In a triangle ABC,if r_1=6, r_2=9, r_3=18,the | vedclass

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(B) We know that $r_1 = \frac{\Delta}{s-a} = 6 \Rightarrow s-a = \frac{\Delta}{6}$   $(i)$$r_2 = \frac{\Delta}{s-b} = 9 \Rightarrow s-b = \frac{\Delta

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$\frac{4}{5}$

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(B) We know that $r_1 = \frac{\Delta}{s-a} = 6 \Rightarrow s-a = \frac{\Delta}{6}$   $(i)$$r_2 = \frac{\Delta}{s-b} = 9 \Rightarrow s-b = \frac{\Delta}{9}$   $(ii)$$r_3 = \frac{\Delta}{s-c} = 18 \Rightarrow s-c = \frac{\Delta}{18}$   $(iii)$Adding $(i), (ii)$ and $(iii)$ gives $3s - (a+b+c) = \Delta(\frac{1}{6} + \frac{1}{9} + \frac{1}{18}) = \Delta(\frac{3+2+1}{18}) = \frac{\Delta}{3}$.Since $a+b+c = 2s$,we have $3s - 2s = \frac{\Delta}{3} \Rightarrow s = \frac{\Delta}{3}$.Substituting $s$ in $(i), (ii), (iii)$:$a = s - \frac{\Delta}{6} = \frac{\Delta}{3} - \frac{\Delta}{6} = \frac{\Delta}{6} = \frac{3\Delta}{18}$.$b = s - \frac{\Delta}{9} = \frac{\Delta}{3} - \frac{\Delta}{9} = \frac{2\Delta}{9} = \frac{4\Delta}{18}$.$c = s - \frac{\Delta}{18} = \frac{\Delta}{3} - \frac{\Delta}{18} = \frac{5\Delta}{18}$.Thus,$a:b:c = 3:4:5$. Let $a=3k, b=4k, c=5k$.Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{(4k)^2+(5k)^2-(3k)^2}{2(4k)(5k)} = \frac{(16+25-9)k^2}{40k^2} = \frac{32}{40} = \frac{4}{5}$.

In a triangle $ABC$,if $r_1=6, r_2=9, r_3=18$,then $\cos A=$

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