The angle of elevation of the top of a tower from two distinct points s and t from foot are complementary. Prove that the height of the tower is `sqrt[st]`.

Explanation

Let the height of the tower is h.
and `angleABC=theta`
Given that, BC=s, PC=t
and angle of elevation on both positions are complementary.
i.e., `angleAPC=90^(@)-theta`

[If two angles are complementary to each other, then the sum of both angles is equal to `90^(@)`]
Now in `DeltaABC`, `tantheta=(AC)/(AB)=h/s` ...................(i)
and in `DeltaAPC`
`tan(90^(@)-theta)= (AC)/(PC)` `[therefore tan(90^(@)-theta)=cottheta]`
`rArr cottheta=h/t`
`rArr 1/(tantheta) = h/t` `[therefore cottheta=1/(tantheta)]`...............(ii)
On multyplying Eqs. (i) and (ii), we get
`tan theta. 1/(tantheta) = (h/s)/(h/t)`
`rArr (h^(2))/(st)=1`
`rArr h^(2)=st`
`rArr h=sqrt(st)`
So, the required height of the tower is `sqrt(st)` Hence Proved.