# 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

A. 25 m B. 50 m C. 75 m D. 100 m 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 $\begin{array}{l}\text{Let AB}=h\\ \text{CE }\parallel\ \text{DB}\\ \therefore\text{CE}=\text{BD}=x\ \text{(consider)}\\ \text{BE}=\text{CD}=25\ m\\ \therefore\text{AE}=h-25\\ \text{Now in right}\triangle\text{BDC,}\\ \text{tan}\theta=\frac{CD}{BD}=\frac{25}{x}……\left(i\right)\\ \text{and in right }\triangle\text{CAE}\\ \text{tan}\theta=\frac{AE}{CE}=\frac{\left(h-25\right)}{x}……\left(ii\right)\\ \text{From (i) and (ii)}\\ \frac{25}{x}=\frac{h-25}{x}\\ \Rightarrow25=h-25\\ \Rightarrow h=25+25=50\\ \therefore\text{Height of tower}=50\ \text{m}\end{array}$