Question

HELP. A vector with <-2,-5>, find the nagnitude and directional angle

Answer 1

magnitude is sqrt((-2)^2+(-5)^2) = sqrt(29)
angle is arctan(5/2) but since its in the 3rd quad we add pi to the answer so
arctan(5/2)+pi

Answer 2

OK, i knew how to find that but my real question is regarding that part where you add pi, how do you know what to add and subtract to get the givn quadrant

Answer 3

well arctan gives and angle between -pi/2 to pi/2 but from -2,-5 we know the point lies in the 3rd quad. but we know our angle will be in the first quad (ask me how if needed) so we add pi to make it be the "same" angle in the 3rd quad

Answer 4

my book says to do <-2,-5>= <|v|cos B, |v|sinB> and so -2=sqrt(29)cosB

so cosB= (-2/sqrt(29))

Answer 5

and to get the answer they did 360-(cos^-1(-2/sqrt(29))

Answer 6

there are many ways but I think they way I did it was easy

Answer 7

Yeah, maybe I'm just overthinking it, but I like to understand why the book does what it does, helps me learn better

Answer 8

arctan(5/2)+pi = 2pi-(cos^-1(-2/sqrt(29))

Answer 9

ok well as long as you get it...

Answer 10

ok and just for it how would you find it for sin? Ik it would be sin^1(-5/sqrt29) now what do i add or subtract to get the answer?

Answer 11

im confused about your question. your question says nothing about sin

Answer 12

ok back to where i said <-2,-5>= <|v|cos B, |v|sinB> well remember how we did cos to find the answer? well i wqas asking how do we do sin the find the answer

Answer 13

so -5=sqrt29sinB
(-5/sqrt29)=sinB
sin^1(-5/sqrt29)=B

Answer 14

pi-sin^-1(-5/sqrt(29))

Answer 15

we will get a negative angle out of the arcsin so we need to add the absolute value of that to pi

Answer 16

pi+|pi-sin^-1(-5/sqrt(29))| = pi-sin^-1(-5/sqrt(29))

Answer 17

ok, thanks