From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is

Answer: Option B

Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at C
Let angle be Q and CD = 25 m

$$\begin{array}{rl}& \text{Let}AB=h\\ & CE||DB\\ & \therefore EC=DB=x\left(\text{suppose}\right)\\ & EB=CD=25\\ & \therefore AE=h-25\\ & \text{Now in right }\Delta CDB,\\ & \tan \theta =\frac{CD}{DB}=\frac{25}{x}......\left(\text{i}\right)\\ & \text{and in right }\Delta CAE\\ & \tan \theta =\frac{AE}{CE}=\frac{h-25}{x}......\left(\text{ii}\right)\\ & \text{From}\left(\text{i}\right)\text{ and }\left(\text{ii}\right)\\ & \frac{25}{x}=\frac{h-25}{x}\\ & \Rightarrow 25=h-25\\ & \Rightarrow h=25+25=50\\ & \therefore \text{Height of tower}=50m\end{array}$$