Correct Answer - C
AB=hill = 200 metre
CD=tower
In `triangleAPC`
`tan30^@=(AP)/(PC)`
`1/sqrt3=(AP)/(PC)implies AP:PC=1:sqrt3`…(i)
In `triangleABD`
`tan60^@=(AB)/(BD)`
`sqrt3=(AB)/(BD)=AB:BD=sqrt3:1`(ii)
PB=CD and PC=BD
Now ,
`{:(AB,:,BD,:,AP),(sqrt3,:,1,:,),(,:,sqrt3,:,1),(3,:,sqrt3,:,1):}`
CD=PB`implies`AB-AP
CD=3-1=2 units
AB=3 units =200 metre
CD=2 units `=200/3 xx 2`
`=133 1/3` metre