In triangle ABC,suppose the radius of the exc | vedclass

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(A) Given,in $	riangle ABC$,$r_1=2, r_2=3, r_3=6$. We know the formulas for exradii: $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delt

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$(3, 4, 5)$

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(A) Given,in $	riangle ABC$,$r_1=2, r_2=3, r_3=6$. We know the formulas for exradii: $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$. Thus,$s-a=\frac{\Delta}{2}, s-b=\frac{\Delta}{3}, s-c=\frac{\Delta}{6}$. Adding these equations: $(s-a)+(s-b)+(s-c) = \Delta(\frac{1}{2}+\frac{1}{3}+\frac{1}{6})$. $3s-(a+b+c) = \Delta(\frac{3+2+1}{6}) = \Delta$. Since $a+b+c=2s$,we have $3s-2s = \Delta$,so $s=\Delta$. Using Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$. Substituting $s=\Delta$ and the expressions for $(s-a), (s-b), (s-c)$: $\Delta = \sqrt{\Delta \cdot \frac{\Delta}{2} \cdot \frac{\Delta}{3} \cdot \frac{\Delta}{6}} = \sqrt{\frac{\Delta^4}{36}} = \frac{\Delta^2}{6}$. Since $\Delta 
eq 0$,we get $1 = \frac{\Delta}{6}$,so $\Delta = 6$. Thus,$s = 6$. Now,$s-a = \frac{6}{2} = 3 \Rightarrow a = 6-3 = 3$. $s-b = \frac{6}{3} = 2 \Rightarrow b = 6-2 = 4$. $s-c = \frac{6}{6} = 1 \Rightarrow c = 6-1 = 5$. Therefore,$(a, b, c) = (3, 4, 5)$.

In $\triangle ABC$,suppose the radius of the excircle opposite to angle $A$ is denoted by $r_1$,similarly $r_2$ for angle $B$,and $r_3$ for angle $C$. If $r_1=2, r_2=3, r_3=6$,then what is $(a, b, c)$?

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