In a triangle,if b=5, c=6,and tan frac{A}{2}= | vedclass

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(A) Given $	an \frac{A}{2} = \frac{1}{\sqrt{2}}$.Using the formula $\cos A = \frac{1-	an^2 \frac{A}{2}}{1+	an^2 \frac{A}{2}}$:$\cos A = \frac{1 - (

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$\sqrt{41}$

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(A) Given $	an \frac{A}{2} = \frac{1}{\sqrt{2}}$.Using the formula $\cos A = \frac{1-	an^2 \frac{A}{2}}{1+	an^2 \frac{A}{2}}$:$\cos A = \frac{1 - (\frac{1}{\sqrt{2}})^2}{1 + (\frac{1}{\sqrt{2}})^2} = \frac{1 - 1/2}{1 + 1/2} = \frac{1/2}{3/2} = \frac{1}{3}$.Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.$\frac{1}{3} = \frac{5^2 + 6^2 - a^2}{2(5)(6)}$.$\frac{1}{3} = \frac{25 + 36 - a^2}{60}$.$20 = 61 - a^2$.$a^2 = 41$.$a = \sqrt{41}$.

In a triangle,if $b=5, c=6$,and $\tan \frac{A}{2}=\frac{1}{\sqrt{2}}$,then $a=$

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