In a triangle XYZ , Let x, y, z  be the lengths of sides opposite to the angles X,Y,Z respectively ,and $2s=x+y+z.\text{\hspace{0.17em}If\hspace{0.17em}}\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}$and area of in circle  of the triangle XYZ is  $\frac{8\pi }{3}$, then 

$\begin{array}{rcl}Given \frac{s-x}{4}& =& \frac{s-y}{3}=\frac{s-z}{2}=k\\ s-x& =& 4k\\ s-y& =& 3k\\ s-z& =& 2k\\ 3s-2s& =& 9k\\ s& =& 9k\\ Therefore, x& =& 5,y=6,z=7, S=9,\end{array}$

$\text{ Area }=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=6\sqrt{6}$

$\begin{array}{rcl}r& =& \frac{∆}{S}\\ & =& \frac{2\sqrt{6}}{3}\end{array}$

$\begin{array}{rcl}xyz& =& 4R∆\\ R& =& \frac{210}{24\sqrt{6}}\\ & =& \frac{35\sqrt{6}}{24}\\ & & \end{array}$

$$\begin{gathered} r=4R\sin \frac{x}{2}\sin \frac{y}{2}\sin \frac{z}{2} \\ \Rightarrow \sin \frac{x}{2}\sin \frac{y}{2}\sin \frac{z}{2}=\frac{r}{4R}=\frac{4}{35} \end{gathered}$$

$\begin{array}{rcl}{\sin }^2\left(\frac{x+y}{2}\right)& =& \frac{1-\cos \left(x+y\right)}{2}\\ & =& \frac{1+\cos z}{2}\\ & =& \frac{1}{2}\left[1+\frac{x^2+y^2-z^2}{2xy}\right]\\ & =& \frac{3}{5}\end{array}$