Answer
The phase angle, $\theta$, of the circuit is $16.22^{\circ}$.
Work Step by Step
Since the total Impedance of the circuit, $Z$, is $110 + 32i$ ohms $(\Omega)$, the phase angle in degrees, $\theta$, will be
$tan \theta = \frac{32}{110}$
$\theta$ = $\arctan$ $(\frac{32}{110})$
$\theta$ = $16.22^{\circ}$
The phase angle, $\theta$, of the circuit is $16.22^{\circ}$.