In a \(\triangle A B C\), it is given the angles \(A=75^{\circ}\) and \(C=60^{\circ}\)
and \(\quad \triangle B A D=\sqrt{3}(\triangle B C D) \Rightarrow(A D)=\sqrt{3}(C D)\)
\(\Rightarrow \quad \frac{A D}{C D}=\sqrt{3}\)


As angle \(B=45^{\circ}\), Now let \(\angle A B D=\alpha\), then \(\angle D B C=45^{\circ}-\alpha\)
Now, by sine law
\(\begin{gathered}
\frac{\sin \alpha}{A D}=\frac{\sin 75^{\circ}}{B D},(\text { in } \triangle A B D) \\
\text { and } \frac{\sin \left(45^{\circ}-\alpha\right)}{C D}=\frac{\sin 60^{\circ}}{B D},(\text { in } \triangle D B C) \\
\therefore \quad \frac{(C D) \sin \alpha}{(A D) \sin \left(45^{\circ}-\alpha\right)}=\frac{\sin 75^{\circ}}{\sin 60^{\circ}} \\
\Rightarrow \frac{1}{\sqrt{3}} \times \frac{\sin \alpha}{\sin \left(45^{\circ}-\alpha\right)}=\frac{\frac{1}{2 \sqrt{2}}+\frac{\sqrt{3}}{2 \sqrt{2}}}{\frac{\sqrt{3}}{2}} \\
\Rightarrow \quad \frac{\sin \alpha}{\sin \left(45^{\circ}-\alpha\right)}=\frac{\sqrt{3}+1}{\sqrt{2}} \Rightarrow \frac{\sin \left(45^{\circ}-\alpha\right)}{\sin \alpha}=\frac{\sqrt{3}-1}{\sqrt{2}} \\
\Rightarrow \frac{1}{\sqrt{2}} \cot \alpha-\frac{1}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}} \Rightarrow \cot \alpha=\sqrt{3} \Rightarrow \alpha=30^{\circ}
\end{gathered}\)