In a triangle ABC , with usual notations, if \(\mathrm{m} \angle \mathrm{A}=45^{\circ}, \mathrm{m} \angle \mathrm{B}=75^{\circ}\), then \(\mathrm{a}+\mathrm{c} \sqrt{2}\) has the

Given that :
\(\mathrm{m} \angle \mathrm{~A}=45^{\circ}, \mathrm{m} \angle \mathrm{~B}=75^{\circ}, \mathrm{m} \angle \mathrm{C}=60^{\circ}\)
we know that
\(\begin{aligned}
& \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\
& \frac{\sin 45^{\circ}}{a}=\frac{\sin 75^{\circ}}{b}=\frac{\sin 60^{\circ}}{c} \\
\therefore \quad & \frac{\sin 45^{\circ}}{a}=\frac{\sin \left(45^{\circ}+30^{\circ}\right)}{b}
\end{aligned}\)
\(\begin{array}{ll}\therefore & \frac{1}{\sqrt{2} a}=\frac{\frac{1}{\sqrt{2}} \times \frac{1}{2}+\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}}{b} \\ \therefore & \frac{1}{\sqrt{2} a}=\frac{1+\sqrt{3}}{2 \sqrt{2} b} \\ \therefore & a=\frac{2 b}{1+\sqrt{3}} \\ \therefore & \frac{\sin 60^{\circ}}{c}=\frac{\sin \left(45^{\circ}+30^{\circ}\right)}{b} \\ \therefore & \sqrt{2} c=\frac{2 \sqrt{3} b}{1+\sqrt{3}} \\ \therefore \quad & a+\sqrt{2} c=\frac{2 b}{1+\sqrt{3}}+\frac{2 \sqrt{3} b}{1+\sqrt{3}}=2 b\end{array}\)