One plane flies straight east at an altitude of 31,000 feet. A second plane is flying west at an altitude of 13,000 feet on a course that lies directly below that of the first plane and directly above the straight road from Thomasville to Johnsburg. As the first plane passes over Thomasville, the second is passing over Johnsburg. At that instant both planes spot a beacon next to the road between Thomasville to Johnsburg. The angle of depression from the first plane to the beacon is 62°, and the angle of depression from the second plane to the beacon is 34°. How far is Thomasville from Johnsburg?

Answer 1

That's quite a long story to state a simple problem
Make a diagram of the side view of the situation.
Let the distance from the "spot beacon" to Thomaville be a, let the distance to the other town be b
the distance between the two towns is a+b

In the Thomaville triangle,
tan 62 = 31000/a
a = 31000/tan62

in the other town triangle
tan34 = 13000/b
b= 13000/tan34

evaluate, add them up, done
Your answer will be in feet, you might need to convert to miles.

( I got 35756.3 ft)