How can I show that $\displaystyle \displaystyle\tan{2\alpha}=\frac{24}{7}\equiv\tan{ \alpha}=\frac{3}{4}\mbox{?}$
$\displaystyle \displaystyle\tan{2\alpha}=\frac{2\tan{\alpha}}{1-\tan^2{\alpha}}=\frac{24}{7}\Rightarrow 7\tan{\alpha}=12-12\tan^2{\alpha}$
$\displaystyle \displaystyle\Rightarrow 12\tan^2{\alpha}+7\tan{\alpha}-12=(3\tan{\alpha}+4)(4\tan{\alpha}-3)=0$
$\displaystyle \displaystyle\tan{\alpha}=\frac{3}{4}, \ \frac{-4}{3}$
These are both 3,4,5 right triangles.
This comes from PDE so it doesn't matter which one I would pick for the overall solution of the PDE.