\(\widehat{\left(SC;\left(ABCD\right)\right)}=\widehat{CS;CA}=\widehat{SCA}\)
Ta có: ΔABC vuông tại B
=>\(BA^2+BC^2=AC^2\)
=>\(AC^2=\left(3a\right)^2+\left(\sqrt{3}a\right)^2=12a^2\)
=>\(AC=2a\sqrt{3}\)
Xét ΔSAC vuông tại A có \(tanSCA=\dfrac{SA}{AC}=\dfrac{2a}{2a\sqrt{3}}=\dfrac{1}{\sqrt{3}}\)
nên \(\widehat{SCA}=30^0\)
=>\(\widehat{SC;\left(ABCD\right)}=30^0\)