Gọi $H$ là trung điểm $AB$

$\Rightarrow \begin{cases}SH\perp AB\\SH = \dfrac{a\sqrt3}{2}\end{cases}$

Lại có: $CM = MD = \dfrac12CD$

$\Rightarrow \begin{cases}HM = BC = AD = a\\HM//BC//AD\end{cases}$

$\Rightarrow HM\perp AB$

Ta có:

$\begin{cases}(SAB)\perp (ABCD)\\(SAB)\cap (ABCD) = AB\\SH\perp AB\\SH\subset (SAB)\end{cases}$

$\Rightarrow SH\perp (ABCD)$

$\Rightarrow SH\perp HM$

Chọn hệ trục tọa độ $Oxyz$ sao cho:

$\begin{cases}H\equiv O\\A\left(-\dfrac{a}{2};0;0\right)\\ B\left(\dfrac{a}{2};0;0\right)\\M(0;a;0)\\C\left(\dfrac{a}{2};a;0\right)\\D\left(-\dfrac{a}{2};a;0\right)\\S\left(0;0;\dfrac{a\sqrt3}{2}\right)\end{cases}$

$\Rightarrow N\left(\dfrac{a}{4};\dfrac{a}{2};\dfrac{a\sqrt3}{4}\right)$

$\Rightarrow \begin{cases}\overrightarrow{MN} = \left(\dfrac{a}{4};-\dfrac{a}{2};\dfrac{a\sqrt3}{4}\right)\\\overrightarrow{SB} = \left(\dfrac{a}{2};0;-\dfrac{a\sqrt3}{2}\right)\end{cases}$

Chọn $\overrightarrow{u} = \left(1;-2;\sqrt3\right)$ cùng phương $\overrightarrow{MN}$

và $\overrightarrow{v} = \left(1;0;-\sqrt3\right)$ cùng phương $\overrightarrow{SB}$

Ta được:

$\quad \cos\widehat{(MN,SB)} = \cos\left(\overrightarrow{u},\overrightarrow{v}\right)$

$\Leftrightarrow \cos\widehat{(MN,SB)} =\dfrac{\left|\overrightarrow{u}.\overrightarrow{v}\right|}{\left|\overrightarrow{u}\right|\cdot\left|\overrightarrow{v}\right|}$

$\Leftrightarrow \cos\widehat{(MN,SB)} = \dfrac{\left|1.1 + (-2).0 + \sqrt3.\left(-\sqrt3\right)\right|}{\sqrt{1^2 + (-2)^2 + \left(\sqrt3\right)^2}\ \cdot \sqrt{1^2 + 0^2 + \left(-\sqrt3\right)^2}}$

$\Leftrightarrow \cos\widehat{(MN,SB)} = \dfrac{\sqrt2}{4}$

$\Rightarrow \widehat{(MN,SB)} = \arccos\left(\dfrac{\sqrt2}{4}\right)\approx 69,3^\circ$