#### Answer

The distance between the battleship and the submarine is 1451.95 feet

#### Work Step by Step

Let the airplane be at point A.
Let the battleship be at point B.
Let the submarine be at point C.
The points ABC form a triangle.
Angle A = $24^{\circ}10'-17^{\circ}30' = 6^{\circ}40'$
Angle B = $17^{\circ}30'$
We can find angle C:
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-6^{\circ}40'-17^{\circ}30'$
$C = 155^{\circ}50'$
We can use the law of sines to find the distance $BC$, which is the distance between the battleship and the submarine:
$\frac{BC}{sin~6^{\circ}40'} = \frac{5120}{sin~155^{\circ}50'}$
$BC = \frac{5120~sin~6^{\circ}40'}{sin~155^{\circ}50'}$
$BC = 1451.95~ft$
The distance between the battleship and the submarine is 1451.95 feet