question_answer

O is the circumventer of the triangle ABC and \[{{R}_{1}},{{R}_{2}},{{R}_{3}}\] are the radii of the circumcircles of the triangles OBA, 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) None

Correct Answer: B

Solution :

[b] \[{{R}_{1}}=\frac{BC}{2\sin (\angle BOC)}=\frac{a}{2\sin 2A}\]

\[\therefore \,\,\,\,\,\,\frac{a}{{{R}_{1}}}=2\sin 2A\]

Similarly, \[\frac{b}{{{R}_{2}}}=2\sin 2B\]

and \[\frac{c}{{{R}_{3}}}=2\sin 2C\]

So, \[\frac{a}{{{R}_{1}}}+\frac{b}{{{R}_{2}}}+\frac{c}{{{R}_{3}}}\]

\[=2(sin2A+sin2B+sin2C)\]

\[=2.4\sin A\sin B\sin C\] \[[\because \,\,\,A+B+C=\pi ]\]

\[=(2sinA)(2sinB)(2sinC)=\left( \frac{a}{R} \right)\left( \frac{b}{R} \right)\left( \frac{c}{R} \right)\]