(SC;(SAB))=(SC;SB)=góc BSC
\(AC=\sqrt{a^2+a^2}=a\sqrt{2}\)
\(SC=\sqrt{SA^2+AC^2}=a\sqrt{5}\)
\(SB=\sqrt{a^2+\left(a\sqrt{3}\right)^2}=2a\)
\(cosBSC=\dfrac{SB^2+SC^2-BC^2}{2\cdot SB\cdot SC}=\dfrac{4a^2+5a^2-a^2}{2\cdot2a\cdot a\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)
=>góc BSC=27 độ