The angle of elevation of the top of a tower from a certain point is `30^@`. If the observer moves 20 m towards the tower, the angle of elevation of the top of the tower increases by `15^@`. The height of the tower is

Options

A. `17.3m`
B. `21.9m`
C. `27.3m`
D. `30m`

Answer

C

Explanation

Let the height of the tower be h.
also, SR=x m, `anglePSR = theta`
Given that, QS = 20m
and `anglePQR = 30^(@)`
Now, in `DeltaPSR`,
`tantheta=(PR)/(SR) = h/x`
`rArr tantheta=h/x`
`rArr x=h/(tantheta)`................(i)
Now, in `DeltaPQR`,
`tan30^(@)=(PR)/(QR) = (PR)/(QS+SR)`
`rArr 20+x = h/(tan 30^(@)) = h/(1//sqrt(3))`
`rArr 20+x = hsqrt(3)`
`rArr 20+h/(tantheta)=hsqrt(3)` ..............(ii) From Eq. (i)
Since, after moving 20 m towards the tower the angle of elevation of the top increases by `15^(@)`.
i.e., `anglePSR = theta= anglePQR + 15^(@)`
`rArr theta=30^(@) + 15 = 45^(@)`
`therefore 20 + (h/tan 45^(@)) = hsqrt(3)`
`rArr 20 + h/1= hsqrt(3)`
`rAr 20 = hsqrt(3)-h`
`rArr h(sqrt(3)-1)=20`
`therefore h=(20)/(sqrt(3)-1).(sqrt(3)+1)/(sqrt(3)+1)` [by rationalisation]
`rAr = (20(sqrt(3)+1))/(3-1) = (20(sqrt(3)+1))/(2)`
`rArr =10(sqrt(3)+1)`m
Hence, the required height lof tower is `10(sqrt(3)+1)`m.