In general the circumscribed radius ($R$) of a truncated icosahedron, having 12 congruent regular pentagonal faces & 20 congruent regular hexagonal faces each with edge length $L$, is given by the generalized expression (derived in Mathematical analysis of truncated icosahedron by HCR) $\bbox[4pt, border: 1px solid blue;]{\color{blue}{R}=\color{red}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}\color{purple}{\approx 2.478018659\space L}}$ Now, join one of end-points & the mid-point of the edge to the center of truncated icosahedron, we get a right triangle with base $\color{blue}{\frac{L}{2}}$ & hypotenuse $\color{blue}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}$
Hence, the angle $\theta$ between the edge & the radius is given as $\cos \theta=\frac{\frac{L}{2}}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}=\frac{2}{\sqrt{58+18\sqrt{5}}}=\sqrt{\frac{29-9\sqrt{5}}{218}}$ $\implies \color{blue}{ \theta=\cos^{-1}\sqrt{\frac{29-9\sqrt{5}}{218}}\approx 78.35927686^{o}}$