$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $ and $ \ x+y+z \ = \ xyz \ $ Consider the system of equations in real numbers $ \ x,y,z \  $ satisfying
$$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \  \frac{y}{5 \ \sqrt{y^2+1}} \ = \  \frac{z}{6 \ \sqrt{z^2+1}} $$
and $ \ x+y+z \ = \ xyz \ $. Find $ \ x,y,z \ . $
I substituted $ \ x,y,z \  $ as $ \ \tan(\theta_1) \ , \ \tan(\theta_2) \ , \ \tan(\theta_3) \  $  , where 
$$ \theta_1 \ + \ \theta_2 \  + \ \theta_3 \ = \  180º \ \ . $$

Starting with $\frac{sint}{4} = \frac{sinu}{5} = \frac{sinw}{6}$, then use the law of sines to get: $\frac{a}{4} = \frac{b}{5} = \frac{c}{6}$. So $a = \frac{2c}{3}$, $b = \frac{5c}{6}$.
So apply the law of cosines to have: $a^2 = b^2 + c^2 - 2bccosA$. Substituting these values of $a$ and $b$ into the above equation we can solve for $cosA$, and then $tanA$, and $tanB$, and $tanC$.
