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Question
An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°; find the uniform speed of the aeroplane in km per hour.
Solution
Let A be the aeroplane and B be the observer on the ground. The vertical height will be AC = 1 km = 1000 m. After 10 seconds, let the aeroplane be at point D.
Let the speed of the aeroplane be x m/sec.
∴ CE = 10x
In ΔABC,
`(AC)/(BC) = tan 60^circ`
`=> 1000/(BC) = sqrt(3)`
`=> BC = 1000/sqrt(3)m`
In ΔBDE,
`(DE)/(BE) = tan 30^circ`
`=> BE = 1000 sqrt(3)`
∴ CE = BE – BC
`=> 10x = 1000sqrt(3) - 1000/sqrt(3)`
`=> x = 100 (sqrt(3) - 1/sqrt(3))`
= 100 × 1.154
= 115.4 m/sec
= `115.4 xx 18/5` km/hr
= 415.44 km/hr
Hence, speed of the aeroplane is 415.44 km/hr.
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