From the top of a building 100 m high, the angles of depression of the bottom and the top of an another building just oppositeto it are observed to be `60^(@) and 45^(@)` respectively. Find the height of the building.

Explanation

Let the height of the building be h m.
Let AC=BD = d m.
From `Delta BDE`,
`tan45^(@)=(ED)/(BD) rArr1= (100-h)/(d)` ,brgt `rArrd=100-h" "(1)`
From `Delta ACE`,
`tan60^(@)=(CE)/(AC)`
`rArrsqrt3=(100)/(d)rArrsqrt3d=100 rArrd=(100)/(sqrt3)`
`rArr100-h=(100)/(sqrt3)`
`rArr h=100-(100)/(sqrt3)`
`=100((sqrt3-1)/(sqrt3))=(100(3-sqrt3))/(3)`.
Hence, the height of the tower is `(100(3-sqrt3))/(3) m`.