Question 258923: An airbus A320 aircraft is cruising at an altitude of 10000 m. The aircraft is flying in a straight line away from Rachel, who is standing on the ground. If she sees the angle of elevation of the aircraft change from 70 degrees to 33 degrees in one minute, what is its cruising speed, to the nearest kilometer per hour?
Found 2 solutions by Greenfinch, rfadrogane:
Answer by Greenfinch(383)   (Show Source):
You can put this solution on YOUR website! Initial position = 10000 cos 70 = 3420.2 metres
Final position = 10000 cos 33 = 8386.7 metres
Difference = 4966.5 metres in one minute
so in 1 hour 60 times 4966.5 metres /hr = 297.99 kmetres / hr
Say 298 km/hr
Answer by rfadrogane(214)  (Show Source):
You can put this solution on YOUR website! An airbus A320 aircraft is cruising at an altitude of 10000 m. The aircraft is flying in a straight line away from Rachel, who is standing on the ground. If she sees the angle of elevation of the aircraft change from 70 degrees to 33 degrees in one minute, what is its cruising speed, to the nearest kilometer per hour?
Sol'n:
let x1= Initial horizontal distance at 70 deg. from observer
x2= Final horizontal distance at 33 deg. from observer
v = velocity
x1 = 10,000 cot 70 deg.= 3,639.702-m
x2 = 10,000 cot 33 deg. = 15,398.65-m
so, the horizontal distance travel in one minute is x2-x1
d = 15,398.65-3,639.702 = 11,758.948-m
then, the velocity (v):
v = d/t
v = 11,758.948/60 sec.
v = 195.98 m/s
v = 195.98 m/s x 3600 s/hr x 1 km/1,000 m
v = 705.54 kph say 706 kph - answer
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