Cho hình chóp S.ABCD có đáy ABCD là

* Hướng dẫn giải

Đáp án A

+) Gọi H là hình chiếu vuông góc của D trên mặt phẳng (SBC)

$\begin{array}{l}\Rightarrow \widehat{\left(SD,\left(SBC\right)\right)}=\widehat{HSD}\Rightarrow \cos \widehat{\left(SD,\left(SBC\right)\right)}=\cos \widehat{HSD}=\frac{SH}{SD}\\ +)S_{SAB}=\frac{1}{2}SA.AB=\frac{1}{2}SA.4a=\frac{8a^2\sqrt{6}}{3}\Rightarrow SA=\frac{4a\sqrt{6}}{3}\\ +)V_{D.SBC}=\frac{1}{3}DH.S_{SBC}vàV_{D.SBC}=V_{S.BCD}=\frac{1}{3}.SA.S_{BCD}=\frac{1}{3}.\frac{4a\sqrt{6}}{3}.\frac{1}{2}.4a.4a=\frac{32a^3\sqrt{6}}{9}\\ \\ \Rightarrow \frac{1}{3}DH.S_{SBC}=\frac{32a^3\sqrt{6}}{9}\Rightarrow DH=\frac{32a^3\sqrt{6}}{3S_{SBC}}\left(1\right)\end{array}$

+) Từ $\left\{\begin{array}{l}BC\perp AB\\ BC\perp SA\end{array}\right.\Rightarrow BC\perp \left(SAB\right)\Rightarrow BC\perp SB\Rightarrow S_{SBC}=\frac{1}{2}BC.SB=\frac{1}{2}.4a.SB=2a.SB$

$+)S{B}^2=S{A}^2+A{B}^2={\left(\frac{4a\sqrt{6}}{3}\right)}^2+16a^2=\frac{80a^2}{3}\Rightarrow SB=a\sqrt{\frac{80}{3}}\Rightarrow S_{SBC}=2a^2\sqrt{\frac{80}{3}}$

Thế vào (1) $\Rightarrow DH=\frac{32a^3\sqrt{6}}{3.2a^2\sqrt{\frac{80}{3}}}=\frac{4a\sqrt{10}}{5}$

$\begin{array}{l}+)S{D}^2=S{A}^2+A{D}^2={\left(\frac{4a\sqrt{6}}{3}\right)}^2+16a^2=\frac{80a^2}{3}\Rightarrow SD=a\sqrt{\frac{80}{3}}\\ \Rightarrow S{H}^2=S{D}^2-H{D}^2=\frac{80a^2}{3}-{\left(\frac{4a\sqrt{10}}{5}\right)}^2=\frac{304a^2}{15}\\ \Rightarrow SH=a\sqrt{\frac{304}{15}}\Rightarrow \cos \widehat{\left(SD;\left(SBC\right)\right)}=\frac{SH}{SD}=\frac{a\sqrt{\frac{304}{15}}}{a\sqrt{\frac{80}{3}}}=\frac{\sqrt{19}}{5}\end{array}$