an airplane when 3000m high passes vertically above another plane at an instant when the angle of elevation of the 2 ariplanes from the same point on the ground are 60 and 45.find the vertical distance b/w the 2 aiplane
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height = AC = 300 m
angle of elevation = tan0= height / base = √3
3000/DC = √3
3000/ √3 = DC
1000√3 = DC
In ΔBCD
tan 45 = 1 = BC/ DC
BC/ 1000√3 = 1
BC = 1000√3
AB = 3000 - 1000√3
AB = 1268m this is the distance between two planes
angle of elevation = tan0= height / base = √3
3000/DC = √3
3000/ √3 = DC
1000√3 = DC
In ΔBCD
tan 45 = 1 = BC/ DC
BC/ 1000√3 = 1
BC = 1000√3
AB = 3000 - 1000√3
AB = 1268m this is the distance between two planes
Answered by
0
Answer:
Step-by-step explanation:
height = AC = 300 m
angle of elevation = tan0= height / base = √3
3000/DC = √3
3000/ √3 = DC
1000√3 = DC
In ΔBCD
tan 45 = 1 = BC/ DC
BC/ 1000√3 = 1
BC = 1000√3
AB = 3000 - 1000√3
AB = 1268m this is the distance between two planes
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