In triangle ABC,if a^2 sin^2 frac{C}{2} + c^2 | vedclass

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(A) Given: $a^2 \sin^2 \frac{C}{2} + c^2 \sin^2 \frac{A}{2} = \frac{b^2}{2}$ Using the half-angle formulas $\sin^2 \frac{C}{2} = \frac{(s-a)(s-b)}{ab}

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$2 : 1$

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(A) Given: $a^2 \sin^2 \frac{C}{2} + c^2 \sin^2 \frac{A}{2} = \frac{b^2}{2}$ Using the half-angle formulas $\sin^2 \frac{C}{2} = \frac{(s-a)(s-b)}{ab}$ and $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$: $a^2 \left( \frac{(s-a)(s-b)}{ab} \right) + c^2 \left( \frac{(s-b)(s-c)}{bc} \right) = \frac{b^2}{2}$ $\Rightarrow \frac{a(s-a)(s-b)}{b} + \frac{c(s-b)(s-c)}{b} = \frac{b^2}{2}$ $\Rightarrow \frac{s-b}{b} [a(s-a) + c(s-c)] = \frac{b^2}{2}$ $\Rightarrow (s-b) [as - a^2 + cs - c^2] = \frac{b^3}{2}$ Since $2s = a+b+c$,we have $s-b = \frac{a+c-b}{2}$. Substituting and simplifying leads to $a+c = 2b$. Thus,$\frac{a+c}{b} = \frac{2}{1}$,so $a+c : b = 2 : 1$.

In $\triangle ABC$,if $a^2 \sin^2 \frac{C}{2} + c^2 \sin^2 \frac{A}{2} = \frac{b^2}{2}$,then $a+c : b =$

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