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In a triangle, if $b=5, c=6, \tan \frac{A}{2}=\frac{1}{\sqrt{2}}$, then $a=$
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Verified Answer
The correct answer is:
$\sqrt{41}$
$\because \cos A=\frac{1-\tan ^2 \frac{A}{2}}{1+\tan ^2 \frac{A}{2}}$
$=\frac{1-\left(\frac{1}{\sqrt{2}}\right)^2}{1+\left(\frac{1}{\sqrt{2}}\right)^2}=\frac{1}{3}$
$\cos A=\frac{b^2+c^2-a^2}{2 b c}$
$\frac{1}{3}=\frac{(5)^2+(6)^2-a^2}{2(5)(6)}$
$\begin{aligned} & 20=25+36-a^2 \\ & a^2=41 \\ & a=\sqrt{41}\end{aligned}$
$=\frac{1-\left(\frac{1}{\sqrt{2}}\right)^2}{1+\left(\frac{1}{\sqrt{2}}\right)^2}=\frac{1}{3}$
$\cos A=\frac{b^2+c^2-a^2}{2 b c}$
$\frac{1}{3}=\frac{(5)^2+(6)^2-a^2}{2(5)(6)}$
$\begin{aligned} & 20=25+36-a^2 \\ & a^2=41 \\ & a=\sqrt{41}\end{aligned}$
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