Question

An aeroplane when flying at a height of 3000m above the ground passes vertically above another aeroplane at an instant when the angles of elevation of two planes from the same point on the ground are 60degree and 45degree respectively. Find the vertical distance between the aeroplanes at that instant.

Answer 1

tan60= AC/BC
√3=3000/b
b√3=3000
b=3000/√3
b=1000√3
BC=1000√3

Now tan45°=DC/BC
1=h/1000√3
h=1000√3

x=AC-DC
x=3000-1000√3
x=1000√3(√3-1)

Answer 2

Answer:

Step-by-step explanation:

Mark brainliest