= Solution
Set $\alpha=\arctan c\in(0,\pi/2)$. In <polar coordinates>, the event is $\sin\Theta>c|\cos\Theta|$, the angular sector
$$
\alpha<\Theta<\pi-\alpha.
$$
Because $\Theta$ has the <continuous uniform distribution> on an interval of length $2\pi$, the <probability> is the sector's angular length divided by $2\pi$:
$$
\boxed{\mathbb P(Y>c|X|)=\frac{\pi-2\arctan c}{2\pi}
=\frac12-\frac1\pi\arctan c.}
$$
The <symmetry> of the <standard normal distribution> in $Y$ makes the lower sector $Y<-c|X|$ equally likely. Their disjoint union is $|Y/X|>c$, apart from $X=0$, an event of <probability> zero. Therefore the requested <cumulative distribution function> is
$$
\boxed{F_{|Y/X|}(z)=\begin{cases}0,&z<0,\\[2pt]
\dfrac2\pi\arctan z,&z\geq0.
\end{cases}}
$$
This is the unit-scale <half-Cauchy distribution>. At $z=0$ there is no atom, and as $z\to\infty$ the <cumulative distribution function> tends to one.
\Image[/past-exam-of-the-probability-course-of-the-university-of-oxford/2015-density-regions.png]
{title=Gaussian angular probability and the support of a transformed exponential pair}
The left panel uses $c=1$, giving a sector of angle $\pi/2$ and <probability> $1/4$. The right panel displays the domain $u<v<2u$ for the next <change of variables for probability densities>; a vertical section through it determines the conditional range.
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