A) \[1\,\,sq\,\,unit\]
B) \[2\,\,sq\,\,units\]
C) \[\frac{\sqrt{3}}{2}sq\,\,units\]
D) \[\sqrt{3}\,\,sq\,\,units\]
Correct Answer: D
Solution :
We have,\[\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\] \[\Rightarrow \] \[\frac{\cos A}{k\sin A}=\frac{\cos B}{k\sin B}=\frac{\cos C}{k\sin C}\] \[\Rightarrow \] \[\cot A=\cot B=\cot C\] \[\Rightarrow \] \[A=B=C={{60}^{o}}\] \[\therefore \,\,\Delta \,\,ABC\]is an equilateral triangle. \[\therefore \]Area of triangle\[=\frac{\sqrt{3}}{4}{{a}^{2}}\] \[=\frac{\sqrt{3}}{4}\times {{2}^{2}}=\sqrt{3}\,\,sq\,\,units\]
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