question_answer

The angle of elevation of the top of a tower from point A due south of the tower is \[\alpha \] and  from B due east of the tower is (5, If AB = d, then height of the tower is

A)  \[\frac{d}{\sqrt{{{\cot }^{2}}\alpha +{{\cot }^{2}}\beta }}\]         

B)  \[\frac{d}{\sqrt{{{\tan }^{2}}\alpha -{{\tan }^{2}}\beta }}\]

C)  \[\frac{d}{\sqrt{{{\tan }^{2}}\alpha +{{\tan }^{2}}\beta }}\]        

D)  \[\frac{d}{\sqrt{{{\cot }^{2}}\alpha -{{\cot }^{2}}\beta }}\]

Correct Answer: A

Solution :

Let h be the height of the tower. \[\therefore \]\[\ln \Delta OBT,\]            \[OB=h\cot \beta \] Also, in triangle OAT, \[OA=h\cot \alpha \] \[\therefore \]\[d=\sqrt{O{{B}^{2}}+O{{A}^{2}}}\] \[\Rightarrow \]\[d=h\sqrt{{{\cot }^{2}}\beta +{{\cot }^{2}}\alpha }\] \[\Rightarrow \]\[h=\frac{d}{\sqrt{{{\cot }^{2}}\alpha +{{\cot }^{2}}\beta }}\]