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In a $\triangle A B C$, if $a=13, b=14$ and $c=15$, then the value of $\sin \left(\frac{A}{2}\right)$ is
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The correct answer is:
$\frac{1}{\sqrt{5}}$
we know that, $\begin{aligned} 2 s & =a+b+c \\ 2 s & =42 \\ s & =21\end{aligned}$
$s-a=8, s-b=7$, and $s-c=6$
$\sin \left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}$
$=\sqrt{\frac{7 \times 6}{14 \times 15}}=\frac{1}{\sqrt{5}}$
$s-a=8, s-b=7$, and $s-c=6$
$\sin \left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}$
$=\sqrt{\frac{7 \times 6}{14 \times 15}}=\frac{1}{\sqrt{5}}$
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