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In a $\triangle A B C$, if $a+3 b=3 c$, then $\sin \frac{A}{2}=$
Options:
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Verified Answer
The correct answer is:
$\frac{a}{3} \sqrt{\frac{2}{b c}}$
Since,
$\sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}, s=\frac{a+b+c}{2}$
$=\sqrt{\frac{\frac{a-b+c}{2} \times \frac{a+b-c}{2}}{b c}}$
$=\sqrt{\frac{\frac{4(c-b)}{2} \times \frac{2(c-b)}{2}}{b c}} \quad[$ as $a=3(c-b)]$
$=(c-b) \sqrt{\frac{2}{b c}} \quad\{\because c>b\}$
$=\frac{a}{3} \sqrt{\frac{2}{b c}}$
Hence, option (b) is correct.
$\sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}, s=\frac{a+b+c}{2}$
$=\sqrt{\frac{\frac{a-b+c}{2} \times \frac{a+b-c}{2}}{b c}}$
$=\sqrt{\frac{\frac{4(c-b)}{2} \times \frac{2(c-b)}{2}}{b c}} \quad[$ as $a=3(c-b)]$
$=(c-b) \sqrt{\frac{2}{b c}} \quad\{\because c>b\}$
$=\frac{a}{3} \sqrt{\frac{2}{b c}}$
Hence, option (b) is correct.
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