# Why is this angle taken as negative?

Question:

The resultant of two concurrent forces, $$\vec{P}$$ and $$\vec{Q}$$, $$[P>Q]$$, trisects the angle between them. Show that the angle between them is $$3\cos^{-1}\left(\frac{P}{2Q}\right)$$ and the magnitude of the resultant is equal to $$\frac{P^2-Q^2}{Q}$$.

My book's attempt:

$$R\cos(0^{\circ})=P\cos(-\alpha)+Q\cos(2\alpha)\tag{1}$$

$$[\text{Breaking the convention,}\ R\cos(0^{\circ})=P\cos(\alpha)+Q\cos(-2\alpha)]$$

$$R\sin(0^{\circ})=P\sin(-\alpha)+Q\sin(2\alpha)\tag{2}$$

$$[\text{Breaking the convention,}\ R\sin(0^{\circ})=P\sin(\alpha)+Q\sin(-2\alpha)]$$

Using $$(2)$$,

$$0=P\sin(\alpha)-Q\sin(2\alpha)$$

$$P\sin(\alpha)=Q\sin(2\alpha)$$

$$P\sin\alpha=2Q\sin\alpha\cos\alpha$$

$$\cos\alpha=\frac{P}{2Q}$$

$$\alpha=\cos^{-1}\left(\frac{P}{2Q}\right)$$

$$...$$

In $$(1)$$ and $$(2)$$, why is $$\alpha$$ positive and $$2\alpha$$ negative? I know that going counterclockwise from the x-axis is positive and going clockwise is negative, but in $$(1)$$ and $$(2)$$, both seem to be going in the same direction with respect to $$OA$$, so why are their signs different?
1. Why are the signs of $$\alpha$$ and $$2\alpha$$ different?