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 $tanC$.
This is a comment that’s too long for the usual format. LAcarguy’s method generalizes to any values $a,b,c$ instead of $\frac14,\frac15,\frac16$.
If we put $$ \phi(a,b,c)=ab+ac-bc, \ \psi(a,b,c)=ab+ac+bc \tag{1} $$
and
$$ G(a,b,c)=\phi(a,b,c)\phi(b,c,a)\phi(c,a,b)\psi(a,b,c) \tag{2} $$
then LAcarguy’s method leads to the following explicit formulas for the solutions :
$$ x=\frac{\varepsilon}{\phi(a^2,b^2,c^2)}\sqrt{G(a,b,c)},\ y=\frac{\varepsilon}{\phi(b^2,c^2,a^2)}\sqrt{G(a,b,c)},\ z=\frac{\varepsilon}{\phi(c^2,a^2,b^2)}\sqrt{G(a,b,c)} \tag{3} $$
where $\varepsilon$ is $+1$ or $-1$. In your initial question where $a=\frac14,b=\frac15,c=\frac16$, one finds $G=\frac{7}{921600}$,
$$ x=\varepsilon\frac{\sqrt{7}}{3}, \ y=\varepsilon\frac{5\sqrt{7}}{9}, \ z=3\varepsilon\sqrt{7} \tag{4} $$
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 $$A=\pm \cos ^{-1}\left(\frac{3}{4}\right)=\pm 0.722734$$
$\frac{\sin(\theta_1)}{4}=\frac{\sin(\theta_2)}{5}=\frac{\sin(\theta_3)}{6} \implies$
$15\sin(\theta_1)=12\sin(\theta_2)=10\sin(\theta_3)$ Not sure where to take it from there though. I haven't messed with trig functions for a while.
– ruler501 Mar 27 '14 at 06:34