$BD=\sqrt{AB^2+AD^2}=4\sqrt{2}\\ AO=DO=\dfrac{BD}{2}=2\sqrt{2}\\ AH=HO=\dfrac{AO}{2}=\sqrt{2}\\ SH=\sqrt{SA^2-AH^2}=\sqrt{2}\\ HD=\sqrt{(2\sqrt{2})^2-\sqrt{2}^2}=\sqrt{10}\\ \tan(SD,(ABCD))=\tan(SD,HD)=\tan(\widehat{SDH})=\dfrac{SH}{HD}=\dfrac{\sqrt{5}}{5}$