Answers

2016-02-20T12:19:46+05:30
First plane is on height = 4000 m
second plane below first plane is on height = X m

AB = 4000 m and BC = X m , height between planes = 4000 - x
D is point on ground 
tan 45 in triangle BCD
tan 45 = BC/CD
     1     =  X/CD
CD = X m.

tan 60 in triangle ACD
Tan 60 = AC/CD
√3 = 4000/X 
√3 x = 4000
x = 4000 / 1.73
x = 4000 × 100 / 173
x = 23.12 × 100
x = 2312 

so second plane is on height of 2312.

distance between planes = 4000 - x                 (from above )
                                         = 4000 - 2312
                                         = 1687 m 

so distance between the planes is 1687 m .

thank u hope it help u
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