[Music] okay so in this question we've got three circles inside a square it says down here that the radius of each circle is 24 millimeters and then wants us to work out the area the rectangle now if you have a look we've actually got a series of radiuses that fit along sideways here from edge to edge of this rectangle so we've got four radiuses there so we know that the width or the base here of this rectangle is for lots of 24 and for lots of 24 we work that out is 96 millimeters okay the problem here that we face is actually moving up at the right angle if we get rid of these and have a look moving upwards we've got a radius here a radius here but there's an overlap with the radius on the circle here so we've got one two top and bottom but then an overlap on these middle two so we got to take a bit of a different approach here to work out what the height of the rectangle is now the way to do this is to have a look at this that the points here between the Centers of the circles and if we join them all up we get this triangle and obviously the the radius is there at the same length so this triangle is an equilateral and if I draw this triangle to the side being an equilateral that we want to work out the height of so essentially we want to work out this distance here the height of that right are the height of this triangle now we know all the sides are 48 millimeters this one here is 48 millimeters the length of two radiuses you've got the 24 and the 24 here and that makes a total length of 48 we also know that in an equilateral triangle this angle here is 60 degrees so if I want to work out the height of that triangle which we'll call X over here we can use a bit of sohcahtoa I can use some trigonometry to work out that height there so X being opposite the 60 and 48 being the hypotenuse if we were to label that up that's the opposite length here that we're looking for and the hypotenuse that we've got so we're going to be using O&H so those of you that use formula triangles we can just think okay s o H and to work out the opposite us times H so to work out the link for that opposite link there we would do 48 the opposite no sorry 48 the hypotenuse multiply it by sine 60 there you go 48 times sine 60 and it gives us a link there a forty one point five six nine and again that's going to be in millimeters so now we've got that length there which I'm going to draw on the side of the rectangle being 41 point five six nine let's label that on four to one point five six nine and we already know then that just below that is another radius of the circle being another 24 millimeters and just above that is another radius of another 24 there and if we add up all those lengths there it gives us a total height of this rectangle of eighty nine point five six nine millimeters there we go and we know the width is 96 so we can times that by the 96 millimeters and that will give us the area of the rectangle so the area of the rectangle is eight thousand five hundred and ninety eight point six two four which is equal to because it says to round to three significant figures so we have eight five the nine rounds up because of the eight so actually it's not going to be eight five we're gonna have to get rid of that and round that up to six hundred so it's gonna be eight thousand six hundred there and that is to three significant figures obviously this is an area so that would be millimeters squared and that's our final answer there eight thousand six hundred millimeters squared for our final answer okay so on this question we have some bounds and it's about density so we've got the we've got a cuboid made out of a block of wood while a block of wood in the shape of a cuboid and it gives us the dimensions of the keyboards it can help to draw this here but it gives us a length the width and the height and hopefully you know the volume of a cuboid is just multiplying those three together it says he measures the masses a 1970 correct to the nearest 5 and then by considering bounds work out the density of the wood to a suitable degree of accuracy now the first thing to note here is obviously density equals mass over volume that's gonna be the formula we're gonna need to work out density so it's always worth writing that down so density equals mass over volume now it's about bounds so we have to find the error intervals for all of these numbers and when it says by considering bounds we've got to work out the upper bound of the density the lower bound of the density and work out what these both round to so before I can do that I'm gonna have to work out all the bounds so moving from the first one downwards I'm gonna just go for I'm gonna stick that number in the middle 13.2 I'm gonna write all the error intervals here so it's correct to the nearest millimeter so that one there is gonna be thirteen point two five for the bigger one and thirteen point one five for the lower bound onto the one below we've got 16 point zero so that's gonna be 16 point zero five for the upper bound and sixteen point nine five no sorry not sixteen point nine five fifteen point nine five fifteen point nine five for the lower bound and then doing the last one very similar fashion here twenty one point seven the upper bound there is going to be twenty one point seven five and the lower bound is going to be twenty one point six five now I've got all my bounds there for the lengths underneath is volume now I'm gonna get the bound for this which is correct to the nearest five so slightly different error interval here but let's have a look we've got not 1970 in the middle and if it's the nearest five if that means we can go to point 5 above and two point five below which is half of five so that would be 1972 point five is the upper bound and 1967 point five as I were lower bound there now we've just got to work out the upper and lower bounds so first of I actually need to work out what's the volume of per bound and what's the volume lower bound so the upper bound for the volume I just timed together all these upper bounds so the upper bound for the volume I'm just going to times all these three together so I'm going to do thirteen point two five multiplied by sixteen point zero five make sure I'll get all the right numbers here multiplied by 21 point seven five and that gives me an upper bound of the volume of it's quite a long number here four thousand six hundred and twenty five point four zero nine three seventy-five there we go so there's my upper bound for the volume and now I need to work out the lower bound for the volume and then we can go about working out both these densities so the lower bound for the volume times in order to gather all these and it's doing a different color dungeon together all of these lower bounds here so that would be thirteen point one five multiplied by fifteen point nine five and that multiplying that by twenty one point six five and that gives me a lower bound here again a large number we have four thousand five hundred and forty point nine two five one two five there we go so that is the lower bound of our volume now we can actually work out the density so densities mass over volume we've got a divide going on there so to get the upper bound we want to do the upper bound of the mass divided by the lower bound of the volume there and to get the lower bound we want to do the lower bound of the mass divided by the upper bound of the volume so if we plug all these numbers in and see what we get so the upper bound of the mass is over there one nine seven two point five and the lower bound of the volume is four thousand five hundred and forty point nine two five one two five and if you type that into your calculator you get naught point four three four three eight two I'm not gonna write any more down for the moment although you should really write them all down and for the lower bound of the density we would do the lower bound of the mass which was one thousand night that's not what it says on the calculator let's look 1967 0.5 and the upper bound of the volume there which we've written down four thousand four thousand six hundred and twenty five point four zero nine three seven five that we go and if we work that out on the calculator you get zero point four two five three six seven I'm not going to write down any more it's going to match it with a matter decimals right to the side and now it says by considering bounds this is our final step give your answer to a suitable degree of accuracy so first things first I can already see that they don't match to three decimal places because one of them would be four three four the other would be four to five so they don't match so if I look at two decimal places so naught point four three and that would be not point four three as well because the next numbers are five some are next my next step is just to write that out so not point four three is my answer that's as accurate as I can get them to match and that would be my final answer and then I just obviously give my reason there so the one that I used was two decimal places I'll just write down two decimal places well that you could have said two significant figures as well okay but I just decided to say decimal places I'm going to stick with two decimal places so not point four three correct to two decimal places okay so we've got some similar shapes here we know it's similar shapes given this information here it says tan a equals tan f meaning that if we were to do the angle on the angle f in there we would get the same value when we use our tan value which means the angles the same so if it's in a right angled triangle and E and F are both the same then all of the angles within these triangles are the same so we know it's a similar triangle and that means there's a scale factor between the lengths so if I was to do the big length here and divide it by the little lengths it would give me the same scale factor as this big lengths divided by this little length here and if I write that as a fraction because obviously it is algebra here so I can't just actually work out a number but I can say 12x plus 31 the big length there divided by 4x minus 1 has to be equal to 6x plus 5 divided by the smaller length over there which is just X now again I've got an algebraic fraction and I just need to cross multiply to solve this soakin times its denominator up here and this denominator up there and if we do so we get X lots of 12x plus 31 and that is now equal to we've got a double bracket here we've got 6x plus 5 brackets 4x minus 1 so we need to expand both of these brackets and then we can go about solving it so if I expand the left hand side I get 12x squared and if I expand that 31 by X we get 31 X expanding the right hand side and I am going to skip a step here I'm not gonna do all the working out 6x times 4x is 24x squared and then we'll get minus 6x with this multiplication add 20x some minus 6 or 20 is plus 14 so plus 14x and then minus 5 and there we go I've got some sort of quadratic here that I need to solve so I need to make it equal 0 so I'm gonna - these two from the left hand side just to keep them positive so to avoid an extra step so I'm going to minus these from those from that side so we'll get 0 equals 24 x squared minus 12x squared is 12x squared 14x takeaway 31 X is negative 17 X and then minus 5 now I've got a quadratic I just need to factorize this so the factors of 5 can only be 1 & 5 there I'm just going to bring the double bracket up to the top to actually factorize this so if we factorize it let's have a look we could have 3 X + 4 X we could have 2x + 6 6 as well then it's going to go 4 3 & 4 just because I've spotted there that actually I can times 5 by 4 to make 20 and then take away the 3 there because I want to make 17 in the middle so ones getting at times by 3 once again 8 times by 4 so if I want the 5 to be multiplied by 4 to make 20 and I want the 3 the 1 to be multiplied by 3 to make 3 and I want to make negative 17 so I want it to be positive 3 so plus 1 over here would make positive 3 I'm negative 5 and that make the negative 20 so 3 take away 20 would make my negative 17 in the middle there and then that gives me 2 solutions so the first one here gives me x equals positive 5 over 3 and I've got this one here where X gives me negative 1 over 4 there we go and obviously from there I've got I've got two solutions but it says find a value of x now X cannot be a negative lengths so X does have to be this positive value so my final answer there is x equals 5/3 as my final answer there we go again we've got quite a lot going on they've got solving an algebraic fraction factorizing the harder quadratics and obviously just getting a final ants are all tied up in this similar shape this question okay so this probability equations question now you might think when you first saw it that you had to create a tree here but actually this can be done without a tree and it doesn't it does actually lend itself very nicely to creating a tree but it says the probability that counter is green is 3/7 and that means the probability that it's red will be 4/7 just thinking about that right from the start it says the counter is put back in the bag and then two more red counters three more green counters are put in the bag after one was taken and now all the probability that the counters green is 6 13 now again that means the probability that it's red after these counters are put in the bag is gonna be 7 over 13 so we've got these sort of equivalent ratios that we can start to write because obviously the ratio and if I'll keep it as red to green because it says find the red common that counters to the number of green counters so initially the ratio of red to green was 4 to 3 and then the ratio of red to green became just write it down here red to green became 7 to 6 now obviously we don't know how many lots of 4 and how many lots of 3 that was so gonna have to use a little bit of algebra here so if I say well the ratio of red to green was for lots of X a number we don't know 2 3 lots of X now we've got the part of the story here where more counters were put in the bags are 2 more red and 3 more green so that means that red was 2 more so we have 4 X plus an additional 2 to 3 X plus an additional 3 so we've got 4 X plus 2 to 3 X plus 3 and it says now that is equal to the ratio okay now it's equal to the ratio 7 to 6 okay that's the ratio that we've got just up here after these counters were added back in the bag and when we've got equivalent ratios like this we can turn these into fractions so we can do the first part of the ratio here over the second part the ratio there is equal to 7 over 6 ok the ratio here so for right as a fraction we get 4x plus 2 over 3 X plus 3 is equal to 7 over 6 so we've got an algebraic fraction which we can now cross multiply so 4 cross multiply this so timesing the 6 up to the top here and this up to the 7 we get a nice single line equation I'm gonna do that in one step you can write this in brackets if you want actually I'll write the bracket so we get six lots of 4x plus two x in the six over and x in the other bracket over we get it's equal to seven lots of 3x plus 3 now we've just got an equation that we need to solve nothing too bad there we just need to expand the brackets so 24x plus 12 equals 21x plus 21 now we can go about solving this so take away 21 X from both sides we'll get 3x and take away 12 from both sides we get 3x equals 9 and therefore X must equal 3 and that is solving our equation for x now we see the question does say find the number of red counters and the MER green counters that were in the bag originally and here's that part of the process here we had 4x + 3 X 4 lots of x + 3 lots of X so for red we have for lots of X so for lots of 3 would equal 12 and for green we had 3 lots of x + 3 lots of 3 is 9 so the number of red counters equals 12 and the number of green counters equals 9 there we go and there's our answer 12 + 9 okay so here we've got some bearings and it says it wants to find the bearing of chalten from Acton looking down here find the bearing of chalten from Acton so from a down to C we want to know what that full bearing is so that 37 adding in this additional angle just here so if we label that let's label that just label it theta okay we're going to look for that little angle there right so looking at what we can do to start with first things first if we have a look at these two North lines we could work out this angle here because these two angles here occur interior they add up to 180 so if we do 180 takeaway 37 it leaves us with a hundred and forty-three degrees for this angle here where that Millette while that letter B is sitting now we can work out all the angles around this point good angles around a point add up to 360 so we can work this one out and that comes out as 67 degrees when we take those away from a 360 150 plus 143 so take that away gives us 67 now we can start to have a look at this triangle here because ultimately if I can work out another length in this triangle being this one here for labeling X then I can use cosine rule to work out that angle there with I've got the theta sitting in so if we have a look at this to work out this letter X here I can use the cosine rule so the cosine rule obviously something you need to know therefore this is a squared equals b squared plus c squared minus 2bc cos a there we go so if we label this up 67 being our a in this scenario let's sort colors so a being our 67 X being our little a and then 8 and 9 being our B and C so labeling those however you like B and C now if we plug all those numbers in we get a squared or x squared equals 8 squared plus 9 squared minus 2 times 8 times 9 cos 67 and if we type all of that in and square root our answer to type it all in then press square root you get a value of x equals 9 point 4 1 9 9 obviously showing all these steps but I'm going to you see amount of writing here just to keep the video lab its idea so we have X and I'm going to label it down here being nine point four one nine nine now we can actually use cosine rule for angles so I want to work out this angle here and I've got a and B well B and C either side now if I'd get rid of my a and B from the previous questions let's get rid of that get rid of that oh we're gonna real able all of this I'll just draw that back in there we go so obviously now we know the length of this X here we know that this is nine point four one nine nine so we can get rid of everything else and work out this angle so to work an angle using cosine rule potentially that could be that could be the case we could use cosine rule or we could actually just use the sine rule because we have a pair of opposites as well so totally up to you here it's government you know you could obviously use cosine rule friend Gore's or you could do sine rule not to use time law just service or any other we're different because we've got opposite the angle we've got the nine and also opposite this 67 now we have this length down here so I'm just gonna use the sine rule so I'm gonna say well so KL sine theta obviously sign a over eight course sine B over B so sine theta over nine equals sine 67 over what is it nine point four one nine nine there we go rearranging that so times in both sides by nine and we get sine theta equals nine sign 67 over nine point four one nine nine there we go now finishing this off if we type that into the calculator so 1967 over nine point four one nine nine and then do the inverse sine so sine minus one of the answer there I'll just write sine minus one hunt sir we get the answer sixty one point six there's some decimal places but it's correct to one decimal place I'm going to leave it at the moment so 61 point six degrees there we go so we can label that into our diagram this will be 61 point six so to find the full bearing of chalten from Acton obviously it's always from north going clockwise so we have the sixty one and we have the thirty seven so to finish it off we do 61.7 at the 37 and we're left with a final angle there of 98.6 degrees okay now that is our final answer almost but obviously it is asking us to give a bearing bearings have three figures even though it's got one decimal place we need to include the third figure so our final answer will be naught 98.6 degrees okay finishing off as a bearing with three figures they're obviously to one decimal place as well so naught 98.6 degrees being our final answer there okay so we've got some 3d trigonometry now in order to work this out just got to read this question very carefully but it says the front length is 15 d2 mm2 a is 2 to 3 so M is somewhere on this line here and that's M and it's in the ratio 2 to 3 so straight away I could split that 15 in the ratio 2 to 3 now that's two fifths and 3/5 2/5 of 15 is 6 centimeters and the other side therefore would be 9 centimeters so I can split that line up there now it wants me to find out the angle between en which is this line here for draw it in and the base or the plane so M if I bring that across I can make another right angle triangle triangle here cuz I can join it over to B so I've got lots of rice angled triangles now the first triangle triangle I need to work out here's the height or the length of e B and if you imagine just on the side there we've got a right-angled triangle and this is a this is e and this is B and we've got that this angle here is 35 and this length here is 15 I guess but our degree symbol him so I can actually go about working the length of e to be out because I can use sohcahtoa there I can use my tan because I've got the opposite in the adjacent so to work that out to work out the opposite I'd do T times a or o times a times tan so that'd be 15 the adjacent side x tan 35 and 15 10 35 equals on the calculator 10.50 3 1 1 3 0 7 but I'll just leave it as up at the moment so 10.50 3 let's just label that on 10.50 3 there we go now we can actually have a look at working out the length of the base of that inside triangle that I've drawn there meb there's a right-angled triangle and I can actually work out the base there I'm gonna do this in a different color this link that's just here if I can get that length I can work out that angle again using sohcahtoa but that actual length there I can do quite simply because there's a little right-angled triangle on the base there and that's going from M where is it going to be and a so this is the flat base on the bottom and we've got this length here is 15 and this length here is 9 after we split up that little length there so we can use Pythagoras to work out the length of X so it's the longest side so x squared equals 9 squared plus 15 squared so 9 squared plus 15 squared which is x squared is 306 and then we can square root that for the length of X so square root that on your calculator I need seventeen point four nine to a so it's quite a lot here 285 five I'm just gonna write some of them down although it does go further than that so seventeen point four nine two eight and again I can label that over there seventeen point four nine two eight and now I can almost finish this question off actually because I've got that inside triangle that I've drawn in there so we've got this big triangle in the middle that's sort of going through the open space there and that stretches from M up to the a and then down to the B which is another right-angled triangle and we've worked out the height of that triangle there is ten point five zero three just worked out using Pythagoras that the base is seventeen point four nine two eight and then we can now obviously work out this angle here which I'll call X again using sohcahtoa again so again it's tan because it's the opposite in the adjacent so using your formula triangle or however you use this we can do tan -1 to work out the angle and that's the opposite over the adjacent so 10.50 3 divided by 17 point four nine to eight how many talk that into your calculator you get the answer it's quite a long one thirty point nine eight one three zero three zero nine and then obviously just round it to have a bit a sweeping around as to write it to one decimal place there's a bit of a weird one actually because when you round this to one decimal place if we chop it there the nine rounds up so it becomes 31 but we need to give it to one decimal place that'd be thirty one point zero degrees and that'd be our final answer there 31 point 0 degrees okay so we've got a circle equation and we've got a point on the circle and they wants us to find the equation of the tangent at that point and the point is a very nice server got fractions and thirds now we could always help here just to think of a little diagram so if we just draw a little basic diagram with a little circle Center zero and we'll think about what this actually looks like now P on this diagram as three halves and root semenova two are both positive it would be in the positive quadrant here so if I was to have a tangent it would look something like this now if I want to find the equation of the tangent I need to find the gradient of that tangent and to find the gradient of a tangent we first have to find the gradient of the radius so I can find the gradient of the radius because it goes from zero zero up to this point here which is 3 over 2 and root 7 over 2 so if I do my change in Y over my change in X I can get the gradient there it would be from 0 to root 7 over 2 so it'd be root 7 over 2 divided by or over 3 over 2 and I write it is divided by so divided by 3 over 2 and that'd be my gradient I'm second just treat this like normal fractions so I can do root 7 over 2 times 2 over 3 so flipping the second one over and timesing them would give us 2 root 7 over 6 and again that can simplify I can divide the top and bottom by 2 so I would end up with root 7 over 3 and that's my gradient there of the air of the radius now I want the gradient of the tangent so I need to do the negative reciprocal so if I do the negative reciprocal of that I put a stick it obviously changed a sign to negative and flip it over so negative root 3 3 over root 7 sorry they were going negative 3 over root 7 so that's my gradient of my tangent keeping that all nice and clear now I can plug it into the equation of a line formula so we know that's y equals MX plus C so it'll be y equals M here being that gradient of negative 3 over root 7 so negative 3 over root 7 X plus C and obviously we just need to plug these values in so we've got the values here 3 over 2 and root 7 over 2 for a value of x and y so if we suppose in Y being group seven over two so root 7 over t equals negative 3 over root 7 x the x value there which is 3 over 2 so times 3 over 2 and then plus the C on the end which is try find out what C is now so if we actually work this out let's have a look so we have root 7 over 2 equals multiply the tops multiply the bottoms we get negative 9 over 2 root 3 I'm not equals see there we go so I've actually put root 3 and there should be a root 7 let's get rid of that don't know why I put that in there there we go negative 3 over root 7 and negative 9 over 2 root 7 okay sorry about that now if we go about actually solving for C here so add that 9 over 2 root 7 to both sides let's have a look so at 9 every root 7 to both sides will have a root 7 over 2 add 9 over 2 root 7 and that equals C not very nice here obviously we need a common denominator so would be nice obviously using your calculator but if we just think about obviously to get a common denominator I just need to times this side here by root 7 over root 7 times 1 by root 7 it's nice and easy for us to add that together there so 4 times top and bottom there by roots then over root 7 over root 7 we get 7 on the top over 2 root 7 and I'm gonna add that to 9 over 2 root 7 so that's going to give us 16 over 2 root 7 there we go and that equals this write it over here so C equals 16 over 2 root 7 and now we have all our pieces we just have to push them into y equals MX plus C and that is our equation done so we will have y equals negative 3 over root 7x okay so I actually write 7 this time add C which is 16 over 2 root 7 there we go and there's the equation of the tangent there using our negative reciprocal for the gradient and getting C there with our common denominator with our root 7s our route seven they're actually just finishing this off we actually could simplify that see a little bit further sixteen to so sixteen on the top two on the bottom we could actually simplify that it's not wrong as it is but we should really simplify that so I'm gonna divide the top and bottom by two so let's change this there we go so divide the top by two you get eight under add the bottom by two we get root seven there we go so we should always look to simplify there just spotted that in the last minute so what's that / - and / - just to simplify that so there's our final answer y equals negative 3 over root 7x plus 8 over root 7 okay and we've got some simultaneous equations here not the nicest because obviously when we've got quadratic simultaneous equations we need to rearrange and sub in to one of the others but when we rearrange this bottom equation it doesn't give us something that looks very nice if I rearrange this bottom one ready to sub it into the top I'm gonna rearrange it for X as the subject so if we take away for way for y to the other side we get 3x equals 7 minus 4y and then divide by 3x equals 7 minus 4y over 3 so not the nicest there but now we can sub that into equation 1 so if we submit into that position where x squared is let's see what we get we get 7 minus 4y over 3 that's all squared because it's x squared minus 4y squared equals 9 now we need to expand this bracket out so we have a double bracket on the top there when we square this out and again I'm gonna skip some steps so I'm going to imagine the double brackets 7 times 7 is 49 then we would have minus 28 x minus another 28 X so minus 50 sorry y so minus 56 Y and then 4y times 4y they're both negative so it becomes plus 16y squared that's all over 9 and then it's minus 4y squared equals 9 okay so we've got a fraction involved so we need to times everything by 9 so if we times by 9 they'll get rid of this fraction so we have 49 minus 56 y plus 16y squared now it's gonna be minus 36 y squared when we times up by 9 and that's now going to equal 81 when it gets x by 9 as well so subtract 81 from both sides so this equals 0 one of a nice quadratic we can solved so 49 take away 81 is negative 30 to still take away 56 Y and then 16 Y squared take away 36 y squared is negative 20 Y squared and that now equals 0 so I'm going to times this all by minus 1 so that we've got a positive Y squared there and I'm just gonna rearrange it as well so we've got the Y squared at the start just prefer that for factorizing so we'll have positive 20 wives squared will have positive 56 Y and will have positive 32 all equaling 0 so I've rearranged it but I've also timed everything by minus 1 there just to make it positive so x minus 1 there we go right now we can actually go about looking at factorizing this it does actually simplify there so we can divide everything by 4 so I'm going to divide everything by 4 and that gives us 5y squared plus 14 y plus 8 equals 0 and now I can go about factorizing that so I'm going to bring that up here or factorize it so let's have a look in the brackets 5y + y and the factors of 8 we can have 4 + 2 or 1 + 8 and we want to make 14 in the middle I'm going to write that down over here I think it's 4 and 2 so if we check that out how do I make 14 I could have 8 + 4 know-hows it's gonna work so xx + 4 10 + 4 10 + 4 works there we go so I could have the five times in five the two to make the 10 and the four not multiplying by anything I want to make positive 14 so I want them both to be positive there we go that's gonna allow it to make 14 and that gives me my two solutions there so y equals negative 4/5 and y equals negative 2 so there's my two Y values the next bit is quite nice and easy cuz unions on the calculator we just need to sub those into this x equals equation up here so sub those in on your calculator so when y equals negative 4/5 we need to do so X will equal 7 take away 4 or lots of negative 4/5 you can type this all in on the calculator take away 4 lots of negative 4/5 and then divide that all by 3 and if we type that in we get x equals 17 over 5 good right as a mixed number as well you could write that as 3 and 2/5 so that's when that's when y equals negative 4/5 and our other solution here which I'll try and fit in what do we get we get X cause and type in on the calculator again 7 minus 4 times minus 2 when X is 1 Y is minus 2 and again divide it all by 3 and you get x equals that's a nicer one x equals 5 there you go so there's a lot going on there you've got when y is minus 4/5 x equals 17 fifths and the other one here when y equals negative 2 x equals 5 and there are your solutions for this quadratic simultaneous equation again we could have rearranged it and start to make it y equals and you could've done this a different way but there's one way of actually getting your cell you shouldn't select question alright this is one of our vector questions here so it says given the X today and I'm reading on a bit here says o to see C to D is K - 1 so to see C to D so the line is extended there we go to D and it's in the ratio K - 1 now it also says a X is the midpoint of AC so when you draw another line in here let's draw this one in a different color so A to C and X is the midpoint let's say that's there and it's given like the letter X and then what it says is it says given the X - D and D is over here the one I've just drawn in it's a straight line or I'll accept it's a straight line from X today and it gives us the vector there so for all that in X - D they're moving in that direction is 3 C minus 1/2 a there we go and it also gives a bit more information which is on the diagram 0:02 a is a you notice CSC and it's a parallelogram so also this would be C on the top and this line here would also be an a moving up that way but let's see what we need so it says find the value of K well to start with we need to actually find this for this root of it from X - D in a slightly different way as it says that these two lines o to see C to D is K - 1 so it's saying that this is K and this is 1 it wants us to found it find that value of K if that's 1 so let's have a look now we can get from X today in a different way we don't actually have to follow that one line that it's giving us there it's just giving us the resultant vector but actually what I can do is I can move down here from ecstasy and then I could move from C to D moving along that line this way and I could see what those vectors are there and see if it matches what they've given me so if we start having a look at this let's think what we could do we could go from A to C and then we can work out half of that line so I'm going to go down from A to C so to get for me to see we go down a and along C so if I do that I'm going to keep the C first so that would be C - a moving backwards through a positive through C so C minus a now if I want to do half of that so to get from ecstasy I would times up by a half so I can open up a bracket there I could just times it by a half straight awakes it's not the most complicated half C minus a half a there we go and that's my victus ecstasy now I could have a look at how I get from X to D now we don't actually know what the distance is there so I can't actually finish off this vector but I can write it like this I could say X 2 D has to be half C minus half a that's how we get from X to C and I'm gonna have to add to that a certain amount of C's because to move straight along the line here like on the diagram you've got to move in that C direction so let's just call it X lots of C X lots of the C okay but we don't know what that is that's what we're gonna try and solve so let's have a look at where we're at right now so we've got the vector X today but half C minus 1/2 I've not written that a in there if I keep missing little things out so minus 1/2 a plus X lots of C whatever that is there now if you have a look at the vector that we were given up here so 3 C minus 1/2 a and the vector we've got we've already got the minus 1/2 a so what we have to do is match that 3 lots of C not at the moment I'm gonna use a different color we've got 1/2 C and another amount of C right here so we just need to figure out well how many C's is that going to be to make it up to 3 and hopefully you can spot that it's 2 and 1/2 C's so I've got really we think of it as a bit of an equation we've got 1/2 C plus X lots of C and that has to equal 3 C cuz that's what's given to us in the equation so X has to be 2 and 1/2 so X has to equal 2.5 when it comes to vectors it's just personal preference I like to leave it as a fraction and a heart 2.5 as a fraction in terms of halves is 5 over 2 so X there all this distance has to be 5 over 2 lots of C 5 over 2 so let's write that in in in our little four in our little ratio that it's asked for so it's asked for the ratio K 2 1 at the moment it's 1 C 2 5 over 2 C or 2.5 C but 5 over 2 C ok so we've got it as 1 2 5 - so we don't need to have those C's in there but the ratio at the moment is one two five over two okay one being this distance and then five over two C being this distance here and that's the ratio of those two lengths at the moment but it wants the ratio K - 1 now the only way that I can actually get that in the ratio K - 1 I have to divide this right-hand side by 5 over 2 all divided by 2.5 so divide it by 5 over 2 and what I need you to the one side you have to do to the other so we divide the other side by 5 over 2 as well and that means we just have to work out 1 divided by 5 over 2 so 1 divided by 5 over 2 again it's just like 1 times 2 over 5 so I flip it over 1 times 2 over 5 is 2/5 there we go so K will be 2/5 so my ratio would be 2/5 - 1 and the final answer there I'll just say K equals 2/5 to finish that off and there's my final answer ok so quite a nasty one there in terms of the way that you have to approach it but it's just making sure you write it in the right way with the ratio there okay and looking at the next vector question so we're given some information here it says that CD is a so we need to label that under diagram - C - D is a we've got D - E is B so d3 is B and that's also split by this midpoint so straight away I'm going to label that as half B here so we've got half B and we've also got another half B here as well it says later in the question that's a midpoint we've also got f - C is a minus B so from F to see there's a minus B there we go now it says Express F to e in terms of a and B that's quite nice and easy to start with so we can go from F to C C to D and then D down to a and follow all that follow all those vectors so f - C is a minus B then we go across from C to D so ad da and then down those two half B's which is plus a full B now the a and the a add together so in total we have 2a and plus B and the minus B cancel each other out so we have 2a as that full vector there quite a nice sort of little question there added on to this one then it says M is the midpoint of de X is the point on F M such that F 2 X X 2 M + n - 1 so we can label it if we want we can say this is N one and then it's a CXC as a straight line so we need to try and draw in a little straight line here obviously using a ruler and a pencil going from E to C so we've got a little straight line there then it says work out the value of n obviously there's a few vectors we can work out here I'm gonna work out the vector of the full line from C to e so how do I get from C to e to start with then we can have a look at maybe C to X or X to e but if we work out the full line so to get from C to e we have to go and there's two ways of doing that we could go right to down to D and then down from D to e and that would be a long gay and then down B so from C to e is a plus B now that's the full length of the line there so any vector now that we look at on that line has to have this a plus B so if we have a look at another one let's just go for a bit of C to X so I'm gonna go for this one that's highlight it rather than doing the full line off from X to am is going to go from C to X now to do that I can go in a few different ways I can go from C down to F and then from F up to X but in order to get from F 2x I'm going to have to know what F 2 m is so separately to this I'm going to work out what F M is so have a look at from air tail to get from F to M to get from F to M I would have to go up the a minus B so from F to C so I would do the a minus B then I would add the a to that sword go along gay and then I would go down the 1/2 B so a dinning adding in half B there we go and if we join this all up with the a plus C a makes 2 a so we have 2a and the minus B plus the 1/2 B leaves me with negative 1/2 B there we go so now that I can move from F to M I can move part way along that but we don't know what the fraction is along the line it just as it's in to one so to get from F 2x let's have a look at this to get from F 2x I would have to do n lots of this so n lots of to a minus 1/2 B obviously if we knew the fraction we could write the fraction there but we don't so we can only do n lots of it so to a minus 1/2 B now if we expand that out obviously we can only expand it by n but we get to a N - half B n there we go just expanding it out and that is my vector f2 X now if we have a look at what we need to do next if we have a think C 2 X so the line that we're looking for is C 2 F plus F 2 X so we've worked that out we have F 2 X but now we need to do C 2 X ok that's the part of the line we're looking at we've got F 2 X and now we can do C 2 X so if I want to do C 2 X let's write this out C 2 X so starting at C we can go down F and that's the reverse of a but minus B so that would have to be the reverse symbols s we'd have minus a and the reverse of minus B which is plus B moving backwards down the line and then we'd add in this F 2 X so adding in this to a n minus 1/2 B n so let's add that in so plus 2 a.m. - half BN and that there is our vector C 2 X now we need to tidy this up and just have a look at how many A's and how many B's we have obviously we've got some ends with some of them but if I group them together so the mine is saying this one and the B and the minus 1/2 be the N part we've got minus a plus 2 a.m. and then we've got plus B and a minus 1/2 BN so grouping the B's and the A's together that's what we are left with now we can do an extra step from here we can actually factorize this and just see what we get when we actually factorize out this a and this B so if we have a look at that if we factorize our a and B and I'm going to do it up the top here I would get so just bringing this up to here so factorizing out a out of the first half what I'm going to do is I'm going to almost chop this into when I factorize a out the first half at trial factorize be out the second half - have a look at what my multiples of a and B are there so if we factorize a out this first bit we get a brackets minus 1 plus 2 n and if a factorize be out the second part so we'd have plus B and we factorize to be out we get 1 minus 1/2 in there we go so we've got this a plus B going on there but we've got these little bracket elements with the a and this little bracket element here with the B now in order to be on the same sir line which says it is it says it's a straight line these numbers here all these brackets with a and B have to be the same because it'd be sort of part of a fraction of a plus B so if we say that they are the same or equal to each other we can write in the equation we can say minus one plus two n has to be equal to this multiple of B which is 1 minus 1/2 then there we go and forego about rearranging this so if I add 1 to both sides to get rid of this minus 1 and maybe add 1/2 n to both sides to get rid of the ends from this side we would have 2 and 1/2 n or 2.5 in all right 2 and 1/2 N equals 2 and then solving that we can divide by 2 and 1/2 + n divided by 2 and 1/2 which you can do any calculator n divided by 2 and a half or two divided by two and a half sorry is 4/5 now we're almost finished here what we've worked out is n is 4/5 and that's the end just here so that part of the line there is 4/5 of the line and then this part of the line here would be 1/5 of the line so in terms of writing that as a ratio fits into one while 4/5 - 1/5 which which we write that down here 4/5 - 1/5 would be the ratio 4 to 1 okay times in both those fractions by 5 so the ratio of the line there is 4 to 1 it does say work out the value of K K being this number here into 1 sorry work out the value of n into 1 so N equals 4 is our final answer therefore that back to question [Music] case there's quite a lot of drawing going on on this one so we've got two congruent parallelograms and it says a b and bc are both equal and they equal x so that's this length here and this length here we'll call those both x as its asked us to do that and then gives us some more information it says about this these P at this P and Q and it says that beta P and beta Q are both ten of which all those in B to B P to Q and they are both 10 centimeters just right 10 and 10 there we go and give us any more information it says ABC equals 30 so that is the full angle there ABC and that is 30 so 30 degrees know again so finding 30 degrees now first things first we've got to have a look at actually what we're looking forward to this proves that cause p BQ and p BQ is i've got lots of drawing here the angle between these two it says prove that Cole's PB Q equals one minus and then just looks like a load of rubbish to be fair but we've got to just figure out how we're gonna get to this point here 1 minus 2 minus root 3 over 200 with an x squared after it so first things first is we can I'm just gonna think about how I'm going to draw this over the top of my Sardu in a slightly different color how we can find the length of AC and just think about AC is actually the same length as PQ which means we've got two triangles there we've got the triangle ABC and we've got the triangle pbq obviously just go back and listen to that again if you're not sure but we've got a B C and pbq so have a look at ABC to start with now it's told us at the top of ABC is 30 degrees and it's told us that these are both X so I could work out an expression for a to C okay so I'm going to work out a little expression for that and I do that using the cosine rule so to do the cosine roll it's a squared equals b squared plus c squared minus 2bc coz a and actually as this is non calculator I'm going to have to know what caused 30 is okay because that's what was my a is gonna be it's gonna be cos 30 so obviously you need to know how to work out your exact values the trigonometry here but cos 30 and I will link that in the description for you for the exact values of trigonometry cos 30 equals root 3 over 2 so I'm going to be using this value root 3 over 2 so moving ahead with this we have and this is AC here I'm gonna call that my little a but I'm gonna say a squared equals x squared plus x squared okay cuz we don't know the values of them so x squared plus x squared minus 2 times X times X and that is times root 3 over 2 the cost 30 sits on root 3 over 2 there we go looks like a bit of a jumbled mess but let's have a look and see what we get so if we tidy this up a bit we get a squared equals we get 2x squared minus 2 lots of x times X so minus 2 x squared which is also being x by root 3 over 2 there we go okay somewhere let's see what we can do with this now now if we actually think about rewriting this bit I can actually just write that all as a as a fraction there 2x squared root 3 over 2 so I'm gonna get rid of that I'm gonna write it as a full fraction 2x squared there we go what was what was it now that it was 2x squared root 3 all over 2 because that all simplifies now that divides by 2 so we can simplify that down divide the top and bottom of that fraction by 2 then we get a squared equals 2x squared minus x squared root 3 there we go and actually we're going to have part of what we're gonna start C part of this question else I can factorize that the factorize x squared I don't remember I'm trying to make it match this 2 minus root 3 here so a squared equals x squared brackets 2 minus root 3 and we've got two of our little pieces appearing now we've got our x squared and our 2 minus root 3 so when you get to this point here you know you're onto something now the next thing to note here is that it's a value of a squared obviously I can't really square root that you know what the calculator it's going to be party complicated but that length AC here it's the same as the length PQ so this is a squared or AC squared but that's also the same as remember this is AC squared I've just worked out the length AC squared and it's the same as P Q squared so if I say that I'll say AC squared is equal to peak you squared so I've actually got an expression for P Q squared the bottom of that of the triangle that we have another can if I draw that triangle out so that's B that's P and that is Q and now we've actually got part of that expression there I've got the bottom squared being let's have a look at P Q squared so P Q squared equals this x squared brackets 2 minus root 3 there we go to minus root 3 now I've actually got to figure out still what this cause of this angle is this is the angle and the work out are some general cos B PB Q which is that angle there let's call that theta okay and it says in the question these side lengths are 10 so I can use the cosine rule to work that out obviously I've just got to use the cosine rule for angles so we have caused a or I could say cos PB Q which is the a in this case so it caused PB Q equals B squared plus C squared minus a squared all over 2 BC so I just need to go ahead and plug all these values in and see what we get now if I start plugging some of the values in let's see what we've got we've got cosby BQ p BQ equals 10 squared plus 10 squared I'm going to sub it all in I'll probably just jump that step and write 200 but 10 squared plus 10 squared minus a squared which is this value here that we've worked out so I don't need to square it because that is a squared so minus x squared brackets 2 minus root 3 and that is all over 2 times 10 times 10 so 2 times B times C 2 times 10 times 10 I need to tidy all this up now it's a right mess but if we tidy this up and I think I'm going to have to start getting rid of some of the working out here it's a bit messy so let's get rid of some of these earlier stages let's get rid of all of this I'm going to start tidying up up here so if we bring this up to the top we have ten square plus 10 squared which is 200 100 plus 100 minus this x squared brackets 2 minus root 3 and that's all over 2 times 10 times 10 which is all over 200 now okay now we just need to think how we can actually make this match what's in the question there cuz it says it's 1 minus that's over 200 now what I can do is I can chop this fraction in half at that minus sign so I can kind of chop it up and put it both over 200 so we can do that because 200 over 200 is going to make our 1 so we have 200 over 200 takeaway and then we have x squared brackets 2 minus root 3 and that's all over 200 as well so I'll just split the fraction into 2 there so we're almost there 200 over 200 is 1 minus and we have x squared brackets 2 minus root 3 all over 200 I'm going to get rid of the rest of this now just leaving us with our final bit of our answer here let's get rid of all of that so our final answer here look if you have a look it's almost matching but we've got the x squared sitting on the outside so rather than having it on the top they've just saying right over just multiplying that fraction there by x squared and that's fine just to drop that x squared on the outside so to finish this off it's just writing it in a different way 1 minus and on the top there we have 2 minus root 3 wants us to keep that in a bracket over 200 instead of having the x squared on the top we've just dropped it onto the outside obviously you need to show all those steps and not rub things out that I have but just to make sure that there wasn't too much on the screen there just show me how we get to the final answer on that question there [Music]