Question

An aircraft is spotted by two observers who are 1000 feet apart. As the airplane passes over the line joining them, each observer takes a sighting of the angle of elevation to the plane, as indicated in the figure. How high is the airplane?

Answer 1

See the attached figure to better understand the problem

we know that
PB+BQ=1000 ft------->PB=1000-BQ------->  equation 1

in the triangle PAB
tan 50=AB/PB----------> AB=PB*tan 50------> equation 2

in the triangle ABQ
tan 25=AB/BQ------> AB=BQ*tan 25-------> equation 3

equals equation 2 and equation 3
PB*tan 50=BQ*tan 25--------> equation 4

substitute equation 1 in equation 4

[1000-BQ]*tan 50=BQ*tan 25-----> 1000*tan 50-BQ*tan 50=BQ*tan 25
BQ*[tan 25+tan 50]=1000*tan 50-----> BQ=1000*tan 50/[tan 25+tan 50]
BQ=718.76 ft
PB=1000-718.76-----> PB=281.24 ft
AB=PB*tan 50-----> 281.24*tan 50------> 335.17 ft

the answer is
335.17 ft