Two sides of a triangle are \(\sqrt{3}+1\) and \(\sqrt{3}-1\) and the included angle is \(60^{\circ}\), then the difference of the remaining angles is

Let \(\mathrm{a}=\sqrt{3}+1, \mathrm{~b}=\sqrt{3}-1, \mathrm{C}=60^{\circ}\)
Using cosine Rule,
\(\mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab} \cos \mathrm{C} \)
\( \mathrm{c}^2=(\sqrt{3}+1)^2+(\sqrt{3}-1)^2-2(\sqrt{3}+1)(\sqrt{3}\) \(-1) \frac{1}{2} \)
\( \mathrm{c}^2=6\)
\(\begin{aligned}
\therefore \quad c^2 & =6 \\
\mathrm{c} & =\sqrt{6}
\end{aligned}\)
Using sine Rule,
\(\begin{aligned}
& \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\
& \text { Consider } \\
& \frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\
& \frac{\sqrt{3}-1}{\sin \mathrm{B}}=\frac{\sqrt{6}}{\frac{\sqrt{3}}{2}}
\end{aligned}\)
\(\begin{array}{ll}
& \frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\
\therefore & \frac{\sqrt{3}-1}{\sin \mathrm{B}}=\frac{\sqrt{6}}{\frac{\sqrt{3}}{2}} \\
\therefore & \sin \mathrm{B}=\frac{\sqrt{3}-1}{2 \sqrt{2}} \\
\therefore & \angle \mathrm{B}=15^{\circ} \\
\therefore & \angle \mathrm{A}=105^{\circ}
\end{array}\)
...[Remaining angle of a Triangle]
\(\therefore \quad\) Difference \(=90^{\circ}\)