Question

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.

Answer 1

Let AB be the building and CD be the tower. 
In ∆CDB,
\(\frac{CD}{ BD} = tan 60°\)
\(\frac{50}{BD} = \sqrt3\)
\(BD = \frac{50}{ \sqrt3}\)
In ∆ABD,
\(\frac{AB}{BD} = tan 30°\)
\(AB = \frac{50}{\sqrt3} \times \frac{1}{ \sqrt3 }= \frac{50}{3} = 16\frac{ 2}3\,m\)
Therefore, the height of the building is \(16\frac{ 2}3\,m\).