#### Question

The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building (use `sqrt3`=1.73)

#### Solution

Let the height of the tower AB be *h* m and the horizontal distance between the tower and the building BC be *x* m.

So,

AE=(h−50) m

In ∆AED,

`tan45^@=(AE)/(ED)`

`=>1=(h-50)/x`

⇒x = h−50 .....(1)

In ∆ABC,

`tan^@=(AB)/(BC)`

`=>sqrt3=h/x`

`=>x=sqrt3=h `

Using (1) and (2), we get

`x=sqrt3x-50`

`=>x=(sqrt3-1)=50`

`=>x=(50(sqrt3+1))/2=25xx2.73=68.25m`

Substituting the value of *x* in (1), we get

68.25 = h −50

⇒ h = 68.25 + 50

⇒h=118.25 m

Hence, the height of tower is 118.25 m and the horizontal distance between the tower and the building is 68.25 m.