Câu hỏi 13

* 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.