If the angle of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with It are complementary, then the height of the tower is

Let AB be the tower and P and Q are two points such that PB = a and QB = b and angles of elevation are $$\theta $$ and (90° – $$\theta $$)
Let height of tower = h

$$\eqalign{ & {\text{Then}}\,{\text{in}}\,{\text{right}}\,\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & = \frac{h}{a}\,............\left( {\text{i}} \right) \cr & {\text{Similarly in right }}\Delta AQB, \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \frac{{AB}}{{QB}} = \frac{h}{b} \cr & \Rightarrow \cot \theta = \frac{h}{b}\,...........\left( {{\text{ii}}} \right) \cr & {\text{Multiplying }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \tan \theta \cot \theta = \frac{h}{a} \times \frac{h}{b} \cr & \Rightarrow 1 = \frac{{{h^2}}}{{ab}} \cr & \Rightarrow {h^2} = ab \cr & \Rightarrow h = \sqrt {ab} \cr & \therefore {\text{Height}}\,{\text{of}}\,{\text{tower}} = \sqrt {ab} \cr} $$