If O is the circumcentre of the Delta ABC and | vedclass

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(C) Let $R$ be the circumradius of $\Delta ABC$. The sides of $\Delta OBC$ are $a, R, R$. The area of $\Delta OBC$ is $\Delta_1$. The circumradius $R_

Vedclass · Accepted Answer

$\frac{4\Delta}{R^2}$

Vedclass · Answer

(C) Let $R$ be the circumradius of $\Delta ABC$. The sides of $\Delta OBC$ are $a, R, R$. The area of $\Delta OBC$ is $\Delta_1$. The circumradius $R_1$ of $\Delta OBC$ is given by $R_1 = \frac{a \cdot R \cdot R}{4\Delta_1} = \frac{aR^2}{4\Delta_1}$.Thus,$\frac{a}{R_1} = \frac{4\Delta_1}{R^2}$.Similarly,$\frac{b}{R_2} = \frac{4\Delta_2}{R^2}$ and $\frac{c}{R_3} = \frac{4\Delta_3}{R^2}$.Adding these,we get $\frac{a}{R_1} + \frac{b}{R_2} + \frac{c}{R_3} = \frac{4}{R^2}(\Delta_1 + \Delta_2 + \Delta_3)$.Since $\Delta_1 + \Delta_2 + \Delta_3 = \Delta$ (the area of $\Delta ABC$),the expression becomes $\frac{4\Delta}{R^2}$.

If $O$ is the circumcentre of the $\Delta ABC$ and $R_1, R_2$ and $R_3$ are the radii of the circumcircles of triangles $OBC, OCA$ and $OAB$ respectively,then the value of $\frac{a}{R_1} + \frac{b}{R_2} + \frac{c}{R_3}$ is equal to:

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