In triangle ABC,if b=6, c=7 and tan frac{A}{2 | vedclass

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(D) Given $b=6, c=7$ and $	an \frac{A}{2}=\frac{1}{\sqrt{6}}$.Using the formula $\cos A = \frac{1-	an^2(A/2)}{1+	an^2(A/2)}$,we get:$\cos A = \frac

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$\frac{2 \sqrt{6}}{3}$

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(D) Given $b=6, c=7$ and $	an \frac{A}{2}=\frac{1}{\sqrt{6}}$.Using the formula $\cos A = \frac{1-	an^2(A/2)}{1+	an^2(A/2)}$,we get:$\cos A = \frac{1-1/6}{1+1/6} = \frac{5/6}{7/6} = \frac{5}{7}$.Using the Cosine Rule $\cos A = \frac{b^2+c^2-a^2}{2bc}$:$\frac{5}{7} = \frac{6^2+7^2-a^2}{2 	imes 6 	imes 7} = \frac{36+49-a^2}{84}$.$60 = 85 - a^2$ $\Rightarrow a^2 = 25$ $\Rightarrow a=5$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{5+6+7}{2} = 9$.The inradius $r$ is given by $r = (s-a) 	an \frac{A}{2}$.$r = (9-5) 	imes \frac{1}{\sqrt{6}} = \frac{4}{\sqrt{6}} = \frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$.

In triangle $ABC$,if $b=6, c=7$ and $\tan \frac{A}{2}=\frac{1}{\sqrt{6}}$,then the inradius of $\triangle ABC$ is

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