$c^2$ = ${a^2+b^2-2ab\cos c}$
= ${a^2+(\sqrt{3}a-a)^2-2a(\sqrt{3}a-a)\cos 30^\circ}$
=${a^2+(3a^2-2\sqrt{3}a^2+a^2)-2a(\sqrt{3}a-a)\frac{\sqrt{3}}{2}}$
=${a^2+(3a^2-2\sqrt{3}a^2+a^2)-(3a^2-\sqrt{3}a^2)}$
=$a^2(2-\sqrt{3})$
$c$ = $a$$\sqrt{2-\sqrt{3}}$
$\frac{\sin C}{c}$ = $\frac{\sin B}{b}$
$\frac{\sin 30^\circ}{a\sqrt{2-\sqrt{3}}}$ = $\frac{\sin B}{a(\sqrt{3}-1)}$
$\frac{1}{2\sqrt{2-\sqrt{3}}}$ = $\frac{\sin B}{(\sqrt{3}-1)}$
$\sin B$ = $\frac{\sqrt{3}-1}{2\sqrt{2-\sqrt{3}}}$
=$\sqrt{2(\sqrt{2-\sqrt{3})}^2-(\sqrt{3}-1)^2}$
=$\sqrt{(4(2-\sqrt{3}))-(4-2\sqrt{3})}$
= $\sqrt{8-4\sqrt{3}-4+2\sqrt{3}}$
= $\sqrt{4-2\sqrt{3}}$
= $\sqrt{3}$-$1$
$B$ มีขนาด $45^\circ$ ดังนั้น $sin 3B$ = $\frac{\sqrt{2}}{2}$