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

$\Rightarrow \tan 2B=\frac{2\tan B}{1-{\tan }^{2}B}=\frac{\frac{2}{3}}{1-\frac{1}{9}}$

$=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4}$

$\Rightarrow \sin A=\frac{1}{5\sqrt{2}}$

Then $\tan A=\frac{1}{7}$

$\tan \left[A+2B\right]=\frac{\tan A+\tan 2B}{1-\tan A.\tan 2B}$

$=\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\times \frac{1}{7}}$

$=\frac{25}{25}=1$

$\tan \left[A+2B\right]=1$

$\therefore A+2B=\frac{\pi }{4}$