The shadow of a tower standing on a level ground is found to be 40 m longer when the Suns altitude is `30^o` than when it is `60^o`. Find the height of the tower.

Explanation

Let the height of the tower be h and RQ=x m
Given that, PR=50 m
and `angleSPQ = 30^(@), angleSRQ=60^(@)`
Now, in `DeltaSRQ`, `tan60^(@)=(SQ)/(RQ)`
`rArr sqrt(3) = h/x rArr x=h/sqrt(3)`……………(i)
and in `DeltaSPQ`, `tan30^(@) = (SQ)/(PQ) = (SQ)/(PR+RQ) = h/(50+x)`

`rArr 1/sqrt(3) = h/(50+x)`
`rArr sqrt(3).h = 50+x`
`rArr sqrt(3).h = 50 + h/sqrt(3)` [From Eq. (i)]
`rArr (sqrt(3)-1/sqrt(3))h=50`
`rArr (3-1)/(sqrt(3))h=50`
`therefore h=(50sqrt(3))/(2)`
`h=25sqrt(3)`m
Hence, the required height of tower is `25sqrt(3)`m