Đáp án đúng là: A

Ta có: \[\overrightarrow {SD} = \left( {\frac{a}{2};a; - \frac{{a\sqrt 3 }}{2}} \right) = a\left( {\frac{1}{2};1; - \frac{{\sqrt 3 }}{2}} \right)\],

\[\overrightarrow {SA} = \left( {\frac{a}{2};0; - \frac{{a\sqrt 3 }}{2}} \right) = a\left( {\frac{1}{2};0; - \frac{{\sqrt 3 }}{2}} \right)\],

\[\overrightarrow {SC} = \left( { - \frac{a}{2};a; - \frac{{a\sqrt 3 }}{2}} \right) = a\left( { - \frac{1}{2};1; - \frac{{\sqrt 3 }}{2}} \right)\].

Vectơ pháp tuyến của mặt phẳng \[\left( {SAC} \right)\] là \[\overrightarrow n = \left[ {\overrightarrow {SA} ,\overrightarrow {SC} } \right]\].

Ta có: \[\overrightarrow n = \left[ {\overrightarrow {SA} ,\overrightarrow {SC} } \right] = \left( {\left| {\begin{array}{*{20}{c}}0&{ - \frac{{\sqrt 3 }}{2}}\\1&{ - \frac{{\sqrt 3 }}{2}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{ - \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\{ - \frac{{\sqrt 3 }}{2}}&{ - \frac{1}{2}}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{\frac{1}{2}}&0\\{ - \frac{1}{2}}&1\end{array}} \right|} \right) = \left( {\frac{{\sqrt 3 }}{2};\frac{{\sqrt 3 }}{2};\frac{1}{2}} \right).\]

Ta có: \[\sin \left( {SD,\left( {SAC} \right)} \right) = \left| {\cos \left( {\overrightarrow {SD} ,{{\overrightarrow n }_{\left( {SAC} \right)}}} \right)} \right|\]

\[ = \frac{{\left| {\frac{{\sqrt 3 }}{2}.\frac{1}{2} + 1.\frac{{\sqrt 3 }}{2} + \left( { - \frac{{\sqrt 3 }}{2}} \right).\frac{1}{2}} \right|}}{{\sqrt {{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} .\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {1^2} + {{\left( { - \frac{{\sqrt 3 }}{2}} \right)}^2}} }}\]

\[ = \frac{{\sqrt {42} }}{{14}}.\]

Suy ra \[\widehat {\left( {SD,\left( {SAC} \right)} \right)} \approx 28^\circ .\]