Given a square $A B C D$ with side $4 \sqrt{2}$. Point $O$ is chosen in the plane of the square such that $O B=10, O D=6$.

Find the angle between the vector $\overrightarrow{O B}$ and the vector directed from point $O$ to the farthest vertex of the square from it.
From the condition, it is clear that $B D=8$. Therefore, triangle $B D O$ is a right triangle.

Let points $O$ and $C$ lie on opposite sides of line $B D$. Then $O C^{2}=(6+8 / 2)^{2}+4^{2}=116>O B^{2}$ (that is, vertex $C$ is the farthest from $O$). By the cosine rule,

$$
\cos \angle B O C=\frac{O B^{2}+O C^{2}-B C^{2}}{2 O B \cdot O C}=\frac{100+116-32}{40 \sqrt{29}}=\frac{23}{5 \sqrt{29}}
$$

## Answer

$\arccos \frac{23}{5 \sqrt{29}}$