From the top of a $7 \,m$ high building,the angle of elevation of the top of a cable tower is $60^{\circ}$ and the angle of depression of its foot is $45^{\circ}$. Determine the height of the tower.

(N/A) Let $AB$ be the building of height $7 \,m$ and $CD$ be the cable tower.
In $\triangle ABD$,$\angle ABD = 90^{\circ}$ and $\angle ADB = 45^{\circ}$.
$\tan 45^{\circ} = \frac{AB}{BD}$
$1 = \frac{7}{BD}$
$BD = 7 \,m$.
Since $ABED$ is a rectangle,$AE = BD = 7 \,m$ and $ED = AB = 7 \,m$.
In $\triangle ACE$,$\angle AEC = 90^{\circ}$ and $\angle CAE = 60^{\circ}$.
$\tan 60^{\circ} = \frac{CE}{AE}$
$\sqrt{3} = \frac{CE}{7}$
$CE = 7\sqrt{3} \,m$.
The total height of the tower $CD = CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \,m$.
Thus,the height of the tower is $7(\sqrt{3} + 1) \,m$.