<s>[INST] 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.
 [/INST] 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 $
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