sir this is the question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances

we have the information that An aeroplane when flying at a heigt of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 300 and 600 respectively.

we are asked to Find the distance between the two planes at the instant.

question no 16 , heights and distances , ICSE board

Consider the distance between two planes = h m

It is given that

AD = 3125 m, âˆ ACB = 60

^{0}Â and âˆ ACD = 30^{0}In triangle ACD

tan 30

^{0}Â = AD/ACSubstituting the values

1/âˆš3 = 3125/AC

AC = 3125âˆš3 â€¦â€¦. (1)

In triangle ABC

tan 60

^{0}Â = AB/ACSubstituting the values

âˆš3 = (AD + DB)/ AC

So we get

âˆš3 = (3125 + h)/ AC

AC = (3125 + h)/ âˆš3 â€¦.. (2)

Using both the equations

(3125 + h)/ âˆš3 = 3125âˆš3

By further calculation

h = (3125âˆš3 Ã— âˆš3) â€“ 3125

h = 3125 Ã— 3 â€“ 3125

h = 9375 â€“ 3125

h = 6250 m

Therefore, the distance between two planes at the instant is 6250 m.