Cho hình lăng trụ tam giác đều ABC.A’B’C’ có...

B

- Hướng dẫn giải

Đáp án B.

Dễ thấy:

$\widehat{\left(AB\text{'}C\text{'}\right);\left(MNP\right)}=\widehat{\left(AB\text{'}C\text{'}\right);\left(MNCB\right)}$

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

Ta có:

$\widehat{\left(MNBC\right);\left(ABC\right)}=\widehat{\left(A\text{'}P;AP\right)}=\widehat{A\text{'}PA}=\arctan \frac{2}{3}.$

Và

$\widehat{\left(MNBC\right);\left(ABC\right)}=\widehat{\left(SP;AP\right)}=\widehat{SPA}=\text{arctan}\frac{4}{3},$ 

với S là điểm đối xứng với A qua A’,

thì $SA=2AA\text{'}=4.$

Suy ra 

$\text{cos}\widehat{\left(AB\text{'}C\text{'}\right);\left(MNP\right)}=c\text{os}\left({180}^0\text{-arctan}\frac{2}{3}-\arctan \frac{4}{3}\right)=\frac{\sqrt{13}}{65}.$