If 4+ 8- 32+ 768 =a 2 ((11 π)/(b)), where a and b are natural...

$$\begin{aligned} & \text { Let } \sqrt{768}=32 \cos \theta \\ & 16 \sqrt{3}=32 \cos \theta \\ & \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} \end{aligned}$$\text { So } \sqrt{4+\sqrt{8-\sqrt{32+32 \cos \frac{\pi}{6}}}}=\sqrt{4+\sqrt{8-8 \cos \frac{\pi}{12}}}=\sqrt{4+4 \sin \frac{\pi}{24}}=\sqrt{4+4 \cos \frac{11 \pi}{24}}=\sqrt{4\left(1+\cos \frac{11 \pi}{24}\right)}=\sqrt{4.2 \cos ^2 \frac{11 \pi}{48}}=2 \sqrt{2} \cos \frac{11 \pi}{48}$So$\mathrm{a}=2, \mathrm{~b}=48\therefore \frac{\mathrm{b}}{\mathrm{a}}=24$