# In this vector problem, what would theta equal?

1. 95 m/s South + 435 m/s at 20 degrees North of East

2. Addition of vectors, and sin, cos, and tang.

3. For vector one I got 0 as my x component, and -95 as my y component. For vector 2 I got 408.8 m/s (x) and 148.8 m/s (y).

cos(20)= x2/435
x2=408.8

sin(20)=y2/435
y2=108.8

The resultant vector is 408.8 (x component) 53.8 (y component).

The hypotenuse of that triangle (r) = square root 408.8^2 + 53.8^2
r=412.3

When I did the math, I got 7.5 degrees as θ, but now I have no idea how I got that, because by using tangent I get .022. How do I get the degree of the angle?

## Answers and Replies

You shouldn't need to worry about the hyp to get the angle of the resultant vector. Once you sum up your y components (148.8-95) you have two vectors, one north at 53.8 and the other east at 408.8. You should be able to just use arctan to find the angle theta measured north from the horizontal. That is likely how you got the 7.5.

Thanks so much QuarkCharmer! I completely forgot how I did it. I was so confused!