Star inside pentagon

Let the point $O$ be the centre of the pentagon, and the point $G$ be the intersection point of sides of triangle with base $EA$ and $BC$ respectively.

Note that $\angle FBG = \frac{3\pi}{5} - \frac{\pi}{3} = \frac{\pi}{15}=12 ^{\circ}$ . This implies that $\angle GBA = \frac{\pi}{3} - \frac{\pi}{15} = \frac{4\pi}{15}=48 ^{\circ}$ and that $\angle OBA = \frac{4\pi}{15} + \frac{1}{2}\times \frac{\pi}{15}=\frac{3\pi}{10}=54 ^{\circ}$

Now, the area of the triangle $AGB$ is $\frac{AB^2\tan 48 ^{\circ}}{4}$ while the area of the triangle $AOB$ is $\frac{AB^2\tan 54 ^{\circ}}{4}$

Hence $\displaystyle \left \lfloor \frac{1000U}{T} \right \rfloor$ $=\left \lfloor 1000 \left(1-\frac{\tan 48 ^{\circ}}{\tan 54 ^{\circ}}\right) \right \rfloor=193$