In a triangle ABC,if a+3b=3c,then sin frac{A} | vedclass

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(B) Given $a+3b=3c$,we have $a=3(c-b)$.Using the formula $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,where $s = \frac{a+b+c}{2}$.$s-b = \frac{a+

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$\frac{a}{3} \sqrt{\frac{2}{bc}}$

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(B) Given $a+3b=3c$,we have $a=3(c-b)$.Using the formula $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,where $s = \frac{a+b+c}{2}$.$s-b = \frac{a+b+c}{2} - b = \frac{a-b+c}{2} = \frac{3(c-b)-b+c}{2} = \frac{4c-4b}{2} = 2(c-b)$.$s-c = \frac{a+b+c}{2} - c = \frac{a+b-c}{2} = \frac{3(c-b)+b-c}{2} = \frac{2c-2b}{2} = c-b$.Substituting these into the formula:$\sin \frac{A}{2} = \sqrt{\frac{2(c-b)(c-b)}{bc}} = \sqrt{\frac{2(c-b)^2}{bc}} = (c-b) \sqrt{\frac{2}{bc}}$.Since $c-b = \frac{a}{3}$,we get $\sin \frac{A}{2} = \frac{a}{3} \sqrt{\frac{2}{bc}}$.Thus,option $B$ is correct.

In a $\triangle ABC$,if $a+3b=3c$,then $\sin \frac{A}{2} =$

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