I'm going to answer this question with an algorithm since it sounds as if it is going to be used in a computer program.
So I'll treat all variables as if they were variables in a computer program stored in memory.
I assume that $\theta_{min} \leq \theta_{max}$ and that $0 \leq \theta_{max} - \theta_{min} \leq 2 \pi$.
If $\theta_{min} = \theta_{max}$ then this is the question of whether the ray intersects a line segment (if $r_{min} < r_{max}$) or whether it intersects a point (if $r_{min} = r_{max}$) where both of these questions have well known solutions. So we will henceforth that this is not the case and therefore assume that $\theta_{min} < \theta_{max}$.
Suppose that the equation of the (bidirectional) line formed by the ray is $a x + b y + c = 0$ with $a \geq 0$ and that the ray starts at $(x_0, y_0)$.
(1) Shift everything by $-P$: Replace $(x_0, y_0)$ with $(x_0 - p_x, y_0 - p_y)$ and replace $c$ with $c + a p_x + b p_y$. We may now assume that $P$ is the origin $P = (0, 0).$
(2a) If the ray is horizontal (i.e. if $a = 0$) then replace $c$ with $\frac{c}{b}$ and replace $b$ with $1$ (i.e. divide the equation $a x + b y + c = 0$ by $b$).
(2b) If the ray is not horizontal (i.e. if $a \neq 0$) then rotate everything so that it is horizontal:
let $\theta_0 = \arctan\left( - \frac{b}{a} \right)$ and $\theta_1 = \frac{\pi}{2} - \theta_0$. Add $\theta_1$ to $\theta_{min}$ and to $\theta_{max}$ (i.e. replace $\theta_{min}$ by $\theta_{min} + \theta_1$ and replace $\theta_{max}$ by $\theta_{max} + \theta_1$). Replace $(x_0, y_0)$ with its rotation about the origin $\left( x_0 \cos \theta_1 - y_0 \sin \theta_1, x_0 \sin \theta_1 + y_0 \cos \theta_1 \right)$. Let $d = \frac{\left| c \right|}{\sqrt{a^2 + b^2}}$, which is the distance from the line to the origin. Replace $a$ with $0$, $b$ with $1$, and $c$ with $d$.
At this point, $a = 0,$ $b = 1,$ and $P = (0, 0)$.
(3) Let $h$ be the smallest $y$-value of the wedge and let $H$ be the largest, which we find as follows:
(3a) Case $\theta_{max} = \theta_{min} + 2 \pi$:
Let $h = -r_{max}$ and $H = r_{max}$.
(3b) Case $\theta_{max} \neq \theta_{min} + 2 \pi$:
Let $$S = \{ r_{min} \sin\left( \theta_{min} \right),\; r_{max} \sin\left( \theta_{min} \right),\; r_{min} \sin\left( \theta_{max} \right),\; r_{max} \sin\left( \theta_{max} \right) \}$$
be the $y$-values of the four extreme points (i.e. "corners") of the wedge.
If $\theta_{min} \leq \frac{\pi}{2} \leq \theta_{max}$ then let $H = r_{max}$ and otherwise let $H = \max S$.
If $\theta_{min} \leq \frac{3 \pi}{2} \leq \theta_{max}$ then let $h = -r_{max}$ and otherwise let $h = \min S$.
(4) The (bidirectional) line formed by the ray intersects the wedge if and only if $h \leq c \leq H$. If this is false then the ray doesn't intersect the wedge and we're done. So assume otherwise. Given that the ray is horizontal, it is now straightforward to determine whether or not the ray intersects the wedge based on the ray's starting point and its direction (if the horizontal ray "points away from the wedge" then it doesn't intersect the wedge, otherwise it does intersect the wedge).