Question

An aeroplane, when flying at a height of 400m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point in the ground are 60° and 45° respectively. Find the vertical distance between the aeroplane at that instant. (√3= 1.73)​

Answer 1

Step-by-step explanation:

Let the height of first plane be AB=4000m and the height of second plane be BC=xm

∠BDC=45°

and ∠ADB=60°

In △CBD(i)

x/y = tan45° =1

⇒x=y

In △ABD, 4000/Y =tan60° = √3

⇒ (x=y) = (4000×√3) ÷ 3

=2306.67m

∴ the vertical distance between two=4000−y=4000−2306.67=1693.33m

⛔I think you're question have some fault⛔

⛔ NONEXISTENT ⛔