Question

An aeroplane when flying at a height of 4000 m from the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 60 degree and 45 degree find the vertical distance between the aeroplanes at that instant
Please give me EXPLANATION

Answer 1

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