Question

An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when angle of elevation of two planes from the same point on the ground is 60 degree and 45 degree respectively find the vertical distance between the aeroplane

Answer 1

Let 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