The angles of depression of the top and bottom of a tower as seen from the top of a 60 `sqrt(3)` m high cliff are 45° and 60° respectively. Find the height of the tower.

#### Solution

Let AD be the tower and BC be the cliff.

We have,

BC = 60 `sqrt(3)` , ∠ CDE = 45° and ∠BAC = 60°

Let AD = h

⇒ BE = AD = h

⇒ CE = BC - BE= 60 `sqrt(3)` - h

In ΔCDE,

` tan 45° = (CE)/(DE)`

`⇒ 1 = (60 sqrt(3) -h)/(DE)`

`⇒ DE = 60 sqrt(3) - h`

`⇒ AB = DE = 60 sqrt(3) - h` ............(1)

Now, in ΔABC

`tan 60° = (BC)/(AB)`

`⇒ sqrt(3)= (60 sqrt(3) )/ (60 sqrt(3) -h)` [ Using (1)]

`⇒ 180 - h sqrt(3) = 60 sqrt(3)`

`⇒ h sqrt(3) = 180- 60 sqrt(3)`

`⇒ h = (108 -60sqrt(3) )/sqrt(3) xx sqrt(3)/sqrt(3)`

`⇒ h = ( 180 sqrt(3)-180) /3`

`⇒ h = (180 sqrt(3)-1) /3`

∴` h = 60 ( sqrt(3)-1)`

= 60 (1.732 -1)

= 60 (0.7.32)

Also, h = 43.92m

So, the height of the tower is 43. 92 m.