A plane flying at an altitude of 13,000 feet finds the angle of depression for a building ahead of the plane to be 10.1°. With the building still straight​ ahead, two minutes later the angle of depression is 25.8°. Find the speed of the plane relative to the ground​ (in feet per​ minute).

Answer 1

Make a sketch showing the two positions of the plane.
Let the speed of the plane be x ft/min
Let the horizontal distance of the plane from the building be d
(along the ground)
tan 10.1° = 13000/d, d = 13000/tan10.1 = 72,981.6 ft
let the horizontal distance of the plane from the building be k for the second position,
tan 25.8° = 13000/k, k = 13000/tan25.8 = 26,891.8 ft
difference = d - k = 46,089.8 ft
rate = distance/time = 46,089.8/2 ft/min = 23,044.9 ft/min
(which would be about 262 mph, making sense)

Check my arithmetic