#### Answer

The distance of the ship from point A is 39.2 km.

#### Work Step by Step

Let A be the starting point.
Let B be the point where the ship turns.
Let C be the final point.
The pints A, B, and C form a triangle.
The length of AB is 18.5 km,
The length of BC is 47.8 km.
We can find the angle of the triangle at the point B.
Angle $B = (90^{\circ}-47^{\circ})+9^{\circ} = 52^{\circ}$
We can use the law of cosines to find the length $AC$, which is the distance of the ship from point A:
$AC^2 = AB^2+BC^2-2(AB)(BC)~cos~B$
$AC = \sqrt{AB^2+BC^2-2(AB)(BC)~cos~B}$
$AC = \sqrt{(18.5~km)^2+(47.8~km)^2-(2)(18.5~km)(47.8~km)~cos~52^{\circ}}$
$AC = \sqrt{1538.23~km^2}$
$AC = 39.2~km$
The distance of the ship from point A is 39.2 km