If in a triangle ABC,a=2,b=3 and c=4,then tan | vedclass

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(D) Given,in a triangle $ABC$,$a=2$,$b=3$,and $c=4$. We know that the semi-perimeter $s = \frac{a+b+c}{2} = \frac{2+3+4}{2} = 4.5$. The formula for $\

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$\sqrt{\frac{1}{15}}$

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(D) Given,in a triangle $ABC$,$a=2$,$b=3$,and $c=4$. We know that the semi-perimeter $s = \frac{a+b+c}{2} = \frac{2+3+4}{2} = 4.5$. The formula for $	an \left(\frac{A}{2}ight)$ is $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$. Substituting the values: $	an \left(\frac{A}{2}ight) = \sqrt{\frac{(4.5-3)(4.5-4)}{4.5(4.5-2)}}$ $= \sqrt{\frac{1.5 	imes 0.5}{4.5 	imes 2.5}}$ $= \sqrt{\frac{0.75}{11.25}}$ $= \sqrt{\frac{75}{1125}} = \sqrt{\frac{1}{15}}$.

If in a triangle $ABC$,$a=2$,$b=3$ and $c=4$,then $\tan \left(\frac{A}{2}\right) = $

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