Diện tích hình thang \(ABCD\) là:

\({{\rm{S}}_{{\rm{ABCD}}}} = \frac{1}{2}{\rm{AD}}\left( {{\rm{AB}} + {\rm{CD}}} \right) = \frac{1}{2}2{\rm{a}} \cdot 3{\rm{a}} = 3{{\rm{a}}^2},\,\,{\rm{CB}} = {\rm{AC}} = {\rm{a}}\sqrt 5 .\)

Độ dài đường cao \({\rm{SI}} = \frac{{3\;{{\rm{V}}_{{\rm{S}}.{\rm{ABCD}}}}}}{{{{\rm{S}}_{{\rm{ABCD}}}}}} = \frac^3}}}{5}}}{{3{{\rm{a}}^2}}} = \frac{{3\sqrt {15} {\rm{a}}}}{5}\).

Vẽ \({\rm{IH}} \bot {\rm{CB}}\) tại \({\rm{H}} \Rightarrow {\rm{BC}} \bot \left( {{\rm{SIH}}} \right) \Rightarrow {\rm{BC}} \bot {\rm{SH}}\).

Ta có \[\widehat {\left( {\left( {{\rm{SBC}}} \right),\left( {{\rm{ABCD}}} \right)} \right)} = \widehat {\left( {{\rm{IH}},\,\,{\rm{SH}}} \right)} = \widehat {{\rm{SHI}}}\].

\({{\rm{S}}_{{\rm{ICB}}}} = {{\rm{S}}_{{\rm{ABCD}}}} - {{\rm{S}}_{{\rm{IDC}}}} - {{\rm{S}}_{{\rm{AIB}}}} = 3{{\rm{a}}^2} - \frac{{{{\rm{a}}^2}}}{2} - {{\rm{a}}^2} = \frac{{3{{\rm{a}}^2}}}{2} \Rightarrow {\rm{IH}}{\rm{.CB}} = 3{{\rm{a}}^2}\)

\( \Rightarrow IH = \frac{{3a\sqrt 5 }}{5},\,\,\tan \widehat {SHI} = \frac = \frac{{3a\sqrt {15} }}{5}:\frac{{3a\sqrt 5 }}{5} = \sqrt 3  \Rightarrow \widehat {SHI} = 60^\circ .\) Đáp án: 60.