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ketta left camp three days ago on a journey into the jungle the three days of his journey can be described by displacement distance and the direction distance and direction vectors or displacement vectors and displacement is distance with the direction and the vectors are d1 d2 and d3 and they list them right over here the distances are given in kilometers how far is Keita from camp at the end of day three so let's just think about what's happening on day one let's say this is his starting point his displacement he starts here and he goes there but if you break it down based on I guess if you call this direction the hole if you call this direction actually drawn them a compass here if you say that this is north this is east this is west and that this is South you can break it down by how much he went in the east direction and how much he went in the north direction so this is saying he went seven in the east direction one two three four five six seven seven in the east direction and he went eight in the north direction one two three four five six seven eight just like that this is 7 and this is 8 and then on day two he went six in the east in the six units I guess these are kilometres six kilometers to the east one two three four five six and he went two by two kilometers to the north one two so he ends up right over here six and two and then finally day three the component of his displacement that is to the east is two and the component of his displacement that's to the north is nine is nine and so to figure out how far he is it from camp at the end of day three we just have to figure out what is his total displacement what is the length of the vector that's the sum of all of these so what is the length of this vector right over here and we could call this let's call this D sub T for total displacement total total displacement and you can see how this is arranged that our total displacement vector is just going to be the sum of d1 let me do those in the appropriate colors it's just going to be the sum of D 1 D 1 D 2 D 2 and D 3 and D 3 and if you're summing these vectors you can just add the corresponding components so for example the total displacement is going to be equal to the sum of the horizontal are not I should say the East the displacements in the east direction so it's going to be 7 plus 6 7 plus 6 plus 2 plus 2 and then in the north direction you have 8 plus 6 or 7 or 8 plus 2 plus 9 plus 9 and so our total displacement vector if we write it in this form is going to be this is 13 plus 2 it's going to be 15 and the to the east or the component of displacement in the eastern direction is 15 and in the northern direction is 10 plus 9 19 so that's this vector right over here it's 15 it's complete its component to the in the east is 15 and to the north is 19 is 19 so let me make that clear so this this distance right over here or this if I were to make this a triangle the length of this side of the triangle here is 19 and that's the total his total displacement in the northern direction and his total displacement in the his total displacement in the eastern direction is that 15 is that 15 so what's going to be the length of our displacement vector or what's the magnitude of our displacement vector the magnitude of our total displacement is let's Pythagorean theorem this is a right triangle 15 squared plus 19 squared is going to be the magnitude squared or we could say that the magnitude is equal to the square root of 15 squared plus 19 squared plus 19 squared so let me get my calculator out all right so I have the square root of 15 squared plus 19 squared gives 24 let's see they want us to round to the nearest tenth twenty four point two so twenty four point two kilometers at the end of day three round to the nearest degree your answer should be between zero and 180 degrees so distance I'm assuming and this would just be the convention they should be a little bit more precise here in their wording or a little bit less ambiguous is the convention is the angle relative to if we're thinking of the coordinate axes the positive x-axis on this it would be you could say the eastern direction so it's really this angle right over here let's call that theta and how could we figure out what theta is we it's part of this right triangle we know this side has like nineteen we know this side has 15 so we know the opposite side to theta and we know the adjacent side to theta so what what what trig function involves opposite in adjacent well tangent involves opposite opposite and adjacent so we could write that the tangent of theta tangent of theta is equal to nineteen over fifteen is equal to opposite over adjacent nineteen over fifteen and to solve for theta we can this say and we got to make sure that our angle if we just take the inverse tangent here is going is actually the angle we're looking for but inverse tangent the convention is it'll give you an angle that is between negative PI over two and pi if we're thinking in radians but since we're thing in degrees it'll give you an angle between negative 90 degrees and 90 degrees and this angle is clearly in that it's it's actually between 0 and 90 degrees so when we take inverse tangent we know that we're going to get the right angle otherwise we would have to make some adjustments so theta is going to be equal to the inverse tangent of 19 over 15 of 19 over 15 which is equal to let's get our calculator out let's make sure that we are in yep we are in degree mode and so let's take the inverse tangent of 19 divided by 15 which is equal to let's see they want us to round to the nearest degree so 52° if we round to the nearest degrees 52 degrees which also looks right right over here this looks like it looks like a little bit more than a 45 degree angle which 52 degrees seems to fit the bill and we are done