In a triangle ABC,if r r_2 = r_1 r_3,then cos | vedclass

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(A) We have $r r_2 = r_1 r_3$. Using the formulas $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta

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(A) We have $r r_2 = r_1 r_3$. Using the formulas $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$: $\left(\frac{\Delta}{s}ight) \left(\frac{\Delta}{s-b}ight) = \left(\frac{\Delta}{s-a}ight) \left(\frac{\Delta}{s-c}ight)$ $\Rightarrow s(s-b) = (s-a)(s-c)$ $\Rightarrow s^2 - sb = s^2 - s(a+c) + ac$ $\Rightarrow s(a+c-b) = ac$ Since $2s = a+b+c$,we have $a+c = 2s-b$. Substituting this: $s(2s-b-b) = ac$ $\Rightarrow s(2s-2b) = ac$ $\Rightarrow 2s(s-b) = ac$. Using the identity $\cos B = \frac{a^2+c^2-b^2}{2ac}$,we know $\sin^2(\frac{B}{2}) = \frac{(s-a)(s-c)}{ac}$ and $\cos^2(\frac{B}{2}) = \frac{s(s-b)}{ac}$. From $s(s-b) = (s-a)(s-c)$,we get $\frac{s(s-b)}{ac} = \frac{(s-a)(s-c)}{ac} \Rightarrow \cos^2(\frac{B}{2}) = \sin^2(\frac{B}{2})$. $\Rightarrow 	an^2(\frac{B}{2}) = 1$ $\Rightarrow \frac{B}{2} = \frac{\pi}{4}$ $\Rightarrow B = \frac{\pi}{2}$. Therefore,$\cos 2B = \cos \pi = -1$.

In a triangle $ABC$,if $r r_2 = r_1 r_3$,then $\cos 2B =$

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