A ship is sailing east. At one point, the bearing of a submerged rock is $48^{\circ} 20^{\prime}$. After the ship has sailed $13.6 \mathrm{mi}$, the bearing of the rock has become $308^{\circ} 40^{\prime}$. Find the distance of the ship from the rock at the latter point.
The distance is approximately $\mathrm{mi}$
(Do not round until the final answer. Then round to the nearest tenth as needed.)

Step 1 ：Set up a coordinate system with the initial position of the ship at the origin. The bearing of the rock is $48^\circ 20^\prime$, which is equivalent to $48 + \frac{20}{60} = 48.33^\circ$ to the north of east. Therefore, the position of the rock is $(x, x\tan(48.33^\circ))$ for some positive $x$.

Step 2 ：After the ship has sailed $13.6$ miles to the east, its position is $(13.6, 0)$. The bearing of the rock has become $308^\circ 40^\prime$, which is equivalent to $360 - 308 - \frac{40}{60} = 51.33^\circ$ to the north of west. Therefore, the position of the rock is also $(13.6 + y, y\tan(51.33^\circ))$ for some positive $y$.

Step 3 ：Equating the two expressions for the position of the rock, we get $x = 13.6 + y$ and $x\tan(48.33^\circ) = y\tan(51.33^\circ)$.

Step 4 ：Solving these two equations, we get $y = \frac{x\tan(48.33^\circ)}{\tan(51.33^\circ) + \tan(48.33^\circ)} - 13.6$.

Step 5 ：Substituting $y$ into the first equation, we get $x = 13.6 + \frac{x\tan(48.33^\circ)}{\tan(51.33^\circ) + \tan(48.33^\circ)} - 13.6$.

Step 6 ：Solving this equation, we get $x = \frac{13.6(\tan(51.33^\circ) + \tan(48.33^\circ))}{1 - \tan(48.33^\circ)}$.

Step 7 ：Therefore, the distance of the ship from the rock at the latter point is $y = \sqrt{x^2 + (13.6 + y)^2} = \sqrt{\left(\frac{13.6(\tan(51.33^\circ) + \tan(48.33^\circ))}{1 - \tan(48.33^\circ)}\right)^2 + (13.6 + \frac{x\tan(48.33^\circ)}{\tan(51.33^\circ) + \tan(48.33^\circ)})^2}$.

Step 8 ：Calculating this expression, we get $y \approx \boxed{8.6}$ miles.