A circle with radius 2 has a regular hexagon $A B C D E F$ inscribed in it. From point $K$, lying on the extension of side $A F$ such that $K A<K F$ and $K A=\sqrt{11}-1$, a secant $K H$ is drawn, intersecting the circle at points $N$ and

H. It is known that the external part of the secant $K N$ is equal to $2(K N=2)$, and angle $N F H$ is obtuse. Find the angle $H K F$.
$K A \cdot F K=K N \cdot K H$.

Let $O$ be the center of the circle; $P, Q$ be the midpoints of chords $N H$ and $A F$. Since $K A \cdot F K=K N \cdot K H$, then

$$
(\sqrt{11}-1)(\sqrt{11}+1)=2(2+N H)
$$

From this, we find that $N H=3, K H=K N+N H=5$.

From the right triangles $H P O$ and $F Q O$, we find that

$$
\begin{aligned}
& O P=\sqrt{O H^{2}-P H^{2}}=\sqrt{4-\frac{9}{4}}=\frac{\sqrt{7}}{2} \\
& O Q=\sqrt{O A^{2}-A Q^{2}}=\sqrt{4-1}=\sqrt{3}
\end{aligned}
$$

We express the segment $P Q$ using the cosine theorem from triangles $K P Q$ and $O P Q$ and solve the obtained equation for the cosine of the desired angle.

Let $\angle P K Q=\angle P O Q=\varphi$. Then

$$
K P^{2}+K Q^{2}-2 K P \cdot K Q \cos \varphi=O P^{2}+O Q^{2}-2 O P \cdot O Q \cos \varphi
$$

Therefore,

$$
\begin{gathered}
\cos \varphi=\frac{K P^{2}+K Q^{2}-O P^{2}-O Q^{2}}{2(K P \cdot K Q-O P \cdot O Q)}= \\
=\frac{\frac{49}{4}+11-\frac{7}{4}-3}{2\left(\frac{7}{2} \cdot \sqrt{11}-\frac{\sqrt{7}}{2} \cdot \sqrt{3}\right)}=\frac{37}{2(7 \sqrt{11}-\sqrt{21})}=\frac{7 \sqrt{11}+\sqrt{21}}{28}
\end{gathered}
$$

Second method.

Let $O$ be the center of the circle; $P, Q$ be the midpoints of chords $N H$ and $A F$. Since $K A \cdot F K=K N \cdot K H$, then

$$
(\sqrt{11}-1)(\sqrt{11}+1)=2(2+N H)
$$

From this, we find that $N H=3, K H=K N+N H=5$. We then sequentially find:

$$
\begin{gathered}
O Q=\sqrt{K Q^{2}+O Q^{2}}=\sqrt{K O^{2}-K P^{2}}=\frac{\sqrt{7}}{2} \\
\sin \angle Q K O=\frac{\sqrt{3}}{\sqrt{14}}, \sin \angle P K O=\frac{1}{2 \sqrt{2}}, \cos \angle Q K O=\frac{\sqrt{11}}{\sqrt{14}} \\
\cos \angle P K O=\frac{\sqrt{7}}{2 \sqrt{2}}
\end{gathered}
$$

Therefore,

$$
\sin \angle H K F=\sin (\angle Q K O-\angle P K O)=
$$

$$
=\frac{\sqrt{3}}{\sqrt{14}} \cdot \frac{\sqrt{7}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}} \cdot \frac{\sqrt{11}}{\sqrt{14}}=\frac{7 \sqrt{3}-\sqrt{77}}{28}
$$

$$
\arccos \frac{37}{2(7 \sqrt{11}-\sqrt{21})}=\arccos \frac{7 \sqrt{11}+\sqrt{21}}{28}=\arcsin \frac{7 \sqrt{3}-\sqrt{77}}{28}
$$