The angles of elevation of the top of from two points P and Q at distance $${{m^2}}$$ and $${{n^2}}$$ respectively, from the base and in the same straight line with it are complementary. The height of the tower is-

A. $${\left( {mn} \right)^{\frac{1}{2}}}$$

B. $$m{n^{\frac{1}{2}}}$$

C. $${m^{\frac{1}{2}}}n$$

D. $$mn$$

Answer: Option D

Solution(By Examveda Team)

$$\eqalign{ & {\text{tan }}\theta = \frac{H}{{{m^2}}} \cr & \Rightarrow \tan \left( {90^ \circ - \theta } \right) = \frac{H}{{{n^2}}} \cr & \Rightarrow \cot \theta = \frac{H}{{{n^2}}} \cr & \Rightarrow \tan \theta .\cot \theta = \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} \cr & \Rightarrow \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} = 1 \cr & \Rightarrow {H^2} = {m^2}{n^2} \cr & \Rightarrow H = mn \cr} $$