OK, so in this question we've got three circles inside a square. It says down here that the radius of each circle is 24mm and it wants us to work out the area of 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 the width or... The base here of this rectangle is 4 lots of 24 and 4 lots of 24, and we work that out, is 96mm.
OK, the problem here that we face is actually moving up the rectangle. 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 1, 2, top and bottom, but then an overlap on these middle two. So we've 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 the points here between the centres of the circles and if we join them all up we get this triangle.
Now obviously 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 this triangle. Now we know all the sides are 48mm, this one here is 48mm. 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 we 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 and h so those of you that use formula triangles we can just think okay s o h and to work out the opposite you do s times h so to work out the length of that opposite length there we would do 48 the opposite or sorry 48 the hypotenuse multiply it by sine 60 there you go 48 times sine 60 and it gives us a length there of 41.5 And again that's going to be in millimetres. So now we've got that length there which I'm going to draw on the side of the rectangle being 41.569.
Let's label that on 41.569. And we already know then that just below that is another radius of the circle being another 24 millimetres. And just above that is another radius so another 24 there.
And if we add up all those lengths there it gives us a total height of this rectangle of 89.569 millimetres. There we go, and we know the width is 96, so we can times that by the 96 millimetres, and that will give us the area of the rectangle. So the area of the rectangle is 8,598.624, which is equal to, because it says to round to three significant figures, so we have 8, 5, the 9 rounds up because of the 8, so actually it's not going to be 85, we're going to have to get rid of that and round that up to 600 so it's going to be 8600 there and that is to three significant figures obviously this is an area so that would be millimetres squared and that's our final answer there, 8600 millimetres squared for our final answer. Okay, so on this question we have some bounds and it's about density. So we've got a cuboid made out of a block of wood, well a block of wood in the shape of a cuboid, and it gives us the dimensions of the cuboid.
So it can help to draw this here, but it gives us the 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 mass as 1970, correct to the nearest five, and then by considering bounds, work out the density of the wood to a suitable degree of accuracy. Okay. Now the first thing to note here is obviously density equals mass over volume. That's going to be the formula we're going to 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're going to 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 going to have to work out all the bounds. So moving from the first one downwards.
I'm going to just go for, I'm going to stick that number in the middle, 13.2. I'm going to write all the error intervals here. So it's correct to the nearest millimetre.
So that one there is going to be 13.25 for the bigger one, and 13.15 for the lower bound. Onto the one below, we've got 16.0. So that's going to be 16.05 for the upper bound, and 16.95. Oh sorry, not 16.95, 15.95. 15.95 for the lower bound.
And then doing the last one, very similar fashion here, 21.7. The upper bound there is going to be 21.75, and the lower bound is going to be 21.65. Now I've got all my bounds there for the lengths I'm going to use for the volume, now I'm going to get the bound for this, which is correct to the nearest five. So a slightly different error interval here, but let's have a look. We've got 1970 in the middle.
And if it's the nearest 5, that means we can go 2.5 above and 2.5 below, which is half a 5. So that would be 1,972.5 as the upper bound, and 1,967.5 as our lower bound there. Now we've just got to work out the upper and lower bounds. So first of all, I actually need to work out what's the volume upper bound and what's the volume lower bound.
So the upper bound for the volume, I just times 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 13.25 multiplied by 16.05, make sure we get all the right numbers here, multiplied by 21.75. And that gives me an upper bound of the volume of, and it's quite a long number here, 4,625.409.375. There we go.
So there's my upper bound for the volume. 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 all together all these, and let's do it in a different colour, times in together all of these lower bounds here, so that would be 13.15, multiplied by 15.95, and multiplying that by 21.65, and that gives me a lower bound here, again a large number, we have 4540. 0.925125. There we go, so that is the lower bound of our volume. Now we can actually work out the density.
So density is 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, 1972.5, and the lower bound of the volume is 4540.925125. And if you type that into your calculator you get 0.434382.
I'm not going to 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 1,000... Oh, not that there.
That's not what it says on the calculator. Let's have a look. 1,967.5.
And the upper bound of the volume there, which we've written down, 4,625.409.375. There we go. And if we work that out on the calculator, you get 0.42...
5, 3, 6, 7, I'm not going to write down anymore, I'm just going to match it with the amount of decimals I wrote to the side. And now it says by considering bounds, so 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 4, 3, 4, the other would be 4, 2, 5, so they don't match.
So if I look at two decimal places, so 0.43, and that would be 0.43 as well, because the next numbers are 5. So my next... My next step is just to write that out. So 0.43 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. So I'll just write down two decimal places. Although you could have said two significant figures as well.
But I just decided to say decimal places. So I'm going to stick with two decimal places. So 0.43, 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 E equals tan F, meaning that if we were to do the angle E or the angle F in there, we would get the same value when we use our tan value, which means the angles are the same. So if it's in a right angle 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 length, it would give me the same scale factor as this big length 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. So I can times this 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, and 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 31x.
Expanding the right-hand side, and I am going to skip a step here. I'm not going to do all the working out. 6x times 4x is 24x squared. And then we'll get minus 6x with this multiplication.
Add 20x, and minus 6 add 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 going to minus 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 12 x squared is 12 x squared 14 x take away 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 and 5 there I'm just going to bring the double bracket up to the top to actually factorise this. So if we factorise it, let's have a look. We could have 3x and 4x. We could have 2x and 6x as well.
I'm just going to go for 3 and 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 1's going to get times by 3, 1's going to get 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 and negative 5 and that would make the negative 20 so 3 take away 20 would make my negative 17 in the middle there and then that gives me two 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 two solutions but it says find the value of x. Now x cannot be a negative length so x does have to be this positive value.
So my final answer there is x equals five thirds as my final answer. There we go again we've got quite a lot going on there. You've got solving an algebraic fraction, factorising the harder quadratics and obviously just getting your final answer all tied up in this similar shapes 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 actually lend itself very nicely to creating a tree. But it says the probability that a counter is green is 3 sevenths, and that means the probability that it's red will be 4 sevenths, 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 the probability that the counter is green is 6 thirteenths.
Now again, that means the probability that it's red after these counters are put in the bag is going to 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 keep it as red to green, because it says find the red 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, let's 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 we're going to have to use a little bit of algebra here. So if I say, well, the ratio of red to green was 4 lots of x, a number we don't know, to 3 lots of x. Now we've got the part of the story here where more counters were put in the bag, so 2 more red and 3 more green.
So that means that red was 2 more, so we have 4x plus an additional 2, to 3x plus an additional 3. So we've got 4x plus 2 to 3x plus 3. And it says now that is equal to the ratio. Now it's equal to the ratio 7 to 6. 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 of the ratio there is equal to 7 over 6. The ratio here. So if I write that as a fraction, we get 4x plus 2. over 3x plus 3 is equal to 7 over 6. So we've got an algebraic fraction which we can now cross multiply.
So if I 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. Now I'm going to do that in one step, you can write this in brackets if you want, so actually I'll write the brackets out, we get 6 lots of 4x plus 2, timesing the 6 over, And times in the other bracket over, we get it's equal to 7 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 21x 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 obviously the question does say find the number of red counters and the number of green counters that were in the bag originally and here's that part of the process here we had 4x and 3x, 4 lots of x and 3 lots of x, so for red we have 4 lots of x so 4 lots of 3 would equal 12 and for green we had 3 lots of x and 3 lots of 3 is 9 so the number of red counters equals 12. equals 12 and the number of green counters equals 9. There we go and there's our answer 12 and 9. Okay, so here we've got some bearings, and it says it wants us to find the bearing of Cholton from Acton. Looking down here, find the bearing of Cholton 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, let's 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 are co-interior, they add up to 180. So if we do 180, take away 37, it leaves us with 143 degrees for this angle here where that letter B is sitting. Now we can work out all the angles around this point, because 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 the 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 if I label it x then I can use cosine rule to work out that angle there that 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 there for 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 swap colours. So a being our 67, x being our little a, and then 8 and 9 being our b and c. So labelling 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, so type it all in, then press square root, you get a value of x equals 9.4199. Obviously showing all these steps, but I'm going to reduce the amount of writing here just to keep the video a little bit tidier. So we have x, and I'm going to label it down here being 9.4199. 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 get rid of my A and B from the previous question. So let's get rid of that, get rid of that.
Oh, we're going to relabel 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 9.4199. So we can get rid of everything else and work out this angle. So to work out 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 come you know you could obviously use cosine rule for angles or you could use sine rule i'm going to opt to use sine rule just so something a little bit 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 going to use the sine rule so i'm going to say well okay well sine theta obviously sine a over a equals sine b over b so sine theta over nine equals sine 67 over, what is it, 9.4199.
There we go. Rearranging that, so timesing both sides by nine, we get sine theta equals nine sine 67 over 9.4199. There we go. Now, finishing this off.
If we type that into the calculator, so 9 sine 67 over 9.4199, and then do the inverse sine, so sine minus 1 of the answer there, which I'll just write sine minus 1 answer, we get the answer 61.6, with some decimal places, but it says correct to one decimal place, so I'm going to leave it there for the moment, so 61.6 degrees. There we go. So we can label that into our diagram, this will be 61.6.
So to find the full bearing of Cholton from Acton, obviously it's always from north going clockwise, so we have the 61 and we have the 37. So to finish it off we do 61.7, add 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 098.6 degrees.
Finishing off as a bearing with three figures there, but obviously to one decimal place as well. So 098.6 degrees being our final answer there. Okay, so we've got some 3D trigonometry.
Now, in order to work this out, we've just got to read this question very carefully. But it says the front length is 15 there, D to M, M to 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 2 fifths and 3 fifths. 2 fifths of 15 is 6 centimetres, and the other side, therefore, would be 9 centimetres. So I can split that line up there.
Now it wants me to find out the angle between EM, which is this line here if I draw it in, and the base or the plane. So M, if I bring that across, I can make another right angle triangle here because I can join it over to B. So I've got lots of right angle triangles. Now the first right angle triangle I need to work out here is the height or the length of EB.
Now if you imagine just on the side there we've got a right angle 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. OK, let's put our degree symbol in. So I can actually go about working the length of E to B out because I can use Sokotoa there. I can use my tan because I've got the opposite and the adjacent.
So to work that out, to work out the opposite, I'd do T times A or A times tan. So that would be 15, the adjacent side, multiplied by tan 35 and 15 tan 35. equals on the calculator 10.50311307 but I'll just leave it as that for the moment so 10.503 let's just label that on 10.503 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 angle triangle and I can actually work out the base there I'm going to do this in a different color this length just here. If I can get that length I can work out that angle again using Sokotoa but that actual length there I can do quite simply because there's a little right angle triangle on the base there and that's going from M where is it going to B 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've split up that little length there so I 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, and you get 17.4928. So there's quite a lot here, 2855. I'm just going to write some of them down, although it does go further than that.
So 17.4928, and again I can label that over there. 17.4928. 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 E and then down to the B, which is another right-angled triangle. And we've worked out the height of that triangle there is 10.503. Just worked out using Pythagoras that the base is 17.4928. And then we can now obviously work out this angle here, which I'll call x again, using Socrates 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 minus 1 to work out the angle. And that's the opposite over the adjacent. So 10.503 divided by 17.4928. And when you type that into your calculator, you get the answer. It's quite a long one.
130 point. 98130309 and then obviously just round it to however we've been asked. We've been asked to round it to one decimal place.
It's a bit of a weird one actually because when you round this to one decimal place, if we chop it there, the 9 rounds up so it becomes 31. But we need to give it to one decimal place so it'll be 31.0 degrees and that'll be our final answer there, 31.0 degrees. Okay, so we've got a circle equation and then we've got a point on the circle and it wants us to find the equation of the tangent at that point and the point isn't very nice so we've got fractions and thirds. Now it could always help here just to think of a little diagram.
So if I just draw a little basic diagram with a little circle centre 0 and we'll think about what this actually looks like. Now p on this diagram as 3 halves and root 7 over 2 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 0, 0 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 i'm going to write it as divided by so divided by 3 over 2 and that would be my gradient obviously i can 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 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 sticker obviously change the sign to negative and flip it over so negative root 3 over root 7 sorry there we go 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 equals m here being that gradient of negative 3 over root 7. So negative 3 over root 7x plus c.
Now 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 our value of x and y. So if we sub those in, y being root 7 over 2. So root 7 over 2 equals negative 3 over root 7 multiplied by the x value there, which is 3 over 2. So times 3 over 2. And then plus the c on the end, we're just trying to 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, and that equals c.
There we go. I've actually put root 3 in there, it should be 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. Negative 9 over 2 root 7. There we go.
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 add 9 over root 7 to both sides, we'll have 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 it 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 so times the top and bottom by root 7 It's nice and easy for us to add that together there So if I times the top and bottom there by root 7 over root 7 we get 7 on the top over 2 root 7 I'm going to 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 let's write it over here So c equals 16 over 2 root 7. And now we have all our pieces, we just have to put them into y equals mx plus c, and that is our equation done.
So we will have y equals negative 3 over root 7 x, make sure 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 sevens, our two root seven there. Actually just finishing this off we actually could simplify that c a little bit further, 16 and 2, so 16 on the top, 2 on the bottom, we could actually simplify that, it's not wrong as it is but we should really simplify that, so I'm going to divide the top and bottom by 2, so let's change this, there we go, so divide the top by 2 you get 8, Undivide the bottom by 2, we get root 7. There we go. So we should always look to simplify there. Just spotted that in the last minute. So what's that?
Divide by 2 and divide by 2 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 into one of the others.
But when we rearrange this bottom equation, it doesn't give us something that looks very nice. So if I rearrange this bottom one, ready to sub it into the top, I'm going to rearrange it for x as the subject. So if we take away 4y to the other side, we get 3x equals 7 minus 4y. And then divide by 3, get x equals 7 minus 4y. over 3. So not the nicest there, but now we can sub that into equation 1. So if we sub it 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 going to skip some steps. So I'm going to imagine the double bracket is 7 times 7 is 49. Then we would have minus another 28x, so minus 56y. 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. OK, so we've got a fraction involved, so we need to times everything by 9. So if we times by 9, it'll get rid of this fraction.
So we have 49 minus... 56y plus 16y squared. Now it's going to be minus 36y squared when we times that by 9. And that's now going to equal 81 when that gets times by 9 as well. So subtract 81 from both sides. So this equals 0. We'll have a nice quadratic we can solve.
So 49 take away 81 is negative 32. Still take away 56y. and then 16y squared take away 36y squared is negative 20y squared and that now equals zero. So I'm going to times this all by minus one so that we've got a positive y squared there and I'm just going to rearrange it as well so we've got the y squared at the start just prefer that for factorising so we'll have positive 20y squared we'll have positive 56y and we'll have positive 32 all equaling zero.
So I've rearranged it, but I've also times everything by minus 1 there, just to make it positive. So times minus 1, there we go. Right, now we can actually go about looking at factorising 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 14y plus 8 equals 0. And now I can go about factorising that. So I'm going to bring that up here.
We'll factorise it. So let's have a look in the brackets 5y and y. And the factors of 8, we can have 4 and 2 or 1 and 8. And we want to make 14 in the middle. I'm just 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 and 4. No, that's how it's going to work. So 20 and 4. 10 and 4, 10 and 4 works, there we go. So I can have the 5 times the 2 to make the 10, and the 4 not multiplying by anything. I want to make positive 14, so I want them both to be positive.
There we go, that's going to allow me to make 14. And that gives me my two solutions there. So y equals negative 4 fifths, and y equals negative 2. So there's my two y values. The next bit's quite nice and easy because you can do this 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 fifths, we need to do... So x will equal 7, take away 4 lots of negative 4 fifths.
You can type this all in on the calculator. Take away 4 lots of negative 4 fifths, and then divide that all by 3. And if we type that in, we get x equals 17 over 5. You could write that as a mixed number as well. You could write that as 3 and 2 fifths. So that's when y equals negative 4 fifths.
And our other solution here, which I'll try and fit in. What do we get? We get x equals, and type it 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 fifths, 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 in the start to make it y equals, and you could have done this a different way, but there's one way of actually getting your solutions for that question. All right, this is one of our vector questions here.
So it says, given that x to d, and I'm reading on a bit here, it says o to c, c to d is k to 1. So o to c, c to d, so the line is extended, there we go, to d. And it's in the ratio k to 1. Now it also says x is the midpoint of ac. So we need to draw another line in here.
Let's draw this one in in a different colour. So a to c, and x is the midpoint, let's say that's there. And it's given that the letter x. and then what it says is it says given that X to D and D is over here the one I've just drawn in is a straight line well it's a straight line from X to D and it gives us the vector there so if I draw that in X to D they're moving in that direction is 3 C minus a half a there we go and it also gives a bit more information which is on the diagram o o to a is a and o to C is C 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 root from x to d in a slightly different way. As it says that these two lines, o to c, c to d is k to 1. So it's saying that this is k and this is 1. And it wants us to find that value of k if that's 1. So let's have a look. Now we can get from x to d in a different way.
We don't actually have to follow that one line that it's given us there. It's just given us the resultant vector. But actually what I can do is I can move down here from X to C, 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 from A to C, 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 minus 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 x to c, I would times that by a half. So I could open up a bracket there, I could just times it by a half straight away, because it's not the most complicated. Half c minus a half a.
There we go. And that's my vector x to c. 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 to d has to be half c minus half a.
That's how we get from x to c, and I'm going to 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. OK, but we don't know what that is.
That's what we're going to try and solve. So let's have a look at where we're at right now. So we've got the vector x to do, we've got half c minus a half I've not written that a in there have I?
I keep missing little things out. So minus a half 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 3c minus half a and the vector we've got we've already got the minus a half a So all we have to do is match that three lots of c. Now at the moment I'm going to use a different color. We've got half 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 2.5 C's. So I've got, really, if we think of it as a bit of an equation, we've got half C plus x lots of C, and that has to equal 3C, because that's what's given to us in the equation. So x has to be 2.5, 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 2.5 as a fraction in terms of halves is 5 over 2. So x there, or this distance, has to be 5 over 2 lots of c, 5 over 2. So let's write that in, in our little ratio that it's asked for. So it's asked for the ratio k to 1. At the moment, it's 1c to 5 over 2c, or 2.5c, but 5 over 2c. Okay, so we've got it as 1 to 5 over 2. So we don't need to have those C's in there, but the ratio at the moment is 1 to 5 over 2. Okay, 1 being this distance, and then 5 over 2C being this distance here, and that's the ratio of those two lengths at the moment.
But it wants the ratio K to 1. Now the only way that I can actually get that in the ratio K to 1, I have to divide this right-hand side by 5 over 2, or divide it by 2.5, so divide it by 5 over 2. And whatever you do 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 flip it over, 1 times 2 over 5 is 2 fifths, there we go, so k will be 2 fifths, so my ratio would be 2 fifths to 1, and the final answer there, I would just say k equals 2 fifths, to finish that off, and there's my final answer. Okay, 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 C to D is A so we need to label that on the diagram So C to D is A we've got D to E is B So D to E is B and that's also split up 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 on in the question that's a midpoint We've also got f to c is a minus b, so from f to c is 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 e and follow all those vectors. So f to c is a minus b. Then we go across from c to d, so add the a, 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 the... the 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 fm such that f to x x to m is n to 1 so we can label that if we want we can say this is n this is 1 and then it says cxe is 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 straight line there and then it says work out the value of n Now obviously there's a few vectors we can work out here. I'm going to 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 along A 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 going to go for this one let's highlight it rather than doing the full line or from x to e I'm just 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 to x I'm going to have to know what f to m is so separately to this I'm going to work out what f to m is so I'll have a look at from how 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, so I'd go along A. And then I would go down the half B, so adding in half B. There we go. And if we join this all up, where the A plus the A makes 2A, so we have 2A.
And the minus B plus the half B leaves me with negative half B. There we go. So now that I can move from f to m, I can move partway along that, but we don't know what the fraction is along the line, it just says it's n to 1. So to get from f to x, let's have a look at this.
To get from f to x, I would have to do n lots of this, so n lots of 2a minus half 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 2a minus half b. Now if we expand that out, obviously we can only expand it by n, but we get 2an minus half bn. There we go, just expanding it out, and that is my vector f to x.
Now if we have a look at what we need to do next, if we have a think, c to x, so the line that we're looking for is c to f plus f to x. So we've worked that out, and we have f to x, but now we need to do c to x. Okay, that's the part of the line we're looking at.
We've got f to x, so now we can do c to x. So if I want to do c to x, let's write this out, c to 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 there so 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 to x so adding in this 2 an minus half bn so let's add that in so plus 2 an minus half bn and that there is our vector c to 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 n's with some of them, but if I group them together, so the minus a in this one and the b in the minus half b, bn part, We've got minus a plus 2an and then we've got plus b and a minus half bn. So grouping the b's and the a's together, that's what we're left with. Now we can do an extra step from here.
We can actually factorise this and just see what we get when we actually factorise out this a and this b. So if we have a look at that, if we factorise out 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 factorising out a out of the first half, what I'm going to do is I'm going to almost chop this in two. I'm going to factorise a out of the first half, factorise b out of the second half to have a look at what my multiples of a and b are there. So if we factorise a out this first bit, we get a brackets minus 1 plus 2n.
And if I factorise b out the second part, so we'd have plus b, and we factorise the b out, we get 1 minus 1 half n. 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 straight line, which it says it is, it says it's a straight line, these numbers here, or these brackets with a and b, have to be the same because it would be sort of part of or fraction of a plus b. So if we say that they are the same or equal to each other, we can write an equation. We can say minus 1 plus 2n has to be equal to this multiple of b, which is 1 minus a half n. There we go.
And if we go about rearranging this, so if I add 1 to both sides to get rid of this minus 1, and maybe add half n to both sides to get rid of the n's from this side, we would have 2.5n equals 2. And then solving that, we can divide by 2.5, and n divided by 2.5, which you can do on your calculator, n divided by 2.5, or 2 divided by 2.5, sorry, is 4 fifths. Now we're almost finished here. What we've worked out is n is 4 fifths, and that's the n just here.
So that part of the line there is 4 fifths of the line, and then this part of the line here would be 1 fifth of the line. So in terms of writing that as a ratio, if it's n to 1, well, 4 fifths to 1 fifth, which if we write that down here, 4 fifths to 1 fifth would be the ratio 4 to 1, timesing 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, n to 1. So we work out the value of n, n to 1. So n equals 4 is our final answer there for that vector question. OK, so 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 b, c are both equal, and they equal x. So that's this length here, and this length here.
We'll call those both x as it's asked us to do that. It then gives us some more information. It says about this p and q, and it says that beta p and beta q are both 10. So if we draw those in, beta p, beta q, and they are both 10 centimetres.
Just write 10 and 10. There we go. And does it 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.
There we go. So finding 30 degrees. Now first things first we've got to have a look at actually what we're looking for it says prove that cos p b q and p b q is and I've got lots of drawing here the angle between these two it says prove that cos p b q equals one minus and it just looks like a load of rubbish to be fair but we've got to just figure out how we're going to get to this point here one minus two minus root three over 200 with an x squared after it. So first things first is we can I'm just going to think about how I'm going to draw this over the top I might have to do it in a slightly different colour.
how we can find the length of AC. And just think about AC as 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 ABC and PBQ. So have a look at ABC to start with.
Now, it's told us that 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, OK? So I'm going to work out a little expression for that, and I'll do that using the cosine rule. So to do the cosine rule, it's a squared equals b squared plus e squared minus 2bc cos a. And actually, as this is non-calculator, I'm going to have to know what cos 30 is, because that's what my a is going to be, it's going to be cos 30. So obviously you need to know how to work out your exact values of trigonometry here, but cos 30, and I'll tell you what, I'll 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 a...
We see here, I'm going to call that my little a, but I'm going to say a squared equals x squared plus x squared. OK, because 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 cos 30, so times root 3 over 2. There we go, looks like a bit of a jumble 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 2x squared, which is also being times by root 3 over 2. There we go. OK, we're getting somewhere. Let's see what we can do with this now. Now, if we actually think about rewriting this bit, I could actually just write that all as a fraction there, 2x squared root 3 over 2. So I'm going to get rid of that. I'm going to write it as a full fraction, 2x squared.
What was it now? 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, and we get a squared equals 2x squared minus x squared root 3. There we go. And actually we're going to have part of, we're going to start to see part of this question now, so I can factorise that. If I factorise x squared out, 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 there.
We've got our x squared and our 2 minus root 3. So when you get to this point here, you know you're on to something. Now, the next thing to note here is that is the value of a squared. Obviously, I can't really square root that, particularly not with a calculator. It's going to be far too complicated.
But that's... length AC here is 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 that is the same as PQ squared. So if I say that I'll say AC squared is equal to PQ squared.
So I've actually got an expression for PQ squared the bottom of that other triangle that we're having a look at 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, or pq squared. So pq squared equals this x squared brackets 2 minus root 3. There we go, 2 minus root 3. Now I've actually got to figure out still what this cos of this angle is. This is the angle I'm trying to work out now, so I'm trying to work...
cos b p b 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 cos a or i could say cos p b q which is the a in this case so cos p b q equals b squared plus c squared minus a squared all over 2bc 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 cos bbq, pbq 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 because 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 because 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 10 squared 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. There we go.
Now we just need to think how we can actually make this match what's in the question there, because it says it's 1 minus that 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 r1, so we have 200 over 200, take away, and then we have x squared brackets 2 minus root 3, and that's all over 200 as well, so I've just split the fraction into two 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 said, right, we'll be 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. It 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 like I have, but just to make sure that there wasn't too much on the screen there, just showing me how we get to the final answer on that question there.