If $O$ is circumcentre of $\triangle ABC$ and ${R}{1},{R}{2},{R}{3}$ are circumradii of triangles $OBC,OCA$ and $OAB$, then $\displaystyle \frac { a }{ { R }{ 1 } } +\frac { b }{ { R }{ 2 } } +\frac { c }{ { R }{ 3 } } =$
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$\displaystyle \frac { abc }{ { R }^{ 3 } } $
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$\displaystyle \frac { 2abc }{ { R }^{ 3 } } $
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$\displaystyle \frac { abc }{ { 2R }^{ 3 } } $
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none of these
A
Correct answer
Explanation
In a triangle, a/sinA = 2R. For OBC, the circumradius R1 = a / (2*sin(angle BOC)). Angle BOC = 2A. So R1 = a / (2*sin(2A)). This leads to a/R1 = 2*sin(2A) = 4*sinA*cosA. Summing these is a known identity related to the circumcenter, resulting in abc/R^3.