# Calculating the components of vectors

1. Sep 20, 2012

### 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?

2. Sep 20, 2012

### Nessdude14

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

3. Sep 20, 2012

### 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.

4. Sep 20, 2012

### kgianqu2

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

5. Sep 20, 2012

### Muphrid

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

6. Sep 20, 2012

### 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.