Question

consider the vector v=(-6,13)
Write v in the form (absolutevalueof v costheta, absolutevalueof v sintheta) Express the angle theta in degrees.

Answer 1

What have you tried so far?

Answer 2

http://www.wolframalpha.com/input/?i=sqrt250%28-6costheta%2B13sintheta%29
@Limitless

Answer 3

How did you get this?

Answer 4

idk i tried plugging in stuff:/
i need help @Limitless

Answer 5

\[v=|v||v|\cos (\theta) \sin (\theta)=|v|^2\cos(\theta) \sin (\theta)\]
I'm not really sure about the angle.. I know the length is just \(\sqrt{v_1^2+v_2^2}\) where \(v_1\) and \(v_2\) are the components of \(v\).

What is the other vector which determines \(\theta\)?

Answer 6

yeah thats the problem, i only finding the length..idk how to get the angle either

Answer 7

There are two possible options that I can see: The angle between \(v\) and \(o_y=(0,13)\) or the angle between \(v\) and \(o_x=(-6,0)\). Let me look this up some more and I'll try to get back to you.

Answer 8

ok...

Answer 9

Wait a moment. Are we supposed to write \(v\) in *this form*:
\[v=(|v|\cos(\theta),|v|\sin(\theta))\]
?????????

Answer 10

yes

Answer 11

Then this is easy, lol.

Answer 12

Just solve these two equations:
\[
|v|\cos(\theta)=-6 \text{ and } |v|\sin(\theta)=13
\]

Answer 13

@Limitless like this?
http://www.wolframalpha.com/input/?i=%28sqrt250%29costheta%3D-6

Answer 14

It is quicker to just take the arccos of \(\frac{-6}{|v|}\). Similarly, take the arcsin of \(\frac{13}{|v|}\). Your approach is correct, but I prefer to do this because W|A gives me the angles immediately and I don't need to convert. By the way, that \(2\pi n\) that is included is just included because of the periodic nature of \(\cos(x)\). It is standard to take the solution when \(n=0\).

Answer 15

so then the answer to part 2 would be (sqrt250cos(112.3)sqrt250sin(34.7)? @Limitless

Answer 16

Careful. You need the comma. You're representing \(v\) as an ordered pair. Also, it should be \(\sin(55.3^{\circ})\). So you have:
\[v=\left(5\sqrt{10}\cos\left(112.3^{\circ}\right), \, 5\sqrt{10}\sin\left(55.3^{\circ}\right)\right)\]

Answer 17

u the bomb @Limitless ! oneLAST question
What are the fourth roots of sqrt 3 + i