$ \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 
$$ \theta_1 \ + \ \theta_2 \  + \ \theta_3 \ = \  180º \ \ . $$

If you have $$15\sin(\theta_1)=12\sin(\theta_2)=10\sin(\theta_3)$$ as suggested by  ruler501, you can extract $\theta_2$ and $\theta_3$ as a function of $\theta_1$. This leads to $$\theta_2=\sin ^{-1}\left(\frac{5 \sin (\theta_1)}{4}\right)$$ $$\theta_3=\sin ^{-1}\left(\frac{3 \sin (\theta_1)}{2}\right)$$ So, the equation to solve is $$\theta_1+\sin ^{-1}\left(\frac{5 \sin (\theta_1)}{4}\right)+\sin ^{-1}\left(\frac{3 \sin (\theta_1)}{2}\right)=\pi$$ which only has three solutions. There is an obvious solution which is $\theta_1=\pi$. The other are $\theta_1=0.722734$ and $\theta_1=5.56045$ which I evaluated using numeric methods.
If you use the more elegant solution proposed by LAcarguy while I was typing my answer, the solution is quite simple since you arrive to $$A=\pm \cos ^{-1}\left(\frac{3}{4}\right)=\pm 0.722734$$
