The angle of depression of the top of a building from the top of a tower is $45^{\circ}$ and that of the base of the building from the top of the tower is $60^{\circ}$. If the height of the building is $7 \, m$,find the height of the tower.

(16.56 M) Let the height of the tower be $H$ and the height of the building be $h = 7 \, m$. Let the distance between the tower and the building be $x$.
From the top of the tower,the angle of depression to the top of the building is $45^{\circ}$,so the angle of elevation from the top of the building to the top of the tower is $45^{\circ}$. Thus,$\tan(45^{\circ}) = \frac{H - 7}{x} \implies 1 = \frac{H - 7}{x} \implies x = H - 7$.
From the top of the tower,the angle of depression to the base of the building is $60^{\circ}$,so the angle of elevation from the base of the building to the top of the tower is $60^{\circ}$. Thus,$\tan(60^{\circ}) = \frac{H}{x} \implies \sqrt{3} = \frac{H}{x} \implies x = \frac{H}{\sqrt{3}}$.
Equating the two expressions for $x$: $H - 7 = \frac{H}{\sqrt{3}}$.
$H\sqrt{3} - 7\sqrt{3} = H \implies H(\sqrt{3} - 1) = 7\sqrt{3}$.
$H = \frac{7\sqrt{3}}{\sqrt{3} - 1} = \frac{7\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{7(3 + \sqrt{3})}{2} = \frac{7(3 + 1.732)}{2} = \frac{7(4.732)}{2} = 7 \times 2.366 = 16.562 \, m$.