The angle of elevation of a jet plane from a point A on the ground is 60∘. After a flight of 20 seconds at the speed of 432km/hour, the angle of elevation changes to 30∘. If the jet plane is flying at a constant height, then its height is
A
1200√3 m
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B
1800√3 m
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C
3600√3 m
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D
2400√3 m
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Solution
The correct option is A1200√3 m Velocity, v=432×100060×60m/sec=120m/sec Distance PQ=v×20=2400m In ΔPAC tan60∘=hAC⇒AC=h√3 In ΔAQD tan30∘=hAD⇒AD=√3h
Now, AD=AC+CD ⇒√3h=h√3+2400 ⇒2h√3=2400 ⇒h=1200√3m
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