#### Answer

The angle is $48.2^{\circ}$.

#### Work Step by Step

Let $R$ be the radius of the circle.
Let $h$ be the change in height of the skier from the point where the skier loses contact to the top of the sphere.
$\frac{1}{2}mv^2 = mgh$
$v = \sqrt{2gh}$
At the moment when the skier loses contact, the component of the weight directed toward the center of the circle is equal in magnitude to the centripetal force.
$mg~cos(\alpha) = \frac{mv^2}{R}$
$g~cos(\alpha) = \frac{2gh}{R}$
$cos(\alpha) = \frac{2h}{R}$
From the diagram, we can see that $cos(\alpha) = \frac{R-h}{R}$. Therefore:
$\frac{2h}{R} = \frac{R-h}{R}$
$R = 3h$
Now we can find the angle $\alpha$:
$cos(\alpha) = \frac{2h}{R} = \frac{2h}{3h} = \frac{2}{3}$
$\alpha = arccos(\frac{2}{3})$
$\alpha = 48.2^{\circ}$
The angle is $48.2^{\circ}$.