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?

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

Aand b is what i think