$\sin\alpha=\dfrac{\sqrt6-\sqrt2}{4}$
$=\dfrac{\sqrt2(\sqrt3-1)}{4}$
$=\dfrac{\sqrt3-1}{2\sqrt2}$
$=\dfrac{1}{\sqrt2}\Big( \dfrac{\sqrt3}{2}-\dfrac{1}{2}\Big)$
$=\sin45^o.(\sin60^o-\sin30^o)$
$=\sin45^o.2\cos45^o.\sin15^o$
$=\sin90^o.\sin15^o$
$=\sin15^o$
$\to \alpha=15^o$