Câu hỏi 25

Ta có:  \(\left. \begin{array}{l}BC \bot AB\\BC \bot SA\end{array} \right\} \Rightarrow BC \bot SB\)

\(\left. \begin{array}{l}\left( {SBC} \right) \cap \left( {ABC} \right) = BC\\SB \bot BC\\AB \bot BC\end{array} \right\} \Rightarrow \widehat {\left( {\left( {SBC} \right);\left( {ABC} \right)} \right)} = \widehat {\left( {SB;AB} \right)} = \widehat {SBA} = {60^0}\)

Vì \(\widehat {SBA} < {90^0}\) nên ta có: \(\tan \widehat {SBA} = \frac{{SA}}{{AB}} \Rightarrow \sqrt 3  = \frac{x}{a} \Rightarrow x = a\sqrt 3 \) .

Chọn D.