11. Ans: 1
$A E=A B \cos 60^{\circ}, \angle A H E=\angle A C B$. Hence
$$
A H=A E / \sin \angle A H E=A B \cos 60^{\circ} / \sin \angle A C B=2 R \cos 60^{\circ}
$$
where $R=1$ is the radius of the circle. Therefore, $A H=1$.
11. A triangle $\triangle A B C$ is inscribed in a circle of radius 1 , with $\angle B A C=60^{\circ}$. Altitudes $A D$ and $B E$ of $\triangle A B C$ intersect at $H$. Find the smallest possible value of the length of the segment $A H$.