Answer
$14366ft$
Work Step by Step
Step 1. Use the figure given in the Exercise, draw a vertical line from the pilot down to the sea surface (Point P), the height of the airplane is $h=35000ft$.
Step 2. In the triangle formed by the pilot, point P and the left ship, assume the horizontal distance is $x1$, we have
$tan52^{\circ}=\frac{h}{x1}$ thus $x1=35000cot52^{\circ}\approx27345ft$
Step 3. In the triangle formed by the pilot, point P and the right ship, assume the horizontal distance is $x2$, we have $tan40^{\circ}=\frac{h}{x2}$ thus $x2=35000cot40^{\circ}\approx41711ft$
Step 4. The distance between the two ships is then $x2-x1=14366ft$
