An observer 2 m tall is $$10\sqrt 3 $$  m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:

Solution (By Examveda Team)

$$\eqalign{ & SR = PQ = 2\,m \cr & PS = QR = 10\sqrt 3 \,m \cr & \tan {30^ \circ } = \frac{{TS}}{{PS}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{TS}}{{10\sqrt 3 }} \cr & TS = \frac{{10\sqrt 3 }}{{\sqrt 3 }} = 10\,m \cr & TR = TS + SR = 10 + 2 = 12\,m \cr} $$