From the top of a building 60m high the angles of depression of the top and the bottom of a tower are observed to be `30^0a n d60^0` . Find the height of the tower.

Explanation

Let AB be the builiding and CD be the tower such that `angleBDE = 30^(@), angleBCA = 60^(@)` and `AB = 60 m`.
Let `CA = DE = x` metres.
From right `DeltaCAB`, we have
`(CA)/(AB) = cot 60^(@) = 1/(sqrt(3))`
`rArr x/60= 1/(sqrt(3)) rArr x = 60 xx 1/(sqrt(3))`
`rArr x = 60 xx 1/(sqrt(3)) xx (sqrt(3))/(sqrt(3)) = 20sqrt(3)`
`rArr CA = DE = 20sqrt(3) m`.
From right `DeltaBED`, we have
`(BE)/(DE) = tna 30^(@) = 1/(sqrt(3) m) rArr (BE)/(20sqrt(3) m) = 1/(sqrt(3))` [using (i)]
`rArrBE = 20sqrt(3) xx 1/(sqrt(3)) m = 20 m`.
`:. CD = AE = AB - BE = 60 m - 20 m - 40 m` .
Hence the height of the tower is `40 m`.