Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$. Let $O$ and $P$ be two points on the plane with $OP = 200$. Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
Noting that $\angle OQP$ and $\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$, if $\overline{QR} \leq 100$, then $\overarc{QR}$ must be less than or equal to $60^{\circ}$.
This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:
Given $a, b$ such that $0<a, b<75$, what is the probability that $|a-b| \leq 30$? 
Through simple geometric probability, we get that $P = \frac{16}{25}$.
The answer is $16+25=\boxed{041}$
~IYN~
Note: The Geometric probability can be easily be found through graphic through the x-y plane.