Question:

Let $d$ be the horizontal distance between the building and the tower, and let the tower's height be $H$. Using the angle of depression to the tower's foot ($30^\circ$), and noting this angle equals the angle of elevation of the top of the building as seen from the tower's foot: $\tan30^\circ=\dfrac{50}{d}\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{50}{d}$. [1.5 Marks]

$d=50\sqrt3$ m. [1.0 Mark]

Using the angle of elevation to the top of the tower ($60^\circ$) from the top of the building, the extra height above the building's own height is $x$: $\tan60^\circ=\dfrac{x}{d}\Rightarrow\sqrt3=\dfrac{x}{50\sqrt3}$. [1.0 Mark]

$x=50\sqrt3\times\sqrt3=150$ m. Total height of tower $H=50+150=200$ m. [1.5 Marks]