Speed of plane flying 5.9 km above ground with an angle elevation of 78.6

Here's a graph representing geno's set up..........The station is located at B....BC is the distance the plane is from the station intially and AC is the height of the plane aboive rhe station.

If you draw a right triangle ABC, with C the right angle. Let point A be the location of the plane and location B be the locaion of the station.  AC will be the height of the plane (5.9 km). The angle at B is 78.6°. Side BC represents the distance from the plane to the station (both in the air and across the ground).

For angle B, the side AC is the opposite side and BC is the adjacent side. If we want to find the length of BC, let's use tangent, because tangent = opposite / adjacent.

tan(78.6)  =  5.9 / BC

Cross-multiply:  BC · tan(78.6)  =  5.9

Divide:              BC  =  5.9 / tan(78.6)     --->     BC  =  1.19 km

Speed of the plane  =  distance / time  =  1.19 km / 10 sec  =  0.119 km/s

0.119 km/s x 60s/1min x 60min/1hr  =  428 km/hr