Question

When an aeroplane was flying at a height of 4000m from the ground passes vertically above another aeroplane at an instant when the angle of elevation of the 2 Planes from the same point on the ground are 60 degree and 45 degree respectively . Find the vertical distances between the aeroplane at that instant (root3 = 1.73)

Answer 1

Let A is an aeroplane when it is flying at height 4000m from Ground passes vertical above another Aeroplane B at an instant angle of elevation of two planes from same point D on the ground are 60° and 45° respectively as shown in figure , here distance between two planes is AB = h
And distance between point D and C is x { Let }

Now, from ∆ACD ,
tan60° = perpendicular/base =4000/x
x = 4000/tan60° = 4000/√3 m ---(1)

Again, from ∆BCD ,
tan45° = BC/CD = (4000 - h)/x
1 =(4000 - h)/(4000/√3) [ from equation (1)]
4000/√3 = 4000 - h
h = 4000( 1 - 1/√3) m

Hence, distance between two planes is 1690.59 m

Answer 2

1690.59 metres is your answer