**Q) **An aeroplane when flying at a height of 3000 m from the ground passes vertically above another aeroplane at an instant when the angles of 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. Also, find the distance of first plane from the point of observation. (Take **√**3 = 1.73)

**Ans: **

Let A be the point of observation and planes are C and D. H is the vertical distance between 2 planes and S is the distance of first plane D from observation point A.

Given that D is at a height of 3,000 m from ground, Hence, height of plane C = 3000 – h m

From Δ ABD, = tan 60^{0}

= √3

R = = 1000√3 ……….. (i)

Next, from Δ ABC, = tan 45^{0}

= 1

3000 – h = R = 1000√3 [from equation (i)]

h = 3000 – 1000√3 = 1000 (3 – √3) = 1000 (3 – 1.73) = 1000 x 1.27 = 1,270 m

**Therefore, distance between 2 planes = 1,270 m**

From Δ ABD, = sin 60^{0}

S = 3000 x

= 2000√3 = 2000 x 1.73 = 3460

**Therefore, distance of first plane from observation point is = 3,460 m**