7. Three parallel lines touch a sphere with center at point O at points $K$, $L$, and $M$. It is known that the radius of the sphere is 5, the area of triangle $O K L$ is 12, and the area of triangle $K L M$ is greater than 30. Find the angle $K M L$.

Answer. $\angle K M L=\arccos \frac{3}{5}$.
Solution If a line is tangent to a sphere, then the plane passing through the center of the sphere perpendicular to this line passes through the point of tangency. Consider the plane $\alpha$, passing through point 0 perpendicular to the three given lines. The section of the sphere by the plane $\alpha$ - a circle of radius 5, passing through points $K, L$ and $M$. Denote the height of the triangle $O K L$, dropped from vertex 0 as $0 \mathrm{H}$, and $\angle \mathrm{KOL}=2 \alpha$. There are two cases.

1) Point $M$ lies on the larger arc $K L$. The area of the triangle is maximal if $M$ lies on the line $\mathrm{OH}$, and it is equal to $S_{\max }=\frac{1}{2} \mathrm{MH} \cdot \mathrm{KL}=(5+\mathrm{OH}) \mathrm{KH}=(5+5 \cos \alpha) 5 \sin \alpha=25(1+\cos \alpha) \sin \alpha$.
2) Point $M$ lies on the smaller arc $K L$. The area of the triangle is maximal if $M$ lies on the line $\mathrm{OH}$, and it is equal to $S_{\max }=\frac{1}{2} \mathrm{MH} \cdot \mathrm{KL}=(5-\mathrm{OH}) \mathrm{KH}=(5-5 \cos \alpha) 5 \sin \alpha=25(1-\cos \alpha) \sin \alpha$.

The area of the triangle $\mathrm{KOL}$ is $S_{\mathrm{KOL}}=\frac{1}{2} \mathrm{OK} \cdot \mathrm{OL} \cdot \sin 2 \alpha=\frac{25}{2} \sin 2 \alpha$. By the condition $S_{\mathrm{KOL}}=12$, so $\sin 2 \alpha=\frac{24}{25}$. Then $\cos 2 \alpha= \pm \sqrt{1-\frac{576}{625}}= \pm \frac{7}{25}$.

If $\cos 2 \alpha=-\frac{7}{25}$, then $\cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2}=\frac{9}{25}, \cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5}$. Then in the first case $\mathrm{S}_{\max }=25 \cdot \frac{8}{5} \cdot \frac{4}{5}=32$, and in the second case $-\mathrm{S}_{\max }=25 \cdot \frac{2}{5} \cdot \frac{4}{5}=8$. Only the first case is possible. Then $\angle \mathrm{KML}=\alpha=\arccos \frac{3}{5}$.

If $\cos 2 \alpha=\frac{7}{25}$, then $\cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2}=\frac{16}{25}, \quad \cos \alpha=\frac{4}{5}, \quad \sin \alpha=\frac{3}{5}$. Then in the first case $S_{\max }=25 \cdot \frac{9}{5} \cdot \frac{3}{5}=27$, and in the second case $-S_{\max }=25 \cdot \frac{1}{5} \cdot \frac{3}{5}=3$. Both options are impossible.

Therefore, $\angle K M L=\arccos \frac{3}{5}$.