Question

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.

Answer 1

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.