\(\left\{{}\begin{matrix}BC\perp SA\subset\left(SAB\right)\\BC\perp AB\subset\left(SAB\right)\end{matrix}\right.\Rightarrow BC\perp\left(SAB\right)\Rightarrow BC\perp SB\)
\(\left\{{}\begin{matrix}BC\perp SB\\BC\perp AB\\\left(SBC\right)\cap\left(ABCD\right)=BC\end{matrix}\right.\Rightarrow\left(\left(SBC\right),\left(ABCD\right)\right)=\widehat{SBA}\)
\(\tan\widehat{SBA}=\dfrac{SA}{AB}=\dfrac{a\sqrt{3}}{3.a}=\dfrac{\sqrt{3}}{3}\Rightarrow\widehat{SBA}=30^0\)
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