Câu hỏi 14

 Lưu ý : \(\widehat{ABD}=\widehat{ACD}={{90}^{0}}\)

\(\left\{ \begin{align}  BD\bot AB \\  BD\bot SA \\ \end{align} \right.\Rightarrow BD\bot SB\)

* Chọn B. Vẽ giả tưởng \(BH\bot \left( SCD \right)\).

Vẽ \(HM\bot \) giao tuyến SD \(\Rightarrow \widehat{\left( \left( SBD \right);\left( SCD \right) \right)}=\widehat{HMB}=\widehat{{{M}_{1}}}\).

* Tam giác vuông SBD : \(\frac{1}{B{{M}^{2}}}=\frac{1}{2{{a}^{2}}}+\frac{1}{3{{a}^{2}}}=\frac{5}{6{{a}^{2}}}\Rightarrow BM=\frac{a\sqrt{6}}{\sqrt{5}}\).

* \(BH=d\left( B;\left( SCD \right) \right)=\frac{1}{2}d\left( A;\left( SCD \right) \right)=\frac{1}{2}AK\)

Tam giác vuông SAC : \(\frac{1}{A{{K}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{3{{a}^{2}}}=\frac{4}{3{{a}^{2}}}\Rightarrow AK=\frac{a\sqrt{3}}{2}\Rightarrow BH=\frac{a\sqrt{3}}{4}\).

* Tam giác BHM : \(\sin \widehat{{{M}_{1}}}=\frac{BH}{BM}=\frac{a\sqrt{3}}{4}:\frac{a\sqrt{6}}{\sqrt{5}}=\frac{\sqrt{10}}{8}\).

Chọn đáp án C.