Finding phase of voltage divider
The question's values for impedances \$Z_{R-L}\$ and \$Z_C\$ are actually the magnitude of those impedances.  I suspect that the approach was mixing time-domain and frequency-domain approaches; I will address both to make the distinction.
Since the inductor voltage depends on changing current and the capacitor current depends on changing voltage, a time domain analysis produces a differential equation.
$$ v_{OUT}(t) = v_C(t) = v_{IN}(t) - L\frac{d}{dt}i_{IN}(t) - Ri_{IN}(t) $$
and since the input current is split between the capacitor and output:
$$   i_{IN}(t) = C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)     $$
putting them together:
$$ \begin{align} v_{OUT}(t) &amp;= v_{IN}(t) - L\frac{d}{dt}\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) - R\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big)  \\
                            &amp;= v_{IN}(t) - LC\frac{d^2}{dt^2}v_{OUT}(t) - L\frac{d}{dt}i_{OUT}(t) - RC\frac{d}{dt}v_{OUT}(t) - Ri_{OUT}(t) \\
\end{align}$$
This is still somewhat manageable, except that \$i_{OUT}\$ likely also depends on \$v_{OUT}\$.  Flipping to the frequency domain, the diff-eq turns to algebra and we get a much better handle on the output load.
Let's show the output load explicitly, with an impedance of \$Z_{LOAD}\$.  Also note that L and C have impedances \$j\omega L\$ and \$\frac{1}{j\omega C}\$, respectively.  See this concise reference for more detail there.
$$ $$

Preserving our output nets, we can lump together L with R (in series) and C with the load (parallel).  This gives:
$$ \begin{align} Z_{LR} &amp;= j\omega L + R \\
Z_{COUT} &amp;= \bigg({\frac{1}{j\omega C}}^{-1}+{Z_{LOAD}}^{-1}\bigg)^{-1} \\
         &amp;= \frac{Z_{LOAD}}{j\omega C Z_{LOAD}+1} \end{align}$$
Now you can use the resistor divider rule to calculate a transfer function:
$$ H = \frac{V_{OUT}}{V_{IN}} = \frac{Z_{COUT}}{Z_{LR}+Z_{COUT}} $$
And finally, the phase shift can be determined by comparing the real and imaginary parts of the transfer function:
$$ \Theta = tan^{-1}\biggl(\frac{\mathfrak{Im}(H)}{\mathfrak{Re}(H)}\biggr)$$