Answer
The distance between the ship and the rock when the ship is at its final position is $~~10.8~miles$
Work Step by Step
Let $A$ be the initial position of the ship.
Let $B$ be the final position of the ship.
Let $C$ be the position of the rock.
Then the points $A,$ $B,$ and $C$ form a triangle.
The angle at point $A$ is $90^{\circ} - 45^{\circ}20' = 44^{\circ}40'$
The angle at point $B$ is $308^{\circ}40'-270^{\circ} = 38^{\circ}40'$
The angle at point $C$ is $180^{\circ}-44^{\circ}40'-38^{\circ}40' = 96^{\circ}40'$
We can find $b$, the side of the triangle subtended by angle $B$:
$\frac{b}{sin~B} = \frac{c}{sin~C}$
$b = \frac{c~sin~B}{sin~C}$
$b = \frac{(15.2)~sin~38^{\circ}40'}{sin~96^{\circ}40'}$
$b = 9.56~mi$
We can use the law of cosines to find $a$, the distance between the ship and the rock when the ship is at its final position:
$a^2 = b^2+c^2-2bc~cos~A$
$a^2 = (9.56)^2+(15.2)^2-2(9.56)(15.2)~cos~44^{\circ}40'$
$a^2 = 115.739$
$a = 10.8~miles$
The distance between the ship and the rock when the ship is at its final position is $~~10.8~miles$
