If the angles A, B,C of a triangle are in A.P. and b : c $=\sqrt{3}:\sqrt{2}$ and the angle A is $\frac{5\mathrm{\pi }}{2\mathrm{k}}$, then k is _______

$\mathrm{A},\mathrm{B},\mathrm{C}\text{ are in A.P }\Rightarrow 2\mathrm{B}=\mathrm{A}+\mathrm{C}\text{; }$
$3\mathrm{B}=\mathrm{A}+\mathrm{B}+\mathrm{C}={180}^{\circ };\mathrm{B}={60}^{\circ }$
$\text{ Given }\frac{\mathrm{b}}{\mathrm{c}}=\frac{\sqrt{3}}{\sqrt{2}}\Rightarrow \frac{\sin \mathrm{B}}{\sin \mathrm{C}}=\frac{\sqrt{3}}{\sqrt{2}}$
$\frac{\sqrt{3}}{2}\times \frac{1}{\sin \mathrm{C}}=\frac{\sqrt{3}}{\sqrt{2}}\Rightarrow \sin \mathrm{C}=\frac{1}{\sqrt{2}}\Rightarrow \mathrm{C}={45}^{\circ }$
$\text{ Now }\mathrm{A}={180}^{\circ }-(\mathrm{B}+\mathrm{C})={180}^{\circ }-{105}^{\circ }={75}^{\circ }=\frac{5\mathrm{\pi }}{12}$
$\mathrm{A}=\frac{5\mathrm{\pi }}{2\mathrm{k}}\Rightarrow \mathrm{k}=6$