From a point on the ground,the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20 \, m$ high building are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower.

(N/A) Let $BC$ be the building,$AB$ be the transmission tower,and $D$ be the point on the ground from where the elevation angles are measured.
In $\triangle BCD$:
$\frac{BC}{CD} = \tan 45^{\circ}$
$\frac{20}{CD} = 1$
$CD = 20 \, m$
In $\triangle ACD$:
$\frac{AC}{CD} = \tan 60^{\circ}$
$\frac{AB + BC}{CD} = \sqrt{3}$
$\frac{AB + 20}{20} = \sqrt{3}$
$AB + 20 = 20\sqrt{3}$
$AB = 20\sqrt{3} - 20$
$AB = 20(\sqrt{3} - 1) \, m$
Therefore,the height of the transmission tower is $20(\sqrt{3} - 1) \, m$.