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(B) Let $a$ be the side length and $b$ be the diagonal length of a regular pentagon inscribed in a circle. In a regular pentagon, the interior angle i

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$2 \cos \frac{\pi}{5}$

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(B) Let $a$ be the side length and $b$ be the diagonal length of a regular pentagon inscribed in a circle. In a regular pentagon, the interior angle is $\frac{3\pi}{5}$. Consider the triangle formed by two sides and a diagonal. By properties of a regular pentagon, the diagonal $b$ subtends an angle of $\frac{2\pi}{5}$ at the center, and the angle between a side and a diagonal is $\frac{\pi}{5}$. Drawing a perpendicular from vertex $C$ to the diagonal $AB$ at point $D$, we form a right-angled triangle $	riangle ACD$. In $	riangle ACD$, the angle $\angle CAD = \frac{\pi}{5}$ and the hypotenuse $AC = a$. Thus, $\cos \frac{\pi}{5} = \frac{AD}{AC} = \frac{AD}{a}$, which implies $AD = a \cos \frac{\pi}{5}$. Since the diagonal $b$ is bisected by the perpendicular from the opposite vertex in a regular pentagon, $b = 2AD$. Therefore, $b = 2a \cos \frac{\pi}{5}$, which gives $\frac{b}{a} = 2 \cos \frac{\pi}{5}$.

If $a$ and $b$ respectively represent the lengths of a side and a diagonal of a regular pentagon that is inscribed in a circle, then $\frac{b}{a}=$

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