Đáp án:

\(\widehat {\left( {SM;\left( {SCD} \right)} \right)} \approx 33,{17^0}\).

\(\widehat {\left( {SM;\left( {SBD} \right)} \right)} \approx 8,{1^0}\).

\(\widehat {\left( {SC;\left( {SMD} \right)} \right)} \approx 14,{83^0}\)

Giải thích các bước giải:

Đặt hệ trục tọa độ như hình vẽ. Chọn a=1 ta có:

\(A\left( {0;0;0} \right);\,\,B\left( {1;0;0} \right);\,\,C\left( {1;2;0} \right);\,\,D\left( {0;2;0} \right);\,\,S\left( {0;0;3} \right)\).

M là trung điểm của AB \( \Rightarrow M\left( {\dfrac{1}{2};0;0} \right) \Rightarrow \overrightarrow {SM}  = \left( {\dfrac{1}{2};0; - 3} \right)\).

* Ta có: \(\overrightarrow {SC}  = \left( {1;2; - 3} \right);\,\,\overrightarrow {SD}  = \left( {0;2; - 3} \right) \Rightarrow {\overrightarrow n _{\left( {SCD} \right)}} = \left[ {\overrightarrow {SC} ;\overrightarrow {SD} } \right] = \left( {0;3;2} \right)\)

\( \Rightarrow \sin \widehat {\left( {SM;\left( {SCD} \right)} \right)} = \dfrac{{\left| {\overrightarrow {SM} .{{\overrightarrow n }_{\left( {SCD} \right)}}} \right|}}{{\left| {\overrightarrow {SM} } \right|\left| {{{\overrightarrow n }_{\left( {SCD} \right)}}} \right|}} = \dfrac{{\left| { - 6} \right|}}{{\sqrt {\dfrac{{37}}{4}} .\sqrt {13} }} \Rightarrow \widehat {\left( {SM;\left( {SCD} \right)} \right)} \approx 33,{17^0}\).

 * Ta có: \(\left\{ \begin{array}{l}\overrightarrow {SB}  = \left( {1;0; - 3} \right)\\\overrightarrow {SD}  = \left( {0;2; - 3} \right)\end{array} \right. \Rightarrow {\overrightarrow n _{\left( {SBD} \right)}} = \left[ {\overrightarrow {SB} ;\overrightarrow {SD} } \right] = \left( {6;3;2} \right)\).

\( \Rightarrow \sin \widehat {\left( {SM;\left( {SBD} \right)} \right)} = \dfrac{{\left| {\overrightarrow {SM} .{{\overrightarrow n }_{\left( {SBD} \right)}}} \right|}}{{\left| {\overrightarrow {SM} } \right|.\left| {{{\overrightarrow n }_{\left( {SBD} \right)}}} \right|}} = \dfrac{{\left| {3 - 6} \right|}}{{\sqrt {\dfrac{{37}}{4}} .\sqrt {49} }} \Rightarrow \widehat {\left( {SM;\left( {SBD} \right)} \right)} \approx 8,{1^0}\).

* Ta có: \({\overrightarrow n _{\left( {SMB} \right)}} = \left[ {\overrightarrow {SM} ;\overrightarrow {SB} } \right] = \left( {6;\dfrac{3}{2};1} \right)\).

\( \Rightarrow \sin \widehat {\left( {SC;\left( {SMD} \right)} \right)} = \dfrac{{\left| {\overrightarrow {SC} .{{\overrightarrow n }_{\left( {SMD} \right)}}} \right|}}{{\left| {\overrightarrow {SC} } \right|.\left| {{{\overrightarrow n }_{\left( {SMD} \right)}}} \right|}} = \dfrac{6}{{\sqrt {14} .\sqrt {\dfrac{{157}}{4}} }} \Rightarrow \widehat {\left( {SC;\left( {SMD} \right)} \right)} \approx 14,{83^0}\)