An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are `60^@` and `45^@` respectively. Find the vertical distance between the aeroplanes at that instant.

Explanation

Let O be the point of observation. Let A and B be the positions of the two planes at the given instant when A is vertically above B.
Let AB when produced meet the ground at C.
Then, `angleCOA) = 60^(@), angleCOB = 45^(@), angleOCB = angleOCA = 90^(@)` and `AC = 3000 m`.
Let AB = x metres. Then, `BC = (3000 -x ) m`.

From right `DeltaOCB`, we have
`(OC)/(BC) = cot 45^(@) = 1 rArr (OC)/((300-x) m) = 1`
`rArr OC = (3000 - x) m."......"(i)`
From right `DeltaOCA)`. we have
`(OC)/(AC)= cot 60^(2) = 1/(sqrt(3))rArr (OC)/(3000 m) = 1/(sqrt(3))`
`rArr OC = ((3000)/(sqrt(3)) xx (sqrt(3))/(sqrt(3))) m =1000sqrt(3) m"........"(ii)`
From (i) and (ii), we get
`(3000-x) = 1000sqrt(3)`
`rArr x= (3000 - 1000sqrt(3)) = (3000 - 1000 xx 1.73)`
`= (3000 - 1730) = 1270`.