3.5. Two submarines, without moving, observe, simultaneously, a ship that is moving uniformly in a straight line at a speed of $30 \mathrm{~km} / \mathrm{h}$ for one minute. If the first submarine has turned its periscope by an acute angle $\alpha$, and the second by an acute angle $\beta$, what is the maximum distance between the submarines?

## fourth grade
3.5. Let $P Q$ be the segment that the ship has traveled in the designated minute, and let $S$ be its midpoint. Submarines $A$ and $B$ are located on the arcs $k_{1}$ and $k_{2}$ from which $P Q$ is seen at angles $\alpha$ and $\beta$ respectively. It is easy to show that the distance $A B$ is greater in the case where $k_{1}$ and $k_{2}$ are on different sides of the line $P Q$. Let $O_{1}$ and $O_{2}$ be the centers of the circles to which $k_{1}$ and $k_{2}$ belong, and let $A_{0}$ be the intersection of the line $S O_{1}$ with $k_{1}$ and $B_{0}$ the intersection of the line $\mathrm{SO}_{2}$ with $k_{2}$.

\[
\begin{gathered}
|A B| \leq|A S|+|S B| \leq\left|A O_{1}\right|+\left|O_{1} S\right|+\left|S O_{2}\right|+\left|O_{2} B\right|= \\
=\left|A_{0} S\right|+\left|S B_{0}\right|=\left|A_{0} B_{0}\right|= \\
=\operatorname{ctg} \frac{\alpha}{2}|P S|+\operatorname{ctg} \frac{\beta}{2}|P S|=\left(\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\beta}{2}\right) 250 \mathrm{~m} \\
\text { fourth grade }
\end{gathered}
\]