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

Answer: Option B

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,

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