[ Intersecting lines, angle between them]

In rectangle $A B C D$, the sides are given as $A B=3, B C=4$. Point $K$ is at distances $\sqrt{10}$, 2, and 3 from points $A, B$, and $C$ respectively. Find the angle between the lines $C K$ and $B D$.
In the plane $A B C D$, consider the rectangle $B C P Q$, symmetric to the given rectangle $A B C D$ with respect to the line $B C$. Let $K Q=x$. The segment $K B$ is the median of triangle $A K Q$. By the formula for the median,

$$
K B 2=\frac{1}{4}(2 K A 2+2 K Q 2-A Q 2)
$$

or

$$
4=\frac{1}{4}(2 \cdot 10+2 \cdot x 2-36)
$$

from which we find that $x=4$. In triangle $K Q C$, it is known that

$$
C Q=B D=5, K C=3, K Q=4 .
$$

Thus, triangle $K Q C$ is a right triangle,

$$
\angle C K Q=90^{\circ}, \sin \angle K C Q=\frac{K Q}{C Q}=\frac{4}{5}
$$

Since $C Q \| B D$, the angle between the lines $C K$ and $B D$ is equal to the angle $K C Q$, i.e., $\arcsin \frac{4}{5}$.

## Answer

 $\arcsin \frac{4}{5}$