Cho hình chóp S.ABCD có đáy ABC vuông tại B,...

A

- Hướng dẫn giải

Chọn A.

$$\begin{gathered} \mathrm{Dựng} \mathrm{SH}\perp \mathrm{AC}, \mathrm{do} \left(\mathrm{SAC}\right)\perp \left(\mathrm{ABC}\right) \\ \mathrm{nên} \mathrm{SH}\perp \left(\mathrm{ABC}\right);\mathrm{AC}=2\mathrm{a}. \\ \mathrm{Dựng} \mathrm{HE}\perp \mathrm{BC};\mathrm{HF}\perp \mathrm{SE} \\ \Rightarrow \mathrm{d}(\mathrm{H};(\mathrm{SBC}\left)\right)=\mathrm{HF}. \mathrm{\Delta SAC}=\mathrm{\Delta BCA} \\ \Rightarrow \mathrm{\Delta SAC} \mathrm{vuông} \mathrm{tại} \mathrm{S}. \end{gathered}$$

$$\begin{gathered} \mathrm{Dễ} \mathrm{thấy} \\ \tan \widehat{\mathrm{ACB}}= \frac{1}{\sqrt{3}} \Rightarrow \widehat{\mathrm{ACB}} = {30}^{\mathrm{o}} = \widehat{\mathrm{SAC}} \\ \mathrm{HC} = \mathrm{SCcos}{60}^{\mathrm{o}} = \frac{\mathrm{a}}{2}; \\ \mathrm{HE} = \mathrm{HCsin}{30}^{\mathrm{o}} =\frac{\mathrm{a}}{4}; \mathrm{SH} = \frac{\mathrm{a}\sqrt{3}}{2}. \\ \mathrm{Do} \mathrm{AC} = 4\mathrm{HC} \\ \Rightarrow {\mathrm{d}}_{\mathrm{A}}=4{\mathrm{d}}_{\mathrm{H}}=4.\frac{\mathrm{SH}.\mathrm{HE}}{\sqrt{{\mathrm{SH}}^2+{\mathrm{HE}}^2}}=\frac{2\sqrt{39}}{13} \\ \mathrm{Do} \mathrm{đó} \mathrm{Sin\alpha } =\frac{{\mathrm{d}}_{\mathrm{A}}}{\mathrm{SA}}=\frac{2}{\sqrt{13}}. \end{gathered}$$