question_answer

The angle of elevation of the top of a tower from a point A due south of the tower is \[\alpha \]and from a point B due east of the tower is \[\beta \]. If AB =d, then the height of the tower is   [Roorkee 1979; Kurukshetra CEE 1998]

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

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

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

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

Correct Answer: C

Solution :

\[OB=h\cot \beta ,\,OA=h\cot \alpha \] \[{{h}^{2}}=\frac{{{d}^{2}}}{{{\cot }^{2}}\beta +{{\cot }^{2}}\alpha }\]\[\Rightarrow \] \[\Delta =\left( \frac{\sqrt{3}{{a}^{2}}}{4} \right)=3\sqrt{3}sq.\,cm,\,\text{ }\therefore \,R=\frac{abc}{4\Delta }=2cm\].