A) 1/24
B) \[\sqrt{3}/24\]
C) 1/8
D) \[1/\sqrt{3}\]
E) \[5/12\sqrt{3}\]
Correct Answer: B
Solution :
Since, \[\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\] \[\Rightarrow \] \[\frac{\cos A}{k\sin A}=\frac{\cos B}{k\sin B}=\frac{\cos C}{k\sin C}\] \[\Rightarrow \] \[\cot A=\cot B=\cot C\] \[\Rightarrow \] \[A=B=C=60{}^\circ \] \[\Rightarrow \] \[\Delta ABC\]is an equilateral triangle. \[\therefore \] \[\Delta =\frac{\sqrt{3}}{4}{{a}^{2}}=\frac{\sqrt{3}}{4}\times \frac{1}{6}\] \[=\frac{\sqrt{3}}{24}sq\,unit\]
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