Chọn: Đáp án A

Ta có:

$\text{cos(}\overrightarrow{SE};\overrightarrow{BC})=\frac{\overrightarrow{SE}.\overrightarrow{BC}}{SE.BC}$

$\begin{align}   & \overrightarrow{SE}.\overrightarrow{BC}=(\overrightarrow{SH}+\overrightarrow{HE}).\overrightarrow{BC}=\overrightarrow{HE}.\overrightarrow{BC}=\frac{9}{25}\overrightarrow{HC}.\overrightarrow{BC}=\frac{9}{25}\overrightarrow{CH}.\overrightarrow{CB} \\  & =\frac{9}{25}CH.CB.c\text{os}\widehat{HCB}=\frac{9}{25}.CH.CB.\frac{CB}{CH}=\frac{9}{25}.C{{B}^{2}}=\frac{144{{a}^{2}}}{25} \\ \end{align}$

Ta chứng minh được HK$\bot $CH tại E

$\begin{align}  & \frac{HE}{HC}=\frac{HE.HC}{H{{C}^{2}}}=\frac{H{{B}^{2}}}{H{{B}^{2}}+B{{C}^{2}}}=\frac{9}{25} \\ & =>HE=\frac{9}{25}.HC=\frac{9}{25}\sqrt{H{{B}^{2}}+B{{C}^{2}}}=\frac{9a}{5} \\ \end{align}$

$\begin{align}  & SE=\sqrt{S{{H}^{2}}+H{{E}^{2}}}=\sqrt{3{{a}^{2}}+\frac{81{{a}^{2}}}{25}}=\frac{2a\sqrt{39}}{5} \\  & =>c\text{os(}\overrightarrow{SE};\overrightarrow{BC})=\frac{144a}{25}.\frac{5}{2a\sqrt{39}.4a}=\frac{18}{5\sqrt{39}} \\ \end{align}$