The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance 'd' towards the foot of the tower the angle of elevation is found to be β. The height of the tower is

The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance 'd' towards the foot of the tower the angle of elevation is found to be β. The height of the tower is Correct Answer $$\frac{d}{{\cot \alpha - \cot \beta }}$$

Let AB be the tower and C is a point such that the angle of elevation of A is α.
After walking towards the foot B of the tower, at D the angle of elevation is β.
Let h be the height of the tower and DB = x Now in ΔACB,

$$\eqalign{ & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{CB}} \cr & \tan \alpha = \frac{h}{{d + x}} \cr & \Rightarrow d + x = \frac{h}{{\tan \alpha }} \cr & \Rightarrow d + x = h\cot \alpha \cr & \Rightarrow x = h\cot \alpha - d\,.......\left( {\text{i}} \right) \cr & {\text{Similarly in right }}\Delta ADB, \cr & \tan \beta = \frac{h}{x} \cr & \Rightarrow x = \frac{h}{{\tan \beta }} \cr & \Rightarrow x = h\cot \beta \,.........({\text{ii}}) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & h\cot \alpha - d = h\cot \beta \cr & \Rightarrow h\cot \alpha - h\cot \beta = d \cr & \Rightarrow h\left( {\cot \alpha - \cot \beta } \right) = d \cr & \Rightarrow h = \frac{d}{{\cot \alpha - \cot \beta }} \cr} $$
∴ Height of the tower $$ = \frac{d}{{\cot \alpha - \cot \beta }}$$