The question's values for impedances \\$Z_{L-C}\\$ 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:
$$ \begin{align} v_{OUT}(t) &= v_{IN}(t) - L\frac{d}{dt}\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) - R\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big)  \\
                            &= v_{IN}(t) - LC\frac{d^2}{dt^2}v_{OUT}(t) - L\frac{d}{dt}i_{OUT}(t) - RC\frac{d}{dt}v_{OUT}(t) - Ri_{OUT}(t) \\
\end{align}$$

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}\\$.  Additionally, note that L and C have impedances \\$j\omega L\\$ and \\$\frac{1}{j\omega C}\\$, respectively.  See [this concise reference](http://hyperphysics.phy-astr.gsu.edu/hbase/electric/impcom.html) for more detail there.
$$ $$
![schematic](https://i.sstatic.net/ngKmc.png)

Preserving our output nets, we can lump together L with R and C with the load.  This gives.

$$ Z_{LR} = j\omega L + R $$
$$ Z_{COUT} = \bigg({\frac{1}{j\omega C}}^{-1}+{Z_{LOAD}}^{-1}\bigg)^{-1} 
             = \big(j\omega C+{Z_{LOAD}}^{-1}\big)^{-1} $$

*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:
$$ \Theta = sin^{-1}\biggl(\frac{\mathfrak{Im}(H)}{\mathfrak{Re}(H)}\biggr)$$