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.

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 $
