Normal point above a 60 degree line given positive Y

Let $X$ and $Y$ be independent standard normal variables. Given that $Y>0$, what is the probability that $Y>\sqrt3\,X$?

$\tfrac23$. The joint density is rotationally symmetric, so probabilities of wedges are proportional to angles. $Y>0$ is the angular range $(0^\circ,180^\circ)$ and $Y>\sqrt3X$ within it is $(60^\circ,180^\circ)$, giving $\tfrac{120}{180}=\tfrac23$.