Question

An aeroplane when 3000m high passes vertically above another aeroplane at an instant when the angles of elevation of the to play aeroplanes from the same point on the ground is 60 degree and 45 degree find the vertical distance between the two planes

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)

Hope it helps

Answer 2

Answer:the all solution is in the picture

Explanation: