Formula for the angle of a line $y = mx$ as a function of $m$. I was wondering if there was a way to calculate the angle made by a line $(\space y=mx)$ in the Cartesian plane using only $m$. I used the Pythagorean theorem in this figure:
$$AO= \sqrt{AB^2+OB^2}=\sqrt{x^2+m^2x^2}=x \sqrt{1+m^2}$$
Now I know that $\alpha = \cos^{-1} (\cos \alpha) $.
$$\cos \alpha = \frac{OB}{OA}=\frac{x}{x \sqrt{1+m^2}}=\frac{1}{\sqrt{1+m^2}}$$
$$\alpha = \cos^{-1} \left(\frac{1}{\sqrt{1+m^2}}\right)$$
Is this correct? Is there an easy way to solve this?

Looks good! Alternatively, notice that:
$$
m = \frac{y}{x} = \tan \alpha
$$
So we have:
$$
\alpha = \tan^{-1}(m) = \tan^{-1}(0.5) \approx 26.57^\circ
$$
