An [angle](https://artofproblemsolving.com/wiki/index.php/Angle) $x$ is chosen at random from the [interval](https://artofproblemsolving.com/wiki/index.php/Interval) $0^\circ < x < 90^\circ.$ Let $p$ be the probability that the numbers $\sin^2 x, \cos^2 x,$ and $\sin x \cos x$ are not the lengths of the sides of a triangle. Given that $p = d/n,$ where $d$ is the number of degrees in $\text{arctan}$ $m$ and $m$ and $n$ are [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) with $m + n < 1000,$ find $m + n.$
Note that the three expressions are symmetric with respect to interchanging $\sin$ and $\cos$, and so the probability is symmetric around $45^\circ$.  Thus, take $0  \sin^2 x + \sin x \cos x\]
This is equivalent to 
\[\cos^2 x - \sin^2 x > \sin x \cos x\]
and, using some of our [trigonometric identities](https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities), we can re-write this as $\cos 2x > \frac 12 \sin 2x$. Since we've chosen $x \in (0, 45)$, $\cos 2x > 0$ so 
\[2 > \tan 2x \Longrightarrow  x < \frac 12 \arctan 2.\]
The [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that $x$ lies in this range is $\frac 1{45} \cdot \left(\frac 12 \arctan 2\right) = \frac{\arctan 2}{90}$ so that $m = 2$, $n = 90$ and our answer is $\boxed{092}$.