The angle of elevation of an aeroplane from a point on the ground is 60° after a flight of 6 secs the angle of elevation becomes 45° if the aeriplane is flying at a constant height of 3000m find the speed of the aeroplane.
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As the angle of elevation is decreasing the aeroplane is going away from the point.
Let A be the point of observation BC be the aeroplane and PQ is after the flight .
In ABC,Tan60=BC/AC
√3=3000/AC
AC=3000√3=1000√3m
In APQ,Tan45=PQ/AQ
1=3000/AQ
AQ=3000m
QC=3000-1000√3
=3000-1000*1.732
=3000-1732
=1268m
Time =15sec. Distance=1268m
Speed=84.53m/sec.....
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