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