Cho hình chóp $S.ABCD$ có đáy ABCD là hình vuông có cạnh bằng 4a...

+) Gọi H là hình chiếu vuông góc của D trên mặt phẳng $\left( SBC \right)$
$\begin{aligned}
& \Rightarrow \widehat{\left( SD,\left( SBC \right) \right)}=\widehat{HSD}\Rightarrow \cos \widehat{\left( SD,\left( SBC \right) \right)}=\cos \widehat{HSD}=\dfrac{SH}{SD} \\
& +) {{S}_{SAB}}=\dfrac{1}{2}SA.AB=\dfrac{1}{2}SA.4a=\dfrac{8{{a}^{2}}\sqrt{6}}{3}\Rightarrow SA=\dfrac{4a\sqrt{6}}{3} \\
& +) {{V}_{D.SBC}}=\dfrac{1}{3}DH.{{S}_{SBC}} v\grave{a}{{V}_{D.SBC}}={{V}_{S.BCD}}=\dfrac{1}{3}.SA.{{S}_{BCD}}=\dfrac{1}{3}.\dfrac{4a\sqrt{6}}{3}.\dfrac{1}{2}.4a.4a=\dfrac{32{{a}^{3}}\sqrt{6}}{9} \\
& \Rightarrow \dfrac{1}{3}DH.{{S}_{SBC}}=\dfrac{32{{a}^{3}}\sqrt{6}}{9}\Rightarrow DH=\dfrac{32{{a}^{3}}\sqrt{6}}{3{{S}_{SBC}}}\left( 1 \right) \\
\end{aligned}$
+) Từ $\left\{ \begin{aligned}
& BC\bot AB \\
& BC\bot SA \\
\end{aligned} \right.\Rightarrow BC\bot \left( SAB \right)\Rightarrow BC\bot SB\Rightarrow {{S}_{SBC}}=\dfrac{1}{2}BC.SB=\dfrac{1}{2}.4a.SB=2a.SB$
$+) S{{B}^{2}}=S{{A}^{2}}+A{{B}^{2}}={{\left( \dfrac{4a\sqrt{6}}{3} \right)}^{2}}+16{{a}^{2}}=\dfrac{80{{a}^{2}}}{3}\Rightarrow SB=a\sqrt{\dfrac{80}{3}}\Rightarrow {{S}_{SBC}}=2{{a}^{2}}\sqrt{\dfrac{80}{3}}$
Thế vào (1) $\Rightarrow DH=\dfrac{32{{a}^{3}}\sqrt{6}}{3.2{{a}^{2}}\sqrt{\dfrac{80}{3}}}=\dfrac{4a\sqrt{10}}{5}$

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