Question

In $△ABC,\frac{r_1}{bc}+\frac{r_2}{ca}+\frac{r_3}{ab}$ is equal to

• A. $\frac{1}{2R}-\frac{1}{r}$
• B. $2R-r$
• C. $\frac{1}{r}-\frac{1}{2R}$
• D. $R-2R$

Answer 1

Answer: (C)

$\frac{1}{r}-\frac{1}{2R}$

$\frac{r_1}{bc}=\frac{4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}{2R\sin B\cdot 2R\sin C}$

$=\frac{\sin \frac{A}{2}}{4R\sin \frac{B}{2}\sin \frac{C}{2}}$

$=\frac{{\sin }^2\frac{A}{2}}{4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}=\frac{{\sin }^2\frac{A}{2}}{r}$

$\therefore \frac{r_1}{bc}+\frac{r_2}{ca}+\frac{r_3}{ab}=\frac{1}{r}({\sin }^2\frac{A}{2}+{\sin }^2\frac{B}{2}+{\sin }^2\frac{C}{2})$

$=\frac{1}{2r}(1-\cos A+1-\cos B+1-\cos C)$

$=\frac{1}{2r}[3-(\cos A+\cos B+\cos C\left)\right]$

$=\frac{1}{2r}[3-(1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\left)\right]$

$=\frac{1}{2r}[2-\frac{r}{R}]=\frac{1}{r}-\frac{1}{2R}$