Answer
$\phi = -45.0^{\circ}$
Work Step by Step
In part (b), we found that $f_d = 159~Hz$
We can find the phase angle:
$tan~\phi = \frac{-X_C}{R}$
$tan~\phi = \frac{-1}{R~\omega_d~C}$
$tan~\phi = \frac{-1}{2\pi~f_d~R~C}$
$tan~\phi = \frac{-1}{(2\pi)(159~Hz)(50.0~\Omega)(20.0\times 10^{-6}~F)}$
$tan~\phi = -1.00$
$\phi = tan^{-1}~(-1.00)$
$\phi = -45.0^{\circ}$
