From the top of a tower $h \, m$ high,the angles of depression of two objects,which are in line with the foot of the tower,are $\alpha$ and $\beta$ (where $\beta > \alpha$). Find the distance between the two objects.

(N/A) Let the tower be $AD$ with height $h \, m$. Let the two objects be at points $B$ and $C$ on the ground,such that $B, C,$ and $D$ are collinear.
Let the distance between the two objects be $BC = x \, m$ and the distance from the second object to the foot of the tower be $CD = y \, m$.
Given that the angles of depression are $\alpha$ and $\beta$,by the property of alternate interior angles:
$\angle ABD = \alpha$ and $\angle ACD = \beta$.
In the right-angled triangle $\triangle ACD$:
$\tan \beta = \frac{AD}{CD} = \frac{h}{y} \implies y = \frac{h}{\tan \beta} \quad \dots(i)$
In the right-angled triangle $\triangle ABD$:
$\tan \alpha = \frac{AD}{BD} = \frac{h}{x + y} \implies x + y = \frac{h}{\tan \alpha} \implies y = \frac{h}{\tan \alpha} - x \quad \dots(ii)$
Equating $(i)$ and $(ii)$:
$\frac{h}{\tan \beta} = \frac{h}{\tan \alpha} - x$
$x = \frac{h}{\tan \alpha} - \frac{h}{\tan \beta}$
$x = h \left( \frac{1}{\tan \alpha} - \frac{1}{\tan \beta} \right)$
$x = h (\cot \alpha - \cot \beta) \, m$
Thus,the distance between the two objects is $h (\cot \alpha - \cot \beta) \, m$.