#### Answer

$\theta=48.2^o$

#### Work Step by Step

The person starts to leave the surface when there is no longer normal force $F_N$, leaving gravitational force to provide the only centripetal force there. From the figure below, the component of $mg$ that provides for $F_c$ is $mg\cos\theta$ $$mg\cos\theta=F_c=\frac{mv^2}{r}$$ $$g\cos\theta=\frac{v^2}{r}$$ $$v=\sqrt {gr\cos\theta}$$
The person slides down, so his KE increases while his PE decreases by the same amount according to the principle of energy conservation.
We have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m(v_f^2-v_0^2)=-mg\Delta h$$ $$(v_f^2-v_0^2)=-2g\Delta h$$
We know $v_0=0$. We take the finish point to be the point the person leaves the surface, so $v_f=\sqrt {gr\cos\theta}$. From the figure below, we also have $\Delta h=h_f-h_0=r\cos\theta-r$
$$gr\cos\theta=-2g(r\cos\theta-r)$$ $$gr\cos\theta=-2gr(\cos\theta-1)$$ $$\cos\theta=-2\cos\theta+2$$ $$3\cos\theta=2$$ $$\cos\theta=\frac{2}{3}$$ $$\theta=48.2^o$$