In triangle ABC,find the value of frac{r_1-r} | vedclass

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(C) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Substituting these into t

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$\frac{c}{r_3}$

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(C) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Substituting these into the expression:$\frac{r_1-r}{a} + \frac{r_2-r}{b} = \frac{\frac{\Delta}{s-a} - \frac{\Delta}{s}}{a} + \frac{\frac{\Delta}{s-b} - \frac{\Delta}{s}}{b}$$= \frac{\Delta}{a} \left( \frac{s - (s-a)}{s(s-a)} \right) + \frac{\Delta}{b} \left( \frac{s - (s-b)}{s(s-b)} \right)$$= \frac{\Delta}{a} \left( \frac{a}{s(s-a)} \right) + \frac{\Delta}{b} \left( \frac{b}{s(s-b)} \right)$$= \frac{\Delta}{s(s-a)} + \frac{\Delta}{s(s-b)}$$= \frac{\Delta}{s} \left( \frac{1}{s-a} + \frac{1}{s-b} \right) = \frac{\Delta}{s} \left( \frac{s-b+s-a}{(s-a)(s-b)} \right)$$= \frac{\Delta}{s} \left( \frac{2s-a-b}{(s-a)(s-b)} \right) = \frac{\Delta}{s} \left( \frac{c}{(s-a)(s-b)} \right)$Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $(s-a)(s-b) = \frac{\Delta^2}{s(s-c)}$.Substituting this:$= \frac{\Delta}{s} \cdot \frac{c \cdot s(s-c)}{\Delta^2} = \frac{c(s-c)}{\Delta} = \frac{c}{\frac{\Delta}{s-c}} = \frac{c}{r_3}$.

In $\triangle ABC$,find the value of $\frac{r_1-r}{a}+\frac{r_2-r}{b}$.

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