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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: $$ \begin{align} v_{OUT}(t) &= v_{IN}(t) - L\frac{d}{dt}\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) - R\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) \\ &= v_{IN}(t) - LC\frac{d^2}{dt^2}v_{OUT}(t) - L\frac{d}{dt}i_{OUT}(t) - RC\frac{d}{dt}v_{OUT}(t) - Ri_{OUT}(t) \\ \end{align}$$

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}\$. 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:

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

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

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: $$ \begin{align} v_{OUT}(t) &= v_{IN}(t) - L\frac{d}{dt}\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) - R\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) \\ &= v_{IN}(t) - LC\frac{d^2}{dt^2}v_{OUT}(t) - L\frac{d}{dt}i_{OUT}(t) - RC\frac{d}{dt}v_{OUT}(t) - Ri_{OUT}(t) \\ \end{align}$$

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}\$. 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:

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

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

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: $$ \begin{align} v_{OUT}(t) &= v_{IN}(t) - L\frac{d}{dt}\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) - R\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) \\ &= v_{IN}(t) - LC\frac{d^2}{dt^2}v_{OUT}(t) - L\frac{d}{dt}i_{OUT}(t) - RC\frac{d}{dt}v_{OUT}(t) - Ri_{OUT}(t) \\ \end{align}$$

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}\$. 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 and C with the load. This gives.

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

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

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: $$ \begin{align} v_{OUT}(t) &= v_{IN}(t) - L\frac{d}{dt}\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) - R\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) \\ &= v_{IN}(t) - LC\frac{d^2}{dt^2}v_{OUT}(t) - L\frac{d}{dt}i_{OUT}(t) - RC\frac{d}{dt}v_{OUT}(t) - Ri_{OUT}(t) \\ \end{align}$$

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}\$. 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:

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

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

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 for more detail there. $$ $$

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

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: $$ \begin{align} v_{OUT}(t) &= v_{IN}(t) - L\frac{d}{dt}\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) - R\big(C\frac{d}{dt}v_{OUT}(t) + i_{OUT}(t)\big) \\ &= v_{IN}(t) - LC\frac{d^2}{dt^2}v_{OUT}(t) - L\frac{d}{dt}i_{OUT}(t) - RC\frac{d}{dt}v_{OUT}(t) - Ri_{OUT}(t) \\ \end{align}$$

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}\$. 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 and C with the load. This gives.

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

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