I have this diagram: And I have done these calculations: Rx=−20sin(30)+30cos(35)+80cos(45)=71.14  lbR_x=-20sin(30)+30cos(35)+80cos(45)=71.14 \; lb Ry=20cos(30)+30sin(35)−80sin(45)=−22.04  lbR_y=20cos(30)+30sin(35)-80sin(45)=-22.04 \; lb The magnitude of this resultant force is then: ∣R∣=71.142+(−22.04)2=74.48  lb|R|=\sqrt{71.14^2+(-22.04)^2}=74.48 \; lb Here's the part I need help with. I can get a theta angle for where the resultant force's angle is, but I want to figure out how one would get this angle measured from the positive x-axis: θ=tan−1(−22.0471.14)=−17.21  deg\theta={tan^{-1}{({-22.04 \over{71.14}})}}=-17.21 \; deg How can I get this angle measured from the + x-axis? I don't get this and would really appreciate some help. This comes up a lot and I do not understand it. I have also posted this angle in the engineering part of StackExchange but not many people browse that area. If that is not allowed please take my other post down, not this one, thank you.