Solving $\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$ where $x$, $y$, $z$ are angles of a triangle Can any one give me a hint to find value of $x$
where: 

$$\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$$
  and $x$, $y$, $z$ are angles of a triangle.

I tried to use sine law but got nothing.
 A: $2R=a/sinA=b/sinB=c/sinC$
$a=4 ,b=3 ,c=2$
Heron's formula 
$S=1/2*3*2sinx=√9/2*1/2*3/2*5/2$
$sinx=√15/4$
A: Unless $x, y, z$ represent the angles of a triangle, there'll be infinitely many solutions.
If three sides of a triangle are given, it's better to use cosine law:
\begin{align*}
  \cos x &= \frac{3^{2}+2^{2}-4^{2}}{2(3)(2)} \\
  \cos y &= \frac{4^{2}+2^{2}-3^{2}}{2(4)(2)} \\
  \cos z &= \frac{3^{2}+4^{2}-2^{2}}{2(3)(4)} \\
\end{align*}
A: My solution involves finding the size of angle x. Then getting it's sine. 
From the sine law we can tell that the triangle has sides 3, 4 and 2. And that angle x is opposite to the side with length 4.
We can employ cosine law to find the size of angle x.
$c^2 = a^2 + b^2 + 2ab\cos\theta$ 
$4^2 = 2^2 + 3 ^2 - 2(3*2)\cos\theta$
This evaluates to 
$\cos\theta = -\frac{3}{12} = -0.25$
To get the size of the angle, feed $- 0.25$ into the inverse cos function
$\cos^{-1} (-0.25) = 104.478^\circ $
Finally, get the sine of the angle
$\sin (104.478^\circ) = 0.968 $
Therefore, $\sin  x = 0.968 $
