Answer 1

Let AB be the building and CD be the tower.

In ΔCDB,

CB/BD = tan 60º

`50/(BD) = sqrt3`

`BD = 50/sqrt3`

In ΔABD,

(AB)/(BD) = tan 30º

`AB = 50/sqrt3 xx 1/sqrt3 = 50/3 = 16 2/3`

Therefore, the height of the building is 16 2/3 m

Answer 2

Let AD be the building of height h m. and an angle of elevation of the top of building from the foot of the tower is 30° and an angle of the top of the tower from the foot of building is 60°.

Let AD = h, AB = x and BC = 50 and ∠DBA = 30°, ∠CAB = 60°

So we use trigonometric ratios.

In a triangle ABC

`=> tan 60^2 = 50/x`

`=> sqrt3 = 50/x`

`=> x = 50/sqrt3`

Again in a triangle ABD

`=> tan 30° = (AD)/(AB)`

`=> 1/sqrt3 xx h/x`

`=> h = x/sqrt3`

`=> h = 50/(sqrt3 xx sqrt3)`

`=> h = 50/3`

Hene the height of building is  `50/3` m