(ii) As follows,

Draw the perpendicular line from $A$ to $DA$,
intersecting the extension of $CB$ at point $M$, extend $BA$
to $C_{1}$, such that $AC_{1}$
$$= AC \text {. }$$

Connect $AC_{1}$,
$C_{1}C$.
Then $\angle MAD=90^{\circ}$,
$$\therefore MAC=90^{\circ}+\frac{5.25^{\circ}}{2}=92.625^{\circ} \text {. }$$

Also, $\angle CAC_{1}=180^{\circ}-5.25^{\circ}=174.75^{\circ}$.
Thus, $\angle MAC_{1}=360^{\circ}-174.75^{\circ}$
$$-92.625^{\circ}=92.625^{\circ} \text {. }$$

Since $\angle MAC_{1}=\angle MAC$,
therefore $\triangle MAC_{1} \cong \triangle MAC$.
Thus, $\angle MC_{1}A=\angle MCA=\angle BCA$.
Also, $BC_{1}=BA+AC_{1}=AB+AC$
$$=BM \text {, }$$

Therefore, $\angle MC_{1}A=\angle C_{1}MB$.
In $\triangle MCC_{1}$,
$$\begin{array}{l} 3 \angle ACB+5.25^{\circ}=180^{\circ}, \\ \angle ACB=58.25^{\circ} . \end{array}$$

Thus, $\angle ABC=180^{\circ}-58.25^{\circ}-5.25^{\circ}$
$$=116.5^{\circ} \text {. }$$