The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is:

Solution(By Examveda Team)

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have ∠DAC = 30°, ∠DBC = 45°
Let DC = h
$$\eqalign{ & \tan {30^ \circ } = \frac{{DC}}{{AC}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{AC}} \cr & \Rightarrow AC = h\sqrt 3 \,.........\,\left( {\text{i}} \right) \cr & \tan {45^ \circ } = \frac{{DC}}{{BC}} \cr & \Rightarrow 1 = \frac{h}{{BC}} \cr & \Rightarrow BC = h\,........\,\left( {{\text{ii}}} \right) \cr & {\text{We know that,}} \cr & AB = \left( {AC - BC} \right) \cr & \Rightarrow 40 = \left( {AC - BC} \right) \cr} $$
$$ \Rightarrow 40 = \left( {h\sqrt 3 - h} \right)$$   [∵ from (1) and (ii)]
$$\eqalign{ & \Rightarrow 40 = h\left( {\sqrt 3 - 1} \right) \cr & \Rightarrow h = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \cr & = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & = \frac{{40\left( {\sqrt 3 + 1} \right)}}{{\left( {3 - 1} \right)}} \cr & = \frac{{40\left( {\sqrt 3 + 1} \right)}}{2} \cr & = 20\left( {\sqrt 3 + 1} \right) \cr & = 20\left( {1.73 + 1} \right) \cr & = 20 \times 2.73 \cr & = 54.6\,m \cr} $$