Why is this angle incorrect for AC current I_norton?
Treat this like a voltage divider:
Now,

\$ R = 10 \Omega \$

\$ X_L = j \omega L = j200 \cdot 50 \cdot 10^{-3} = j10 .\Omega \$

\$ X_C = {{-j} \over { \omega C} } = {{-j} \over {200 \cdot 500 \cdot 10^{-6}} } = {{-j} \over {10^{-1}} } = -j10 . \Omega \$

\$ V_{ab} = V_C = V_{Thevenin} = V_T , \$

\begin{eqnarray*}
 V_T &amp;=&amp; V_S ∙{{X_C} \over {R + X_L + X_C}}\\
&amp;=&amp; 162 ∠ 60^∘ \cdot {{-j10} \over {10 + j10 + -j10}}\\
&amp;=&amp; 162 ∠ 60^∘ \cdot {{-j10} \over {10}}\\
&amp;=&amp; -j \cdot 162 ∠ 60^∘ \\
&amp;=&amp; 162 ∠ 60^∘ \cdot 1 ∠ {-90}^∘\\
&amp;=&amp; 162 ∠ (60^∘ - 90^∘)\\
&amp;=&amp; 162 ∠ {-30^∘} . V\\
\end{eqnarray*}

\$ Z_N = Z_{Norton} = Z_{Thevenin}  = Z_T \$ is the impedance looking into the terminals ab, and shorting any voltage sources, and opening any current sources. 

So,   

\begin{eqnarray*}
Z_N = Z_T &amp;=&amp; {X_C \parallel ({R + X_L}) }\\
&amp;=&amp; {-j10} \parallel (10 + j10)\\
&amp;=&amp; \lbrace (-j10)^{-1} + (10+j10)^{-1} \rbrace ^{-1} \\
&amp;=&amp; \lbrace{ {10}\over {100 - j100} }\rbrace   ^{-1}  \\
&amp;=&amp; 10 - j10 . Ω\\
&amp;=&amp; 10√2 ∠ {-45^∘}.Ω \\
\end{eqnarray*}

So,

\$ I_N = {{V_T} \over {Z_N}} = 11.46 ∠ 15^∘.A \space   \blacksquare   \$