Answer
$2568ft$
Work Step by Step
Step 1. We illustrate the configuration as shown in the figure, where A is the airplane at $h=5150ft$, the angle of depression to car B is $38^\circ$, the angle of depression to car C is $52^\circ$, the distance between B and C is $x$ ft.
Step 2. As $AD\parallel BC$, we have $\angle ABC=38^\circ, \angle ACE=52^\circ$
Step 3. In right triangle $\Delta ABE$, $tan38^\circ=\frac{h}{BE}, BE=5150cot38^\circ\approx6591.7ft$
Step 4. In right triangle $\Delta ACE$, $tan52^\circ=\frac{h}{CE}, CE=5150cot52^\circ\approx4023.6ft$
Step 5. The distance between the two cars $x=BC=BE-CE\approx2568ft$
