 66 degrees therefore that opposite angle so that there is 66 let's get rid of this arrow there we go and we can label that 66 degrees now whether we'll need it or not we can find obviously this missing angle now on the straight line so I can have together the 54 and 66 and if we have those together let's write that down 54 plus 66 is 120 so I'm a missing angle down the straight line is gonna be 60 degrees there goes that 60 degrees actually despite that's gonna be quite a key one and in fact let's just write the working out down 180 miles hundred 20 and that's 60 degrees there we go so actually we can get to our finance here straight away we can use the officer in a segment theorem so we can go through this angle of 60 up the line and there we go it meets the angle there so that's also gonna be 60 degrees now we asked that's our final answer actually 60 degrees okay so what reasons did we use there so we started off with opposite angles in a cyclic quadrilateral being supplementary or I think it's 180 then we had angles on a straight line and then we use the alternate segment theorem and to be honest that's probably I don't know what there is under the route just looking at the moment but that's probably the easiest route they're only a couple of steps don't you just making sure you write down those reasons all right so look at one more right okay now this is a great question actually because you actually don't even have to use any circle theorems within this one I mean you can do if you like maybe I'll run through this in two different ways because I think I'm believe you can get to here with maybe only just the one circle theorem for this one now it says work out the size of angle C AO now all the other angles given to us in the information are on the diagram so don't draw anything in but see AO is this angle here and that's just high like that and we do need to use one circle theorem here that's where the radius meets the tangent okay and we can see that Oh a is the radius it says this in there in there in the the writing for the question there but we've got 56 just over here so if we take that away from 90 so 9656 and that leaves us where's 34 degrees there we go that's gonna be 34 just here now once we've done that we can actually work our way around using lots of myself' least triangles here okay so if I have a look I've got an isosceles triangle just here where 34 is one of our base angles let's work that up to here 34 and then I can actually find this angle just here at the top of the isosceles there we go so you can always redraw this to the side if you want but there we go we've got 34 34 we're gonna work out that top angle so add those base together is 68 now if we take that away from hundred 80 that gives us a five leaves the top angle there what's that when we take that away he's at 112 there we go 112 degrees so that angle there is 112 now it's up to you can actually split this up into different isosceles triangles now I'll show you I mean I could actually just draw this little line in here and I'm actually just created three isosceles triangles we've obviously got the one that I've highlighted in yellow you got another isosceles triangle here and I can just repeat the process that I've just done all well not all but once you've done that you've also then got another isosceles triangle here and again you can just keep repeating the process for three isosceles triangles and you can get down to that final answer it's a little bit slower but there is a process that you can do with using very minimal circle theorems in this but the next one I've spotted which i think is probably the best way to go about doing this if you want to do them nice and quick is to go for this angle center here and then we've got up angle at the circumference so that angle at the circumference there is half the size of the wall at the center so if we do that / - 112 divided by 2 let's see what does that give us that gives us 56 degrees no so this one up here must be 56 degrees let's see if that actually helps us I'm not sure where we're actually going with this but have a look and now we've got to get down to that angle X there now there's a couple of ways that you could do this we have got I think I have it easier just to do the all those isosceles triangles actually we still quite a lot of work to be done so we've got this shape here there we go which is a quadrilateral isn't it's got four sides so it's a quadrilateral obviously angles in a quadrilateral add up to 360 so if we go about just finding this reflex angle 112 here so we could do 360 take away one hundred and twelve and that gives us two hundred and forty-eight there I guess that one there is two hundred and forty-eight and now we've got three of the four angles in the urn in the quadrilateral there so we can actually work this one out straight away now we can have AB those all together so 248 plus 56 plus 35 up days all of them see what's missing up to 360 so if we have those all at what do we get 14:19 carry the one four nine ten eleven twelve thirteen and three so three hundred and thirty nine degrees at the moment so if we take that away from 360 to finish that off 360 take away three three nine and that leaves us with 21 degrees there we go so our final answer there will be 21 degrees and there's just highlight that right there we go right obviously I did mention there was another way of doing that though so maybe we should have a look at ax I'm not too sure whether that was the most simple way of doing this or not was just a way that came to me as always going along but let's have a look let's get rid of this and have a little look at how he helps we could have done that so for do split that up as I said before I Steve got an isosceles triangle there so this is a base angle also 35 so we can work out the top angle in that triangle so thinking about this to the side let's have a look I've got a base angle of 35 and another one of 35 and I add up to 70 so 180 sake where 70 keeps us 110 for that top angle at 110 there we go now we can work out the angle at the top of this little right angle because those three angles around the point after add up to 360 so if we add those two together you've got 110 I've just got writing everywhere now 110 at 112 is 222 and if we take that away from 360 what does that leave us with a hundred and thirty-eight is it 138 there we go and then again we've got that final isosceles triangle there which if I draw out over here we've got a top angle er 138 so if we take that away from 180 and split it into let's see what we get so take it away from 180 is 42 and 42 divided by 2 gives us 21 in both of those base angles there we go so we get 21 there and obviously 21 for that value of x that we'd already found so different ways that you could do it just depends on which one you prefer really and how you actually spot it in the moment but I'll do it I'll be honest with a lot of these circle theorems you don't have to just spend a little bit of time just looking at them thinking what can you find where can you go unless without thinking actually just trying to get straight to that final answer because a lot of the time as you can see with this one I've pretty much found every single angle in sight before I actually got to that final answer so always just try and take that approach and just think about that as you're going through some of these questions okay but that's the end of the video there I'm gonna do the next video looking at algebra within circle theorems then I'm also going to pick out some of the hardest circles theorem questions and go over some of those but again the link for those will be in the description this was just a bit a bit of a brief overview overview just running through some of the exam style questions on this topic okay but hopefully that was helpful if it was useful please like please come and please subscribe and I'll see you for the next one [Music] [Music]

Oh Oh [Music] [Music] right in this video a selection of sirs was their own questions number one this episode slightly differently I&#39;m gonna run through quite a lot of questions here so what I would suggest is obviously having a go make sure you&#39;ve watched the previous video on all the circle theorem see if you can apply any of that knowledge see how if I can get with each question and then obviously watch me go through the go through like the explanation of all of these different questions now I&#39;m gonna run through about five questions of just sort of what I would call slightly easier questions and then I&#39;m gonna have a look at three specific or relevant exam questions towards the end as well so we&#39;re gonna get kick start over this question but obviously if you want to try and draw this out and make some notes and have a go Eunice obviously pause the video at the start of each question and then run through the example with me so this question here that and there&#39;s normally quite a lot of language I&#39;ve tried to reduce the language as much as possible but it says be OD is the diameter of the circle so there&#39;s obviously this line just here beta o to D is the diameter I says BC and AC are tangents to the circle and we want to work out the size of angle D oh a Saudi okay let&#39;s have a look that is here so we&#39;re trying to work out this one if I label that with an X and it says you must give reasons now for these questions I&#39;m not going to be writing down the reasons but we will sort of speak of all the reasons as we go along obviously the bits that you do need to make sure that you&#39;re writing down within an exam okay so have a look at this question now let&#39;s see where we can go with it now it says Bo D is a diameter and normally when there&#39;s a diameter given we&#39;re looking for a right angle somewhere but this is actually not the case in this question which is quite interesting that is actually more just telling us so that&#39;s a straight line I think because obviously there&#39;s no triangle being made via that diameter within that semicircle so I think that&#39;s kind of giving us a hint there that there&#39;s a straight line going on we&#39;re gonna have to use that at some point but if you ever look we have got tangents involved and they are meeting the radius so we&#39;ve got some 90-degree angles here that we can draw on the straightaway and obviously see if that helps it&#39;s actually moving forward now just looking at that now we&#39;ve got a triangle here I&#39;m just gonna highlight that we&#39;ve got a triangle here where we&#39;ve got one miss so we can work out that other angle in the Triangle and that&#39;s what these sort of circle theorem questions are all about so just forget about what you&#39;re looking for for a minute and just see what you can find so actually looking at that question there if we do 180 degrees take away the 19 to 30 for obviously starting that off we could do 90 plus 34 which gives us 124 and take that away from hundred and eighty so hundred eighty take 124 that&#39;s 524 okay take 124 leaves us with 56 degrees so that&#39;s gonna be 56 just there there we go and actually what we&#39;ve got is two tangents we can meet and equal in there as well so these are going to be at equal lengths which actually means that that length o to see splits that angle there in half at the end so this is also 34 and this is also 56 there we go so actually looking at right now with an and if I highlight that we&#39;ve got three angles on a straight line we&#39;ve got the X we&#39;ve got the 56 and we&#39;ve got the other 56 making a straight line there so if we have to give those 250 sixes so 56 plus 56 that gives us a total of 112 and if we take that away from 180 for the straight line so 180 minus hundred and twelve gives us our final answer here what is it 68 degrees so it goes to 68 degrees would be our final answer obviously there&#39;s a couple of reasons that we would have given along along the route there that I&#39;ve sort of mentioned as we were going we have tangent to a radius at 90 degrees obviously giving us our first right angles we&#39;ve got two tangents meeting equal lengths and obviously forms that little by so angle bisector there that would give us 34 and 34 and let me hand goes on a straight line to finish it off there we go right all those reasons down along with it but I said look at our next question okay so this question says work out the size of angle AEC and you must give your reasons now ADC is this angle just here and all right that will put Alex in there now what we&#39;ve got if we sort of ignore that line CD and ad is we&#39;ve got that sort of Arrowhead of all this these if we&#39;ve got this angle at the centre here made via ANC and we&#39;ve also got this angle at the circumference made over there so I&#39;ve got a little Arrowhead sort of circle theorem going on and actually you can also see that there&#39;s a quadrilateral formed inside this Glantz we&#39;ve got a cyclic quadrilateral going on as well now first things first we can just find this angle here cuz we&#39;ve got 168 so we can have that for this angle and that would be 84 degrees okay and that is obviously you&#39;ve got your angles at the center angles at the circumference let&#39;s get rid of that okay what was that 18 let&#39;s just get rid of all of that and rewrite it without that arrow this highlighter no egos that&#39;d be 84 degrees there we go angles at the center being double angles that are circumference there now we can have a look at this in terms of a cyclic quadrilateral because these two are opposite angles within that cyclic quadrilateral so we can actually just obviously take that away 480 it&#39;s these opposite angles are supplementary they add up to 180 so takeaway 84 and that leaves us with 96 and there we go our reason for that would be angles in a cyclic quadrilateral add up 280 and there&#39;s our final answer there are different ways of doing this and to be honest with a lot of these questions there are more than one way of actually solving these problems but there&#39;s just one way and obviously you might find your own way of doing that that&#39;s absolutely fine but I&#39;m not going to talk through all the different ways I&#39;m just gonna go through the first way that sort of pops into my head is that as I&#39;m doing them okay so there&#39;s one question let&#39;s have a look our next one okay so this question says AP and BP are tangents to the circle work up with sides of angle X and you must keep your reasons so straight away obviously it&#39;s mentioned that it&#39;s tangents so we know there&#39;s some right angle so we&#39;ve got a right angle here which is made up of those two different angles and we&#39;ve got a right angle down here which is made up of that angle X and the other little angle within this triangle also again got tangents meet and equal length so we&#39;ve got this sort of idea here which means that obviously that triangle there APB is an isosceles triangle and if it is an isosceles triangle then we can work out the two base angles so for switch colors we can work out this angle here and this angle here just by using isosceles triangle rules and obviously and to be honest these isosceles triangles are going to be very very relevant in these circle theorems we&#39;re going to see a lots of isosceles triangles throughout different questions so I would actually always be looking for isosceles triangles and I think we&#39;re gonna have more than one in this question but we&#39;ll see where we get to so if we take away 86 180 180 take away the 86 that leaves us with 94 degrees Mele that&#39;s gonna be split between two base angles in the isosceles so if we divide that by two so 94 divided by 2 gives us 47 so we&#39;ve got 47 degrees that&#39;s gonna go in both our base angles here 47 or 47 I always make sure you just label that all over the diagram and now we should actually be able to find the size of angle X straight away there because these two angles that are made up we&#39;ve got the 47 and the X those two are gonna add up to 90 obviously because there&#39;s a tangent to the radius there so to finish that off friend of X here we&#39;re gonna Stu 90 takeaway 47 and that leaves us with 43 degrees there we go 43 all right there we go that&#39;s our final answer for that one we could actually be left under look at different angles I&#39;m just I think if we could have done that in a slightly different way because obviously something that I noticed straight away was that we had another isosceles triangle here this one here because that length o2a and owed to be are both the radius so that was also an isosceles so there&#39;s probably another way of actually going about solving this berthing that&#39;s quite a nice easy way forming that isosceles in the first triangle and they&#39;re just taking that base angle there away from the 90s for our reasons were tangent to radius meeting at 90 and also we have two tangents meeting at equal length as well as well as isosceles base angles being equal there we go there&#39;s an answer to that one some look our next question okay so this question says ABC is a tangent to the circle B II is a diameter of the circle that&#39;s just highlight that so BT is a diameter and work up the size of angle abd and DB so we&#39;ve got two questions within this one and it says you must give reasons again so let&#39;s start with abd where&#39;s that abd is just here and I should be nice and easy for us to do there because that&#39;s where that tangent meets the radius so if the tangent meets the radius there the total angle here is 90 so we can do 90 take away that 35 that&#39;s going to tell us what&#39;s left over so 90 take away 35 is 55 degrees and there we go there&#39;s our final answer for that first part so 55 degrees that&#39;s just how it like that there we go now the next one is asking us to find out the size of de Bie so de Bie is this angle over here so we can find out the size of that as well there&#39;s a couple of ways that we can go about this actually um let&#39;s have a look at this two ways we could get it in one step or we could get it into no what we have got is we&#39;ve got this semicircle angle so we&#39;ve got this angle here being 90 because angles in a semicircle equal 90 at the circumference so we can do that and we can add them together and take it away from hundred and eighty and it will tell us the size of this angle here but actually we&#39;ve also got our alternate segment theorem that we could use in this one as well if you have a look as we walk through this angle and go up the line okay it was here so it&#39;s equal so that we&#39;ve got our alternate angles I&#39;ve got rid of my over there we&#39;ve got alternate segment theorem saying that that um angle there is also 55 degrees as well okay so we could use two different rules here so we can use angles and a semicircle adding up to 90 or we can have a look at our alternate segment theorem we&#39;re all there to say that that one&#39;s 55 so a couple of different ways to approach that that one&#39;s a little bit quicker there and it requires one step rather than to the both ways absolutely fine so I&#39;ve looked under the question okay so this one looks a little bit more complicated we&#39;ve got a TB is a tangent to this circle calculate the size of angle otq and you must keep your reasons now otq is this angle just here so it&#39;s this little angle in between now we&#39;ve got these lines on this triangle here something it&#39;s it&#39;s an isosceles that they must be the same length and we know we&#39;ve got a tangent so we&#39;ve got a 90-degree angle just here and if we&#39;ve got anything else nothing else that stands out to me at the moment so let&#39;s just have a look what we can find so in between the 58 and the radius there we can obviously find that out so this angle here so we can do 90 takeaway 58 as obviously tangent metre radius at 90 so 90 minus 58 and what does that leave us with 32 degrees there we go so that angle there is 32 and we can draw that in there we go now there&#39;s nothing really that massively stands out that you can actually go for on this next movement now obviously there is a couple there are a couple of ways of doing this they&#39;re just spotted to do little things we could draw another line in here and perhaps making little isosceles triangle but I&#39;m just thinking now that any of the circle theorems that we can look for we haven&#39;t got a diameter we haven&#39;t got any angles in the same segment we&#39;ve not got a little our little angles at the centre or angles of circumference circle theorem but we have got like a cyclic triangle here for a meeting a tangent so we can actually apply also in a signal theorem again so actually if I go through this angle this 58 and I go up the length of the triangle it makes this angle here so that there is going to be equal to the 58 so this angle here is going to be 58 again there&#39;s our alternate signal theorem now looking at that obviously we&#39;ve got an isosceles triangle and we know base angles and the most awfully sorry cool so if we work out the size of these two base angles here we can do 180 take away 58 and that would leave us with a hundred and twenty-two degrees there we go 122 now that is needing to be split in half obviously cuz it&#39;s two base angles so if we divide that by two that would give us 61 degrees so both the base angles in this isosceles this are 61 no I&#39;m just about label that one there is 61 there we go and now looking at this value of x look we know that part of that angle is 32 and the following all there is 61 so a little value of X must be what makes 32 up to 61 so we can finish this off by doing 61 to take away taking away 32 and what does that leave us with that is 429 degrees there we go so 29 degrees would be our final answer that one okay a couple of different ways that we could have approached that actually I think it could have also done a little bit of an isosceles triangle here had two base angles of 32 then found the angle at the center here and half the angle at the circumference and worked your way through it that way but there&#39;s always a couple of ways to solve these questions I think that way obviously just involved a little bit less drawing and having to sort of think outside the box but there we go there&#39;s a way of doing that question as well and someone go another one right okay now after this question we&#39;re gonna have a look at three exams style questions as well but this is our last little question on some of these what I would say easier to type the questions but it says PT is a tangent to the circle and it says POS is a straight line again so thinking about using our straight line rules as well in this question so work out the size of the angle marked X and you must give a reason for now at the moment we&#39;ve got a sir I&#39;m gonna highlight it we&#39;ve got this right angled triangle here and we know it&#39;s right and cooled because we&#39;ve got a tangent meeting the radius sir so straight away I&#39;m going to draw that right angle him and then we can obviously work out this missing angle in the Triangle here I&#39;m just thinking about what I can work out without even looking at the X for the moment now that is just gonna be angles in angles no triangles we&#39;ve got the 1932 at the moment or we could just do 90 take away 32 is this 90 degrees left once we get rid of that right angle so 90 take away 30 tea leaves with 58 degrees there we go so there&#39;s our first bit of working out we&#39;ve got 58 degrees just there now we&#39;ve got angles on a straight line so the angle next to that we can work out straight away so it can get this one here and that&#39;s gonna be a hundred and eighty takeaway 58 I&#39;m Stan goes on a straight line equaling a hundred a/c writing down those reasons so 180 minus 58 leaves me with let&#39;s have a look 122 I believe double-check that I didn&#39;t I - yep absolutely fine now this is again one of these isosceles triangles slip coming in because we&#39;ve got this triangle here made by these two loads that are the radius so that forms a little isosceles and again we know base angles and rice off so these are equal so we can find the size of X along with this one when we find out those base angles so if we do 180 take away the hundred and twenty-two obviously again leaves us with 58 it was just the opposite way around of what we did a second ago and then we can split that into so dividing that by to these is 429 degrees for that one there we go sweet 29 degrees for this question as well there we are so that&#39;s our answer there 29 degrees finishing that off and just obviously let&#39;s just have a look at what reasons we use there so the first reason that we had was angles made by the radius in the tangent being 90 then we had angles in a triangle the male angles on a straight line and then we have base angles in our sauce leaves being equal so obviously not a huge amount of circle theorem is actually going on in a lot of these questions it&#39;s more just two one circles there and possibly two and then a lot of triangles and straight lines and why sort of normal angles that were used to using all right let&#39;s look at some example questions right okay so a lot more language in these questions here&#39;s I&#39;m not shortened anything down just here it is as it but as it would on exam so a B and D are points on the circumference of a circle Center zero e DC is a tangent B DC is 57 that&#39;s not given to us on the diagram so let&#39;s label that B to D to C there we go that&#39;s that one there and that&#39;s 57 degrees find the size of angle AOB so it&#39;s over like a OB is that on there now again let&#39;s not worry about sort of how we get to our angle of the members have a think about we can find I&#39;ve spotted two things that we can find so one thing I&#39;m thinking about is obviously where the radius meets the tangent is 90 so I&#39;m straight away thinking we can find this one here so if we do 90 take away 57 that&#39;s going to give us that one so 90 minus 57 and again our reason here is angles where the tangent meets the radius is 90 so let&#39;s take that away and we are left with 33 degrees and there we go we can label that in the diagram let&#39;s get rid of that and label that in there 33 degrees there we go that&#39;s just in there now the next thing I&#39;ve also spotted is that this angle here that we&#39;ve started with given to us in the question is the angle where that tangent meets that triangle and that triangle there let&#39;s have a look does meet on all points on the circumference of the circle so we can do our alternate segment theorem we can go through the angle up the line and there we go it makes this angle here so if we get rid of that we know that this angle here has to be equal to about 57 due to our alternate segment theorem so angles in alternate segment theorems are equal sorry angles alternate segments are equal there we go so that must be 57 degrees so we&#39;ve got a couple of angles there let&#39;s look at what we can find now moving from this so we&#39;ve got I&#39;m gonna highlight this we&#39;ve got that this triangle here as two lengths are the radius so therefore they&#39;re equal so actually that has to be an isosceles triangle we do actually also have that on the other side as well so just high like this we&#39;ve got the radius there the radius there so I&#39;m actually on earth an isosceles triangle here so we could probably find our way through the triangle fire that route but actually us we&#39;ve now got this 57 and we know the base angles and my sources are the same this here must also be 57 so actually we&#39;ve just got wine left in an isosceles triangle and I can always draw that out to the side if you&#39;d like you can always redraw some of these triangle 5757 and we&#39;re trying to find that top angle there so if we add them together that&#39;s 114 degrees and we want to take that away from hundred eighty so 180 take one hundred and fourteen leaves us with 66 degrees and that&#39;s our final answer for this question so that&#39;s there is going to be 66 degrees there we go 66 degrees final answer but again we could have taken some different routes here just have a quick look at this one so if it didn&#39;t take that route so if we get rid of a little bit of this let&#39;s get rid of that and forget we found those so we&#39;re trying to find this one come and get there with another another route now we go now I hope that this one was 33 and I said obviously this is another isosceles so we&#39;ve got a base angle going on down there so that&#39;s also 33 so we could find the top angle in that triangle instead we could add together the two 33 so that would give us 66 and take that away from hundred eighty and that would leave us with that&#39;s if I think one one ball there we go and that means that if this one&#39;s one one four now we&#39;ve got angles on a straight line so angles on a straight line you could do a hundred eighty take away the hundred and fourteen which obviously again leaves us without 66 there so you can actually do that with just the one circle theorem we didn&#39;t even have to use the alternate segment theorem you just do the tangent meets the radius at 90 and then base angles into isosceles and then angles on a straight line okay so that&#39;s prob maybe an eaglet may be a bit of an easier route for some people it may be struggle with using the alternate segment theorem you could have actually done that without it there we go let&#39;s have a look another question all right okay so this question says work out the size of angle C ad so C ad is this one here and we&#39;ve got a few angles are given to us with what angle ad e equals 54 so ad equals 54 this one here there we go we&#39;ve got angle ABC equals hundred 14 so A to B to C that&#39;s that one there one one four there we go and we need to try and find out this angle here now let&#39;s have a think about will become fine and we haven&#39;t got any radius meeting the tangent so we can&#39;t do any 90 degrees but we have got a triangle meeting that tangent so we can apply the alternate segment through we have a cyclic quadrilateral within there so I think we just thinking about those two circle theorems the alternate signal theorem and the cyclic quadrilateral obviously those opposite angles being equals 180 so that&#39;s one of the easiest ones find straight where and go opposite this hundred and fourteen I can do 180 take away the hundred and fourteen and that leaves us with 66 degrees therefore that opposite angle so that there is 66 let&#39;s get rid of this arrow there we go and we can label that 66 degrees now whether we&#39;ll need it or not we can find obviously this missing angle now on the straight line so I can have together the 54 and 66 and if we have those together let&#39;s write that down 54 plus 66 is 120 so I&#39;m a missing angle down the straight line is gonna be 60 degrees there goes that 60 degrees actually despite that&#39;s gonna be quite a key one and in fact let&#39;s just write the working out down 180 miles hundred 20 and that&#39;s 60 degrees there we go so actually we can get to our finance here straight away we can use the officer in a segment theorem so we can go through this angle of 60 up the line and there we go it meets the angle there so that&#39;s also gonna be 60 degrees now we asked that&#39;s our final answer actually 60 degrees okay so what reasons did we use there so we started off with opposite angles in a cyclic quadrilateral being supplementary or I think it&#39;s 180 then we had angles on a straight line and then we use the alternate segment theorem and to be honest that&#39;s probably I don&#39;t know what there is under the route just looking at the moment but that&#39;s probably the easiest route they&#39;re only a couple of steps don&#39;t you just making sure you write down those reasons all right so look at one more right okay now this is a great question actually because you actually don&#39;t even have to use any circle theorems within this one I mean you can do if you like maybe I&#39;ll run through this in two different ways because I think I&#39;m believe you can get to here with maybe only just the one circle theorem for this one now it says work out the size of angle C AO now all the other angles given to us in the information are on the diagram so don&#39;t draw anything in but see AO is this angle here and that&#39;s just high like that and we do need to use one circle theorem here that&#39;s where the radius meets the tangent okay and we can see that Oh a is the radius it says this in there in there in the the writing for the question there but we&#39;ve got 56 just over here so if we take that away from 90 so 9656 and that leaves us where&#39;s 34 degrees there we go that&#39;s gonna be 34 just here now once we&#39;ve done that we can actually work our way around using lots of myself&#39; least triangles here okay so if I have a look I&#39;ve got an isosceles triangle just here where 34 is one of our base angles let&#39;s work that up to here 34 and then I can actually find this angle just here at the top of the isosceles there we go so you can always redraw this to the side if you want but there we go we&#39;ve got 34 34 we&#39;re gonna work out that top angle so add those base together is 68 now if we take that away from hundred 80 that gives us a five leaves the top angle there what&#39;s that when we take that away he&#39;s at 112 there we go 112 degrees so that angle there is 112 now it&#39;s up to you can actually split this up into different isosceles triangles now I&#39;ll show you I mean I could actually just draw this little line in here and I&#39;m actually just created three isosceles triangles we&#39;ve obviously got the one that I&#39;ve highlighted in yellow you got another isosceles triangle here and I can just repeat the process that I&#39;ve just done all well not all but once you&#39;ve done that you&#39;ve also then got another isosceles triangle here and again you can just keep repeating the process for three isosceles triangles and you can get down to that final answer it&#39;s a little bit slower but there is a process that you can do with using very minimal circle theorems in this but the next one I&#39;ve spotted which i think is probably the best way to go about doing this if you want to do them nice and quick is to go for this angle center here and then we&#39;ve got up angle at the circumference so that angle at the circumference there is half the size of the wall at the center so if we do that / - 112 divided by 2 let&#39;s see what does that give us that gives us 56 degrees no so this one up here must be 56 degrees let&#39;s see if that actually helps us I&#39;m not sure where we&#39;re actually going with this but have a look and now we&#39;ve got to get down to that angle X there now there&#39;s a couple of ways that you could do this we have got I think I have it easier just to do the all those isosceles triangles actually we still quite a lot of work to be done so we&#39;ve got this shape here there we go which is a quadrilateral isn&#39;t it&#39;s got four sides so it&#39;s a quadrilateral obviously angles in a quadrilateral add up to 360 so if we go about just finding this reflex angle 112 here so we could do 360 take away one hundred and twelve and that gives us two hundred and forty-eight there I guess that one there is two hundred and forty-eight and now we&#39;ve got three of the four angles in the urn in the quadrilateral there so we can actually work this one out straight away now we can have AB those all together so 248 plus 56 plus 35 up days all of them see what&#39;s missing up to 360 so if we have those all at what do we get 14:19 carry the one four nine ten eleven twelve thirteen and three so three hundred and thirty nine degrees at the moment so if we take that away from 360 to finish that off 360 take away three three nine and that leaves us with 21 degrees there we go so our final answer there will be 21 degrees and there&#39;s just highlight that right there we go right obviously I did mention there was another way of doing that though so maybe we should have a look at ax I&#39;m not too sure whether that was the most simple way of doing this or not was just a way that came to me as always going along but let&#39;s have a look let&#39;s get rid of this and have a little look at how he helps we could have done that so for do split that up as I said before I Steve got an isosceles triangle there so this is a base angle also 35 so we can work out the top angle in that triangle so thinking about this to the side let&#39;s have a look I&#39;ve got a base angle of 35 and another one of 35 and I add up to 70 so 180 sake where 70 keeps us 110 for that top angle at 110 there we go now we can work out the angle at the top of this little right angle because those three angles around the point after add up to 360 so if we add those two together you&#39;ve got 110 I&#39;ve just got writing everywhere now 110 at 112 is 222 and if we take that away from 360 what does that leave us with a hundred and thirty-eight is it 138 there we go and then again we&#39;ve got that final isosceles triangle there which if I draw out over here we&#39;ve got a top angle er 138 so if we take that away from 180 and split it into let&#39;s see what we get so take it away from 180 is 42 and 42 divided by 2 gives us 21 in both of those base angles there we go so we get 21 there and obviously 21 for that value of x that we&#39;d already found so different ways that you could do it just depends on which one you prefer really and how you actually spot it in the moment but I&#39;ll do it I&#39;ll be honest with a lot of these circle theorems you don&#39;t have to just spend a little bit of time just looking at them thinking what can you find where can you go unless without thinking actually just trying to get straight to that final answer because a lot of the time as you can see with this one I&#39;ve pretty much found every single angle in sight before I actually got to that final answer so always just try and take that approach and just think about that as you&#39;re going through some of these questions okay but that&#39;s the end of the video there I&#39;m gonna do the next video looking at algebra within circle theorems then I&#39;m also going to pick out some of the hardest circles theorem questions and go over some of those but again the link for those will be in the description this was just a bit a bit of a brief overview overview just running through some of the exam style questions on this topic okay but hopefully that was helpful if it was useful please like please come and please subscribe and I&#39;ll see you for the next one [Music] [Music]

Transcript for:Circle Theorem Questions Lecture Notes

Transcript for:
Circle Theorem Questions Lecture Notes