Question

$\frac{R_1}{r_1}+\frac{R_2}{r_2}+\frac{R_3}{r_3}=$

• A. $R^2[\frac{sinA}{{\Delta }_1}+\frac{sinB}{{\Delta }_2}+\frac{sinC}{{\Delta }_3}]$
• B. $R^2[\frac{cosA}{{\Delta }_1}+\frac{cosB}{{\Delta }_2}+\frac{cosC}{{\Delta }_3}]$
• C. $R^2[\frac{tanA}{{\Delta }_1}+\frac{tanB}{{\Delta }_2}+\frac{tanC}{{\Delta }_3}]$
• D. $R^2[\frac{cotA}{{\Delta }_1}+\frac{cotB}{{\Delta }_2}+\frac{cotC}{{\Delta }_3}]$

Answer 1

Answer: (C)

$R^2[\frac{tanA}{{\Delta }_1}+\frac{tanB}{{\Delta }_2}+\frac{tanC}{{\Delta }_3}]$

We know that $\Delta =abc/4R$

so ${\Delta }_1=\frac{OB\times OC\times BC}{4R_1}=\frac{a{R}^2}{4R_1}$

$\Rightarrow \frac{a}{R_1}=\frac{4\Delta }{R^2}$ similarly
$\frac{b}{R^2}=\frac{4{\Delta }_2}{R^2},\frac{c}{R_3}=\frac{4{\Delta }_3}{R^2}$

Hence $\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=\frac{4}{R^2}({\Delta }_1+{\Delta }_2+{\Delta }_3)=\frac{4\Delta }{R^2}$

$BOD=\frac{1}{2}BOC=A$

From $\Delta OBD,\frac{a/2}{2r_1}=tanA$

$\Rightarrow \frac{a}{r_1}=4tanA$, similarly $\frac{b}{r_2}=4tanB,\frac{c}{r_3}=4tanC$

Hence $\frac{a}{r_1}+\frac{b}{r_2}+\frac{c}{r_3}=4(tanA+tanB+tanC)$

$=4tanAtanBtanC[\because A+B+C={180}^o]$

$a=4r_{1}tanA=\frac{4{\Delta }_1{R}_1}{R^2}\Rightarrow \frac{R_1}{r_1}=\frac{R^{2}tanA}{{\Delta }_1}$
similarly $\frac{R_2}{r_2}=\frac{R^{2}tanB}{{\Delta }_2},\frac{R_3}{r_3}=\frac{R^{2}tanC}{{\Delta }_3}$

Hence
$\frac{R_1}{r_1}+\frac{R_2}{r_2}+\frac{R_3}{r_3}=R^2[\frac{tanA}{{\Delta }_1}+\frac{tanB}{{\Delta }_2}+\frac{tanC}{{\Delta }_3}]$