If A and B are acute angles and sinA=$\frac{1}{5\sqrt{2}}$ and tanB = $\frac{1}{3}$ then A+2B=

$\begin{array}{rcl}\sin \mathrm{A}& =& \frac{1}{5\sqrt{2}}, \tan \mathrm{B}=\frac{1}{3}\\ \tan 2\mathrm{B}& =& \frac{2\tan \mathrm{B}}{1-{\tan }^2\mathrm{B} }\\ & =& \frac{2\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}{1-{\displaystyle \frac{1}{9}}}=\frac{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{\displaystyle \frac{9-1}{9}}}\\ & =& \frac{6}{8}=\frac{3}{4}\\ \tan \mathrm{A}& =& \frac{1}{7}\\ \tan (\mathrm{A}+2\mathrm{B})& =& \frac{\tan \mathrm{A}+\tan 2\mathrm{B}}{1-\tan \mathrm{A} \tan 2\mathrm{B}}\\ & =& \frac{{\displaystyle \frac{1}{7}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{1}{7}}·{\displaystyle \frac{3}{4}}}\\ & =& \frac{4+21}{28-3}=\frac{25}{25}=1\\ \mathrm{A}+2\mathrm{B}& =& \frac{\mathrm{\pi }}{4}\end{array}$