The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Let AB be the building and CD be the tower.

In △BCD,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{CD}{BD} \\[1em] \Rightarrow \sqrt{3} = \dfrac{50}{BD} \\[1em] \Rightarrow BD = \dfrac{50}{\sqrt{3}} \text{ meters}.$

In △ABD,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{\dfrac{50}{\sqrt{3}}} \\[1em] \Rightarrow AB = \dfrac{50}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{50}{3} = 16\dfrac{2}{3} \text{ meters}.$

Hence, height of building = $16\dfrac{2}{3}$ meters.