Đáp án đúng: B
Giải chi tiết:
Ta có: \(SA \bot \left( {ABCD} \right) \Rightarrow SA \bot CD\)
Lại có: \(CD \bot AD\)
\( \Rightarrow CD \bot \left( {SAD} \right) \Rightarrow CD \bot SD.\)
Mà \(\left( {SCD} \right) \cap \left( {ABCD} \right) = \left\{ {CD} \right\}\)
\(\begin{array}{l} \Rightarrow \angle \left( {\left( {SCD} \right),\,\,\left( {ABCD} \right)} \right) = \angle \left( {SD,\,\,AD} \right) = \angle SDA\\ \Rightarrow \cos SDA = \dfrac{{AD}}{{SD}} = \dfrac{{AD}}{{\sqrt {S{A^2} + A{D^2}} }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{a}{{\sqrt {{{\left( {a\sqrt 2 } \right)}^2} + {a^2}} }} = \dfrac{a}{{a\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}.\end{array}\)
Chọn B.