# Statics problem: Finding resultant force angle measured counterclockwise from positive x-axis

I have this diagram:

And I have done these calculations: \begin{align} R_x&=-20\sin(30)+30\cos(35)+80\cos(45)=71.14\text{ lb} \\ R_y&=20\cos(30)+30\sin(35)-80\sin(45)=-22.04\text{ lb} \end{align}

The magnitude of this resultant force is then: $$|R|=\sqrt{71.14^2+(-22.04)^2}=74.48\text{ lb}$$

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 counterclockwise from the positive x-axis: $$\theta=\tan^{-1}\left(\frac{-22.04}{71.14}\right)=-17.21°$$

How can I get this angle measured counterclockwise 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.

• What do you mean by "angle measured from the + x-axis"? That result is already the angle between the horizontal (x) axis and the resultant force. – Wasabi Oct 1 '18 at 23:28
• Sorry, an important piece of info missing "counterclockwise." Please reread when I edit my post, thank you. – JustHeavy Oct 1 '18 at 23:56

$$-17.21 + 360 = 342.79°$$