The angle of elevation of the top of a tall building from the points M and N at the distances of 72 m and 128 m, respectively, from the base of building and in the same straight line with it, are complementary. The height of the building (in m) is:

Solution (By Examveda Team)

$$\eqalign{ & {\text{In }}\Delta ABM, \cr & \tan \left( {90 - \theta } \right) = \frac{P}{B} = \frac{{AB}}{{72}} \cr & \frac{1}{{\cot \theta }} = \frac{{72}}{{AB}} \cr & \tan \theta = \frac{{72}}{{AB}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & {\text{In }}\Delta ABN, \cr & \tan \theta = \frac{P}{B} = \frac{{AB}}{{128}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{72}}{{AB}} = \frac{{AB}}{{128}} \cr & A{B^2} = 128 \times 72 \cr & AB = \sqrt {16 \times 4 \times 2 \times 36 \times 2} \cr & AB = 4 \times 4 \times 6 \cr & AB = 96 \cr} $$