Question

an engineer standing on the ground level AT certain distance from the Tower observed that angle of elevation is 60°. by moving further distance of 30m away from his original position he observed that the angle of elevation is 30° Help him to find the height of the tower

Answer 1

Let the height of tower be h 
initial distance of engineer from the foot of tower = x
A/q
tan60° =h/x
⇒√3 = h/x
⇒h = x√3_____(1)
also, tan30° = h/(x+30)
⇒1/√3 = x√3/(x +30)
⇒x +30 = 3x
⇒2x =30
⇒x =15 
so from (1)
height ,h =15√3 m

Answer 2

Let the height of tower be h 
initial distance of engineer from the foot of tower = x
A/q
tan60° =h/x
⇒√3 = h/x
⇒h = x√3_____(1)
also, tan30° = h/(x+30)
⇒1/√3 = x√3/(x +30)
⇒x +30 = 3x
⇒2x =30
⇒x =15 
so from (1)
height ,h =15√3 m