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Answer:
3000 m/sec
Step-by-step explanation:
the distance covered by the aircraft during the two observations = d
d = 36000√3(tan60° - tan30°) = 36000√3(√3 - 1/√3)
d = 36000√3(2/√3) = 72000 m
the velocity of the aircraft = 72000 m ÷ 24 sec = 3000 m/sec
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