# Calculating the components of vectors

kgianqu2
Vector A is in the direction 41.0 degrees clockwise from the y-axis. The x component of A is = -15.0 .

A)What is the y component of vector A?

B)What is the magnitude of vector A?

I got -13.0m for part a, and 19.8m for part b, but mastering physics says they are wrong. Any ideas?

Nessdude14
Can you post your attempt at the problem? It makes it easier to see where you might have made a mistake.

kgianqu2
I kind of worked backwards since I knew that the angle is 229 degrees (41 degrees clockwise from the -y axis). And I know that the x=-15. I took tan(229) which = 1.15, which should be the fraction of the y length over the x length. Since I know the x=-15 I just solved for y to get -13.0. And then I calculated the magnitude from that, which give me 19.8.

I thought this method would work, and I am not sure where I went wrong.

kgianqu2
Correction: my first post should say, 41 degrees clockwise from the -y-axis)

Muphrid
If the tangent is greater than 1, how can the y component be smaller than the x component?

Nessdude14
So you did:
$\displaystyle tan(229°)=\frac{A_y}{A_x}$

$\displaystyle A_y=A_x tan(229°)$

and you're claiming that:
$\displaystyle A_y=-15 tan(229°)=-13.0$

right? however, I get:
$\displaystyle A_y=A_x tan(229°)=-17.25$

You should try re-calculating your value for Ay.