Question

If $O$ is the circum-centre of $\Delta ABC$ and $R_1,R_2,R_3$ are the radii of the circum-circle of 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. $\frac{abc}{R^2}$

Answer 1

Answer: (B)

$\frac{abc}{R^3}$

In $\Delta BOC$
$\angle BOC=2A$
From sine rule:

$\frac{BC}{\sin \angle OCB}=2R_1$

$\Rightarrow \frac{a}{R_1}=2\sin 2A$

Similarly,

$\frac{b}{R_2}=2\sin 2B$

and

$\frac{c}{R_3}=2\sin 2C$
Now,

$y=\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=2(\sin 2A+\sin 2B+\sin 2C)$

$\Rightarrow y=4(\sin (A+B)\cos (A-B)+\sin C\cos C)=4\sin C(\cos (A-B)-\cos (A+B))$

$\Rightarrow y=8\sin A\sin B\sin C=\frac{abc}{R^3}$

Hence, option B.