Cho lăng trụ đứng ABC.A’B’C’ có đáy ABC là tam giác cân

Đáp án A

$\begin{array}{l}{S}_{\Delta ABC}=\cos \alpha .S_{\Delta AB\text{'}I}\Rightarrow \cos \alpha =\frac{S_{\Delta ABC}}{S_{\Delta AB\text{'}I}}.\\ S_{\Delta ABC}=\frac{1}{2}.AB.AC.\sin 120°=\frac{\sqrt{3}}{4}{a}^2.\end{array}$

AB’ là đường chéo hình vuông $A\text{'}B\text{'}BA\Rightarrow A\text{'}B=a\sqrt{2}$

$\begin{array}{l}AI=\sqrt{A{C}^2+I{C}^2}=\sqrt{a^2+{\left(\frac{a}{2}\right)}^2}=\frac{\sqrt{5}}{2}a\\ B\text{'}I=\sqrt{C\text{'}{I}^2+B\text{'}C{\text{'}}^2}=\sqrt{C\text{'}{I}^2+B{C}^2}=\sqrt{C\text{'}I{\text{'}}^2+A{C}^2+A{B}^2-2AB.AC.\cos 120°}\\ =\sqrt{{\left(\frac{a}{2}\right)}^2+a^2+a^2-2\left(-\frac{1}{2}\right).a.a}=\frac{\sqrt{13}}{2}a\end{array}$ .

Theo định lý Pi-ta-go đảo ta thấy $\Delta AB\text{'}I$  vuông tại  $A\Rightarrow S_{\Delta AB\text{'}I}=\frac{1}{2}.AI.AB\text{'}=\frac{1}{2}a\sqrt{2}.a\frac{\sqrt{5}}{2}=\frac{\sqrt{10}{a}^2}{4}$ .

Vậy $\cos \alpha =\frac{S_{\Delta ABC}}{S_{\Delta AB\text{'}I}}=\frac{\frac{\sqrt{3}{a}^2}{4}}{\frac{\sqrt{10}{a}^2}{4}}=\frac{\sqrt{30}}{10}.$