The angle of elevation of the top of a vertical tower from a point on the ground is $60^{\circ}$. From another point $10 \, m$ vertically above the first,its angle of elevation is $45^{\circ}$. Find the height of the tower.

(N/A) Let the height of the vertical tower be $OT = H \, m$.
Let the distance from the point on the ground to the base of the tower be $OP = x \, m$.
Given that $AP = 10 \, m$,and the tower is vertical,we have $AB = OP = x \, m$ and $OB = AP = 10 \, m$.
Thus,$TB = OT - OB = (H - 10) \, m$.
In $\triangle TPO$,$\tan 60^{\circ} = \frac{OT}{OP} = \frac{H}{x}$.
Since $\tan 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{H}{x}$,which implies $x = \frac{H}{\sqrt{3}}$ ....$(i)$.
In $\triangle TAB$,$\tan 45^{\circ} = \frac{TB}{AB} = \frac{H - 10}{x}$.
Since $\tan 45^{\circ} = 1$,we have $1 = \frac{H - 10}{x}$,which implies $x = H - 10$ ....$(ii)$.
Equating $(i)$ and $(ii)$,we get $\frac{H}{\sqrt{3}} = H - 10$.
$H = \sqrt{3}(H - 10) \Rightarrow H = H\sqrt{3} - 10\sqrt{3}$.
$10\sqrt{3} = H(\sqrt{3} - 1)$.
$H = \frac{10\sqrt{3}}{\sqrt{3} - 1}$.
Rationalizing the denominator: $H = \frac{10\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10(3 + \sqrt{3})}{3 - 1} = \frac{10(3 + \sqrt{3})}{2} = 5(3 + \sqrt{3}) \, m$.
Thus,the height of the tower is $5(3 + \sqrt{3}) \, m$.