Câu hỏi 8

Dễ thấy \(\widehat{\left( A{B}'{C}' \right);\left( MNP \right)}=\widehat{\left( A{B}'{C}' \right);\left( MNCB \right)}\)

         \(\begin{align}  & ={{180}^{0}}-\widehat{\left( A{B}'{C}' \right);\left( {A}'{B}'{C}' \right)}-\widehat{\left( MNBC \right);\left( {A}'{B}'{C}' \right)} \\ & ={{180}^{0}}-\widehat{\left( {A}'BC \right);\left( ABC \right)}-\widehat{\left( MNBC \right);\left( ABC \right)}. \\\end{align}\)

Ta có \(\widehat{\left( {A}'BC \right);\left( ABC \right)}=\widehat{\left( {A}'P;AP \right)}=\widehat{{A}'PA}=\arctan \frac{2}{3}.\)

Và \(\widehat{\left( MNBC \right);\left( ABC \right)}=\widehat{\left( SP;AP \right)}=\widehat{SPA}=\arctan \frac{4}{3},\) với \(S\) là điểm đối xứng với \(A\) qua \({A}',\) thì \(SA=2\,A{A}'=4.\)

Suy ra \(\cos \widehat{\left( A{B}'{C}' \right);\left( MNP \right)}=\cos \left( {{180}^{0}}-\arctan \frac{2}{3}-\arctan \frac{4}{3} \right)=\frac{\sqrt{13}}{65}.\)

Chọn B.