From a window ( h meter above ground) the angle of elevation of other house is `theta` and angle of depression is `phi`. Find the height of the house.

Explanation

Let the height of the other house =OQ=H
and OB=MW=xm
Given that, height of the first house = WB=h = MO
and `angleQWM=alpha, angleOWM = beta= angleWOB` (alternate angle)
Now, in `DeltaWOB, tanbeta= (WB)/(OB) =h/x`
`rArr = x=(h)/(tanbeta)`...........(i)
And in `DeltaQWM, tanalpha= (QM)/(WM) = (OQ-MO)/(WM)`
`rArr tanalpha= (H-h)/(x)`
`rArr x=(H-h)/(tanalpha)`

From Eqs. (i) and (ii),
`h/(tanbeta)=(H-h)/(tanbeta)`
`rArr htanalpha=(H-h)tanbeta`
`rArr htanalpha=(H-h)tanbeta`
`rArr htanalpha=H(tanbeta-htanbeta)`
`rArr Htanbeta= h(tanalpha + tanbeta)`
`therefore H= h((tanalpha+tanbeta))/(tanbeta)`
`=h(1+tanalpha.1/(tanbeta))=h(1+tanalpha.cotbeta)` `[therefore cottheta=1/(tantheta)]`
Hence, the required height of the other house is `h(1+tanalpha.cotbeta)` Hence proved.