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 tower and the CD be the building. Then, `CD = 60 m`. Ket `CAX` be the horizontal ground.
Draw `DE_|_AB`. Now , `angleEDB= 30^(@)` and `angleCAD = angleADE = 60^(@)`.
Also, `AE = CD = 60m`
Let `BE= h` metres.
From right `DeltaACD`, we have
`(CA)/(CD) = cot 60^(@) = 1/(sqrt(3))`.
`rArr (CA)/(60m) = 1/(sqrt(3)) rArr CA = 60/(sqrt(3)) "........."(i)`
From right `DeltaEDB`, we have
`(DE)/(BE) = cot30^(@) = sqrt(3) rArr (DE)/(h m) = sqrt(3)`
`rArr DE = h sqrt(3) m "......"(ii)`
But , `CA= DE`.
`:.` from (i) and (ii), we get
`60/(sqrt(3)) = h sqrt(3) hArr 3h = 60 rArr h = 20`.
Hence, the difference between the heights of the towards and building is `20` metres.
From (i), we get `CA = ((60)/(sqrt(3)) xx (sqrt(3))/(sqrt(3))) m = 20sqrt(3)` metres.
Hence the distance between the tower and the building is `20sqrt(3) m`.