# Homework Help: Vector Degrees Problem

1. Jan 20, 2009

### Bottomsouth

1. The problem statement, all variables and given/known data

Ax= -4.00 , Ay= -8.00

2. Relevant equations

Theta = ______ degrees

3. The attempt at a solution

I tried inverse of tangent (-8/-4) = 63.4 degrees for theta, but its telling me I am wrong. Tried the degree in negative and the computer showed up wrong also. Any Ideas?

2. Jan 20, 2009

### gabbagabbahey

Hmmm... well just looking at Ax and Ay, what quadrant does A lie in? Does theta=63.4 degrees lie in that quadrant?

The problem is that the inverse tangent of (Ay/Ax) always returns a value between 0 and 180 degrees. But the tangent function has a period of 180 degrees, so tan(theta)=tan(theta+n*180)=Ay/Ax implies that theta=ArcTan(Ay/Ax)-n*180 for some integer value of n and not just theta=ArcTan(Ay/Ax).

3. Jan 20, 2009

### cordyceps

Remember the unit circle. The angle is in the third quadrant.

4. Jan 20, 2009

### Bottomsouth

It is in the 3rd quadrant making it positive for tangent. Just what would the variable n be?

Thanks for the help, appreciate it

5. Jan 20, 2009

### Bottomsouth

Got it now, since its int he 3rd quadrant we add 180 degrees. I knew it was something so simple. so its 243.4 degrees for theta.

Thanks,

6. Jan 20, 2009

### cordyceps

Choose n such that your angle is in the correct quadrant. In this case, what are the limits of n?

7. Jan 20, 2009

### gabbagabbahey

Exactly; A is in the 3rd quadrant.

The tangent is positive in both the third quadrant and the first quadrant, and the arctangent always returns a value in the first to quadrants. That is why you are getting an angle that is in the first quadrant.

Since A is in the 3rd quadrant, 'n' will be any integer that gives a theta in between 180 degrees and 270 degrees. In this case, n=-1 should do nicely.

8. Jan 20, 2009

Welcome