Question

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

• A. $\frac{abc}{R}$
• B. $\frac{abc}{R^3}$
• C. $\frac{abc}{R^4}$
• D. $abc$

Answer 1

The correct option is B $\frac{abc}{R^3}$

$R_1$ be the circum radius of $△BOC$
$\Rightarrow R_1=\frac{BC}{2\sin \angle BOC}=\frac{a}{2\sin 2A}$
$\Rightarrow \frac{a}{R_1}=2\sin 2A\cdots \left(i\right)$
Similarly,
$\Rightarrow \frac{b}{R_2}=2\sin 2B\cdots \left(ii\right)$
Similarly,
$\Rightarrow \frac{c}{R_3}=2\sin 2C\cdots \left(iii\right)$
Adding $\left(i\right),\left(ii\right),\left(iii\right)$ we get
$\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=2(\sin 2A+\sin 2B+\sin 2C)$
$=4\left(\sin \right(A+B\left)\cos \right(A-B)+\sin C\cos C)$
$=4\sin C\left(\cos \right(A-B)-\cos (A+B\left)\right)$
$=8\sin A\sin B\sin C=\frac{abc}{R^3}$