I encountered the following definition:
$\forall tD_t=\{(x,y)\in R^2:\theta(x,y) \in [-\pi,t]\}$ where $\theta(x,y)$ is the angle in $[-\pi,\pi)$ that the vector $(x,y)$ forms with the x-axis.
Now, the goal is to calculate the area $A$ of $D_t$ relatively to a circle with radius $r$, given $t$:
$A = \frac{(t-(-\pi))\frac{r^2}{2}}{\pi r^2}=\frac{t+\pi}{2\pi}$
Well, something here is really odd to me. Why for god sake, is $t-(-\pi)$ represents this angle? I mean, aren't some details missing in order to come up with this conclusion?
Why isn't this angle is simply $t$?