Answer
Ans.$h=d\frac{(tanβtanα)}{(tanβ-tanα)}$
Work Step by Step
Given:-
In ∆ABD
$tanα=BC/AB$
$tanα=\frac{h}{d+x}$ eq. ~1
In∆ABC
$tanβ=BC/BD$
$tanβ=\frac{h}{x}$
h=x$tanβ$
Putting the value of x in eq. 1
We get
$h=d\frac{(tanβtanα)}{(tanβ-tanα)}$
Ans.
