Answer
$9.89^{\circ}$
Work Step by Step
We can find the phase angle $\phi$ as follows:
$\omega_n=\frac{K}{m}$
$\implies \omega_n=\frac{75\times 32.2}{7}$
$\implies \omega_n=18.5742 rad/s$
Now $\phi=tan^{-1}[\frac{2(c/c_{\circ})(\omega_{\circ}/ \omega_n)}{1-(\omega_{\circ}/\omega_n)^2}]$
We plug in the known values to obtain:
$\phi=tan^{-1}[\frac{2\times 0.8\times (2/18.5742)}{1-(2/18.5742)^2}]$
$\implies \phi=9.89^{\circ}$
