Cho lăng trụ tam giác ABC.A'B'C' có BB' = a , góc giữa đường thẳng BB' và mặt phẳng (ABC) bằng 60 độ

Đáp án C

Ta có:  $V_{A^{\text{'}}.ABC}=V_{B^{\text{'}}.ABC}=\frac{1}{3}{B}^{\text{'}}G.S_{ABC}$(*).

Ta có:  $\left\{\begin{array}{l}{B}^{\text{'}}G=B{B}^{\text{'}}\sin 60°=\frac{a\sqrt{3}}{2}\left(1\right)\\ BG=B{B}^{\text{'}}\cos 60°=\frac{a}{2}\Rightarrow BM=\frac{3}{2}BG=\frac{3a}{4}\end{array}\right.$.

Đặt $BC=x\Rightarrow AC=BC.\cot 60°=\frac{x}{\sqrt{3}}\Rightarrow MC=\frac{x}{2\sqrt{3}}$.

Ta có: $B{M}^2=B{C}^2+M{C}^2\iff \frac{9{\text{a}}^2}{16}=x^2+\frac{x^2}{12}\iff x=\frac{3a\sqrt{39}}{26}$.

Hay $BC=\frac{3a\sqrt{39}}{26};AC=\frac{3\text{a}\sqrt{13}}{26}\Rightarrow S_{ABC}=\frac{1}{2}AC.BC=\frac{1}{2}.\frac{3a\sqrt{39}}{26}.\frac{3\text{a}\sqrt{13}}{26}=\frac{9{\text{a}}^2\sqrt{3}}{104}$  (2).

Thay (1), (2) vào (*) ta được: $V_{A^{\text{'}}.ABC}=\frac{1}{3}.\frac{a\sqrt{3}}{2}.\frac{9{\text{a}}^2\sqrt{3}}{104}=\frac{9a^3}{208}$.