If R : r_1 : r = 5 : 12 : 2,then r + r_3 + r_ | vedclass

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(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.We need to evaluate $r +

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$\cos A$

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(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.We need to evaluate $r + r_3 + r_2 - r_1$.Substituting the formulas,we get:$r + r_3 + r_2 - r_1 = \frac{\Delta}{s} + \frac{\Delta}{s-c} + \frac{\Delta}{s-b} - \frac{\Delta}{s-a}$$= \Delta \left( \frac{1}{s} - \frac{1}{s-a} \right) + \Delta \left( \frac{1}{s-c} + \frac{1}{s-b} \right)$$= \Delta \left( \frac{s-a-s}{s(s-a)} \right) + \Delta \left( \frac{s-b+s-c}{(s-c)(s-b)} \right)$$= \Delta \left( \frac{-a}{s(s-a)} \right) + \Delta \left( \frac{2s-b-c}{(s-c)(s-b)} \right)$Since $2s = a+b+c$,we have $2s-b-c = a$.$= \Delta \left( \frac{-a}{s(s-a)} + \frac{a}{(s-c)(s-b)} \right)$$= \Delta a \left( \frac{-(s-c)(s-b) + s(s-a)}{s(s-a)(s-b)(s-c)} \right)$Using $\Delta^2 = s(s-a)(s-b)(s-c)$,we get:$= \frac{\Delta a}{\Delta^2} (-(s^2 - (b+c)s + bc) + (s^2 - as))$$= \frac{a}{\Delta} (-s^2 + (b+c)s - bc + s^2 - as)$$= \frac{a}{\Delta} ((b+c-a)s - bc)$$= \frac{a}{\Delta} ((b+c-a)(\frac{a+b+c}{2}) - bc)$$= \frac{a}{2\Delta} ((b+c)^2 - a^2 - 2bc) = \frac{a}{2\Delta} (b^2 + c^2 + 2bc - a^2 - 2bc) = \frac{a(b^2+c^2-a^2)}{2\Delta}$Since $\cos A = \frac{b^2+c^2-a^2}{2bc}$,we have $b^2+c^2-a^2 = 2bc \cos A$.So,the expression becomes $\frac{a(2bc \cos A)}{2\Delta} = \frac{abc \cos A}{\Delta}$.Since $\Delta = \frac{abc}{4R}$,we have $\frac{abc}{\Delta} = 4R$.Thus,the expression is $4R \cos A$. However,given the options and the specific ratio $R:r_1:r = 5:12:2$,we observe that for a right-angled triangle at $A$,$r = s-a$,$r_1 = s$,$r_2 = s-c$,$r_3 = s-b$. The expression $r+r_3+r_2-r_1 = (s-a) + (s-b) + (s-c) - s = 2s - (a+b+c) = 0$. Since $\cos 90^\circ = 0$,the correct option is $\cos A$.

If $R : r_1 : r = 5 : 12 : 2$,then $r + r_3 + r_2 - r_1 = $

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