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?