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I have this diagram:

enter image description here

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.

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  • $\begingroup$ 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. $\endgroup$
    – Wasabi
    Oct 1 '18 at 23:28
  • $\begingroup$ Sorry, an important piece of info missing "counterclockwise." Please reread when I edit my post, thank you. $\endgroup$
    – JustHeavy
    Oct 1 '18 at 23:56
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I haven't bothered checking your math since that doesn't seem relevant to your question.

Now, assuming that your answer of -17.21° is correct, that is in fact the counter-clockwise angle. It means that you need to go negative 17.21° counterclockwise, which is equivalent to rotating 17.21° clockwise.

Now, if what you actually mean is how to get the positive angle equivalent to -17.21°, well, that's just basic geometry: a circle has 360°. So to go from a negative angle to the equivalent positive angle, just add 360°.

$$-17.21 + 360 = 342.79°$$

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