The angleof elevation of the top of a tower from a certain point is `30o`. If theobserver moves 20 m towards the tower, the angle of elevation of the top ofthe tower increases by `15o`. Theheight of the tower is(a) 17.3m (b) 21.9 m (c) 27.3 m (d) 30 m

The angleof elevation of the top of a tower from a certain point is `3

1.	The angleof elevation of the top of a tower from a certain point is `30o`. If theobserver moves 20 m towards the tower, the angle of elevation of the top ofthe tower increases by `15o`. Theheight of the tower is(a) 17.3m (b) 21.9 m (c) 27.3 m (d) 30 m
Answer» 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.