Cho hình lăng trụ đứng có đáy ABC là tam giác vuông tại B. Biết AB = a, BC = $a\sqrt{3}$ , góc giữa hai mặt phẳng ( $C^{\text{'}}AB$ ) và (ABC) bằng 60°.

Tính $V_{ABC.A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}}$

Ta có:

+) $\left\{\begin{array}{l}AB\perp BC\\ AB\perp CC\text{'}\end{array}\right.\Rightarrow AB\perp \left(BCC\text{'}\right)\Rightarrow AB\perp BC\text{'}.$

+) $\left\{\begin{array}{l}\left(ABC\text{'}\right)\cap \left(ABC\right)=AB\\ AB\perp BC\\ AB\perp BC\text{'}\end{array}\right.$

$\Rightarrow \left(\left(ABC\text{'}\right),\left(ABC\right)\right)=\left(BC,BC\text{'}\right)=\widehat{CBC\text{'}}=60°.$

$\Rightarrow CC\text{'}=BC.\tan 60°=a\sqrt{3}.\sqrt{3}=3a$

$\Rightarrow V_{ABC.A\text{'}B\text{'}C\text{'}}=S_{ABC}.CC\text{'}=\frac{1}{2}.AB.BC.CC\text{'}=\frac{1}{2}.a.a\sqrt{3}.3a=\frac{3\sqrt{3}{a}^3}{2}.$