\(\text { With usual notations, in triangle } \mathrm{ABC}, a=\sqrt{3}+1, b=\) \(\sqrt{3}-1 \text { and } \mathrm{m} \angle \mathrm{C}=60^{\circ},\ \text { then } A-B=\)

\(\text {Given } \mathrm{a}=\sqrt{3}+1, \mathrm{~b}=\sqrt{3}-1, \mathrm{~m} \angle \mathrm{C}=60^{\circ} \)
\( \therefore \quad \cos 60^{\circ}=\frac{(\sqrt{3}+1)^{2}+(\sqrt{3}-1)^{2}-\mathrm{c}^{2}}{2(\sqrt{3}+1)(\sqrt{3}-1)} \)
\( \qquad \frac{1}{2}=\frac{8-\mathrm{c}^{2}}{2(2)} \Rightarrow \mathrm{c}^{2}=6 \Rightarrow \mathrm{c}=\sqrt{6} \)
\(\text {By sine rule,} \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \Rightarrow \frac{\sqrt{3}+1}{\sin \mathrm{A}}=\frac{\sqrt{6}}{\sin 60^{\circ}} \)
\( \therefore \quad \frac{\sqrt{3}+1}{\sin \mathrm{A}}=\frac{\sqrt{6}}{\left(\frac{\sqrt{3}}{2}\right)} \Rightarrow \sin \mathrm{A}=\frac{\sqrt{3}}{2} \times \frac{(\sqrt{3}+1)}{\sqrt{6}} \Rightarrow\) \(\sin \mathrm{A}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \)
\( \therefore \mathrm{A}=75^{\circ} \text { or }\left(180^{\circ}-75^{\circ}\right)=105^{\circ} \)
\( \therefore \mathrm{B}=180^{\circ}-\left(75^{\circ}+60^{\circ}\right)=45^{\circ} \text { or } 180^{\circ}-\) \(\left(105^{\circ}+60^{\circ}\right)\) \(=15^{\circ} \)
\( \text { By } \sin \mathrm{e} \text { rule, } \frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \Rightarrow \frac{\sqrt{3}-1}{\sin \mathrm{B}}=\frac{\sqrt{6}}{\sin 60^{\circ}} \)
\( \therefore \sin \mathrm{B}=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}-1}{\sqrt{6}}=\frac{\sqrt{3}-1}{2 \sqrt{2}} \Rightarrow \mathrm{B}=15^{\circ} \text { or }\) \(\left(180-15^{\circ}\right)\) \(=165^{\circ} \)
\( \text { Since } \mathrm{B} \neq 165^{\circ} \text { as } \mathrm{C}=60^{\circ} \text { given, we take } \mathrm{B}=15^{\circ} \)
\( \therefore \angle \mathrm{B}=15^{\circ}, \angle \mathrm{A}=105^{\circ} \Rightarrow \mathrm{A}-\mathrm{B}=90^{\circ}\)