Question

Answer 1

Gọi D là trung điểm BC \(\Rightarrow SD\perp BC\)

\(\Rightarrow BC\perp\left(SAD\right)\)

\(\Rightarrow\widehat{ADS}=\left[S,BC,A\right]\)

\(SD=\dfrac{1}{2}BC=\dfrac{1}{2}\sqrt{SB^2+SC^2}=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow AD=\sqrt{SA^2+SD^2}=\dfrac{\sqrt{6}}{2}\)

\(cos\alpha=cos\widehat{ADS}=\dfrac{SD}{AD}=\dfrac{1}{\sqrt{3}}\)