The three angles $\pi /18,5\pi /18,7\pi /18$ are $10^{\circ },50^{\circ },70^{\circ }$. Use the identity

$\sin \theta \sin (60^{\circ }-\theta )\sin (60^{\circ }+\theta )=\frac{1}{4}\sin 3\theta$

with $\theta =10^{\circ }$ (so the angles are $10^{\circ },50^{\circ },70^{\circ }$):

$K=\frac{1}{4}\sin 30^{\circ }=\frac{1}{8}$

Then $\frac{10K\pi }{3}=\frac{10\pi }{24}=\frac{5\pi }{12}$, which equals $75^{\circ }$.

$\sin 75^{\circ }=\sin (45^{\circ }+30^{\circ })=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}$.