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
