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The sides of a $\triangle A B C$ are given by $a=3, b=5$ and $c=3$. Then, $\cos A=$
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$5 / 6$
$\begin{aligned} & a=3, b=5, c=3 \\ & \cos A=\frac{b^2+c^2-a^2}{2 b c}\end{aligned}$
$\begin{aligned} & =\frac{25+9-9}{2 \times 5 \times 3} \\ & =\frac{25}{30}=\frac{5}{6}\end{aligned}$
$\begin{aligned} & =\frac{25+9-9}{2 \times 5 \times 3} \\ & =\frac{25}{30}=\frac{5}{6}\end{aligned}$
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