What is the inclination of the line joining $(3, 0)$ and $(2, \sqrt{3}) $
Answer: $\frac{2\pi}{3}$
$m = \tan(\alpha)$
So,
$$m = \frac{y_1 - y_2}{x_1 - x_2}$$
$$\Rightarrow m = \frac{0 - \sqrt{3}}{ 3 - 2} = -\sqrt{3}$$
$$\Rightarrow -\sqrt{3} = \tan(\alpha)$$
How to get angle from negative slope?
There's no real problems here, we take \begin{align*} \theta &= \arctan(-\sqrt{3}) \\ &= -\arctan(\sqrt{3}) & \text{since } \arctan(x)=-\arctan(-x) \text{ for all } x \in \mathbb{R} \\ &= -\tfrac{\pi}{3}. \end{align*}
And, as we can see from the picture below, the angle should be negative:
The answer of $2\pi/3=\pi-|\theta|$ is instead the angle marked $\varphi$; this is the angle of inclination.