# Get vector components from from magnitude and angle

I am given the length and the direction of a vector, and I need to get the the X,Y components. I can go one way, but going the other has me a little lost.

Example:

A man walks 3.50 m south, then 8.20 m at an angle 24.6 degrees north of east, and finally 15.0 m west. What is the magnitude of the man's total displacement (m)? What is the direction of the man's total displacement where directly east is taken as zero degrees and counter-clockwise is positive (degrees)?

I'm pretty sure that the answer will be $$\langle 0, -3.5\rangle + \langle \text{?}, \text{?}\rangle + \langle-15, 0\rangle$$ where $\langle\text{?},\text{?}\rangle$ is the components of $\lVert v\rVert = 8.2$ and $\text{angle} = 24.6$. From there I can find the magnitude and angle of the final vector which should be the answers to the 2 questions. I'm just struggling with getting the components.

-
You'll want to look up the conversion formulae between Cartesian and polar coordinates. – J. M. Apr 25 '11 at 16:59
Ah, thank you! That was it :) – joe_coolish Apr 25 '11 at 17:01

Here is a method to find the components. Sketch the northeast vector on coordinate axes with initial point at the origin, with east being the positive $x$-axis. Drop a vertical line segment from the end of this vector to the $x$-axis. Then you have a right triangle; the base of the triangle lies on the positive $x$-axis, the hypotenuse is $8.2$, and the angle from the base to the hypotenuse is $24.6^\circ$. The base length and height of this triangle are your components, and you can write these down using sine and cosine.