From point $A$ to the right, a road branches off at an angle of $\alpha=38^{\circ} 23^{\prime}$, leading straight to bridge $C$. From point $B$, which is $5.3 \mathrm{~km}$ further along the road, a road branches off to the right, leading straight to the same bridge $C$ after $6 \mathrm{~km}$. From point $B$, a straight road branches off to the left at an angle of $\beta=48^{\circ} 21.5^{\prime}$, and after traveling $7 \mathrm{~km}$, we arrive at bridge $D$. The same river flows under both bridges, and between $C$ and $D$, the river forms a circular arc with a radius of $9 \mathrm{~km}$. A hiker starts from point $A$ at 3:55 AM, walking straight to bridge $C$ at a speed of $6 \mathrm{~km}$ per hour, and then continues by water at a speed of $10 \mathrm{~km}$ per hour to bridge $D$. When should the hiker start from $A$ to arrive at bridge $D$ at the same time as the segment, which travels through $B$ at a speed of $5 \mathrm{~km}$ per hour?
The figure shows the layout.

![](https://cdn.mathpix.com/cropped/2024_05_02_20acc02613a69218a414g-1.jpg?height=507&width=458&top_left_y=186&top_left_x=823)

In $\triangle ABC$, based on the Law of Sines,

$$
\sin \gamma = \frac{5.3 \sin 38^\circ 23'}{6}, \text{ from which } \gamma = 33^\circ 15.5'
$$

In the same triangle, using the Law of Sines again,

$$
AC = \frac{6 \sin \left[180^\circ - (\alpha + \gamma)\right]}{\sin \alpha} = \frac{6 \cdot \sin 71^\circ 38.5'}{\sin 38^\circ 23'} = 9.17 \text{ km}
$$

In $\triangle BCD$, $\angle DBC = \delta + \alpha + \gamma = 48^\circ 21.5' + 71^\circ 38.5' = 120^\circ$, and thus, by the Law of Cosines,

$$
\begin{gathered}
CD^2 = 6^2 + 7^2 - 2 \cdot 6 \cdot 7 \cos 120^\circ = \\
= 36 + 49 - 84(-0.5) = 36 + 49 + 42 = 127
\end{gathered}
$$

Let $\omega$ be the central angle corresponding to the arc $DC$,

$$
\sin \frac{\omega}{2} = \frac{\sqrt{127}}{2 \cdot 9}, \quad \text{ from which } \quad \frac{\omega}{2} = 38.76^\circ, \quad \omega = 77.52^\circ
$$

and thus

$$
DC = \frac{2r \pi \omega}{360} = \frac{18 \pi 77.52}{360} = \frac{77.52 \pi}{20} = 3.876 \pi = 12.18 \text{ km}
$$

The travel time is obtained by dividing the distance by the speed.

The travel time for the segment is therefore

$$
\frac{9.17 \text{ km}}{6 \text{ km/h}} + \frac{12.18 \text{ km}}{10 \text{ km/h}} = 1.528 \text{ hours} + 1.218 \text{ hours} = 2.746 \text{ hours},
$$

while the travel time for the battalion is

$$
\frac{5.3 \text{ km} + 7 \text{ km}}{5 \text{ km/h}} = \frac{12.3}{5} \text{ hours} = 2.46 \text{ hours}
$$

The difference in travel time is 0.286 hours $\approx 17$ minutes.

Therefore, the battalion should start 17 minutes later, at 4:12.