Gắn hệ trục tọa độ như hình vẽ, giả sử ABCD là hình vuông cạnh l,

chiều cao hình chóp SH = h.

Khi đó \(A\left( {0;0;0} \right),B\left( {1;1;0} \right),D\left( {0;1;0} \right),C\left( {1;1;0} \right)\).

Gọi tọa độ \(H\left( {a;b;0} \right) \Rightarrow S\left( {a;b;h} \right)\) 

Ta có: \(\overrightarrow {AS}  = \left( {a;b;h} \right),\overrightarrow {AD}  = \left( {0;1;0} \right) \Rightarrow {\overrightarrow n _{\left( {SAD} \right)}} = \left[ {\overrightarrow {AS} ;\overrightarrow {AD} } \right] = \left( { - h;0;a} \right)\) 

\(\overrightarrow {BS}  = \left( {a - 1;b;c} \right),\overrightarrow {BC}  = \left( {0;1;0} \right) \Rightarrow {\overrightarrow n _{\left( {SBC} \right)}} = \left[ {\overrightarrow {BS} ;\overrightarrow {BC} } \right] = \left( { - h;0;a - 1} \right)\) 

\(\overrightarrow {AB}  = \left( {1;0;0} \right),\overrightarrow {AS}  = \left( {a;b;h} \right) \Rightarrow {\overrightarrow n _{\left( {SAB} \right)}} = \left[ {\overrightarrow {AB} ;\overrightarrow {AS} } \right] = \left( {0; - h;b} \right)\)

\({\overrightarrow n _{\left( {ABCD} \right)}} = \overrightarrow k  = \left( {0;0;1} \right)\) 

Do \(\left( {SAD} \right) \bot \left( {SBC} \right) \Rightarrow {\overrightarrow n _{\left( {SAD} \right)}}.{\overrightarrow n _{\left( {SBC} \right)}} = 0 \Leftrightarrow {h^2} + a\left( {a - 1} \right) = 0 \Leftrightarrow {h^2} + {a^2} = a\,\,\left( 1 \right)\) 

Góc giữa (SAB) và (SBC) bằng \({60^0} \Rightarrow \cos {60^0} = \frac{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}.{{\overrightarrow n }_{\left( {SBC} \right)}}} \right|}}{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}} \right|.\left| {{{\overrightarrow n }_{\left( {SBC} \right)}}} \right|}} \Leftrightarrow \frac{1}{2} = \frac{{\left| {b\left( {a - 1} \right)} \right|}}{{\sqrt {{h^2} + {{\left( {a - 1} \right)}^2}.\sqrt {{h^2} + {b^2}} } }}\)  

\( \Leftrightarrow \frac{1}{2} = \frac{{b\left( {a - 1} \right)}}{{\sqrt {1 - a} \sqrt {{h^2} + {b^2}} }} \Leftrightarrow \frac{{b\sqrt {1 - a} }}{{\sqrt {{h^2} + {b^2}} }} = \frac{1}{2} \Leftrightarrow \frac{b}{{\sqrt {{h^2} + {b^2}} }} = \frac{1}{{2\sqrt {1 - a} }}\) 

Góc giữa (SAB) và (SAD) là \({45^0} \Rightarrow \cos {45^0} = \frac{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}.{{\overrightarrow n }_{\left( {SAD} \right)}}} \right|}}{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}} \right|.\left| {{{\overrightarrow n }_{\left( {SAD} \right)}}} \right|}} \Leftrightarrow \frac{{\sqrt 2 }}{2} = \frac{{\left| {ab} \right|}}{{\sqrt {{h^2} + {a^2}} \sqrt {{h^2} + {b^2}} }}\) 

\( \Leftrightarrow \frac{{\sqrt 2 }}{2} = \frac{{ab}}{{\sqrt a .\sqrt {{h^2} + {b^2}} }}\) 

Suy ra \(\frac{{ab}}{{\sqrt a .\sqrt {{h^2} + {b^2}} }}:\frac{{b\sqrt {1 - a} }}{{\sqrt {{h^2} + {b^2}} }} = \frac{{\sqrt 2 }}{2}:\frac{1}{2} \Leftrightarrow \frac{{\sqrt a }}{{\sqrt {1 - a} }} = \sqrt 2  \Leftrightarrow a = \frac{2}{3}\)

Gọi góc giữa (SAB) và (ABCD) là \(\alpha  \Rightarrow \cos \alpha  = \frac{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}.{{\overrightarrow n }_{\left( {ABCD} \right)}}} \right|}}{{\left| {{{\overrightarrow n }_{\left( {SAB} \right)}}} \right|\left| {{{\overrightarrow n }_{\left( {ABCD} \right)}}} \right|}} = \frac{b}{{\sqrt {{h^2} + {b^2}} }} = \frac{1}{{2\sqrt {a - \frac{2}{3}} }} = \frac{{\sqrt 3 }}{2}\)