$A$ statue,$1.6 \, m$ tall,stands on the top of a pedestal. From a point on the ground,the angle of elevation of the top of the statue is $60^{\circ}$ and from the same point the angle of elevation of the top of the pedestal is $45^{\circ}$. Find the height of the pedestal.

(N/A) Let $AB$ be the statue,$BC$ be the pedestal,and $D$ be the point on the ground from where the angles of elevation are measured.
In $\triangle BCD$,we have:
$\frac{BC}{CD} = \tan 45^{\circ}$
$\frac{BC}{CD} = 1$
$BC = CD$
In $\triangle ACD$,we have:
$\frac{AC}{CD} = \tan 60^{\circ}$
$\frac{AB + BC}{CD} = \sqrt{3}$
Since $CD = BC$,we can write:
$\frac{1.6 + BC}{BC} = \sqrt{3}$
$1.6 + BC = BC \sqrt{3}$
$1.6 = BC(\sqrt{3} - 1)$
$BC = \frac{1.6}{\sqrt{3} - 1}$
Rationalizing the denominator:
$BC = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$
$BC = \frac{1.6(\sqrt{3} + 1)}{3 - 1}$
$BC = \frac{1.6(\sqrt{3} + 1)}{2}$
$BC = 0.8(\sqrt{3} + 1) \, m$
Thus,the height of the pedestal is $0.8(\sqrt{3} + 1) \, m$.