Phương pháp giải:

Lời giải chi tiết:

* Vẽ $DH\bot \left( SBC \right)$.

Vẽ $HM\bot SC\Rightarrow \widehat{\left( \left( SCD \right);\left( SBC \right) \right)}=\widehat{M}$.

* Nối $AD\cap BC=I$. Ta có :

$DH=d\left( D;\left( SBC \right) \right)=\frac{1}{2}d\left( A;\left( SBC \right) \right)=\frac{1}{2}AK$ $\begin{align}+\,\,\frac{1}{A{{K}^{2}}}=\frac{1}{S{{A}^{2}}}+\frac{1}{A{{C}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{2{{a}^{2}}} \\  \Rightarrow AC=\frac{a\sqrt{6}}{3}\Rightarrow DH=\frac{a\sqrt{6}}{6} \\ \end{align}$.

+ Tam giác vuông SCD :

$\begin{align}\frac{1}{D{{M}^{2}}}=\frac{1}{S{{D}^{2}}}+\frac{1}{C{{D}^{2}}}=\frac{1}{2{{a}^{2}}}+\frac{1}{{{a}^{2}}}=\frac{3}{2{{a}^{2}}}\Rightarrow DM=\frac{a\sqrt{6}}{3} \\  \Rightarrow \sin\widehat{M}=\frac{DH}{DM}=\frac{\frac{a\sqrt{6}}{6}}{\frac{a\sqrt{6}}{3}}=\frac{1}{2}\Rightarrow \widehat{M}={{30}^{0}} \\ \end{align}$

Chọn đáp án A.