12.303. The base of an oblique prism is a rectangle with sides $a$ and $b$. Two adjacent lateral faces form angles $\alpha$ and $\beta$ with the base plane. Find the volume of the prism if the lateral edge is equal to $c$.
Solution.

Let the rectangle $ABCD$ be the base of the oblique prism $ABCD A_1B_1C_1D_1$, where $AD = a$, $CD = b$, $DD_1 = c$, and $D_1O$ is the height of the prism (Fig. 12.165).

Draw perpendiculars $D_1E$ to $AD$ and

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1058.jpg?height=551&width=735&top_left_y=74&top_left_x=317)

Fig. 12.166

$D_1F$ to $CD$ in the faces $AA_1D_1D$ and $CC_1D_1D$. Then $OE \perp AD$, $OF \perp CD$, $\angle D_1EO$ and $\angle D_1FO$ are the angles of inclination of the faces $AA_1D_1D$ and $CC_1D_1D$ to the base plane, $\angle D_1EO = \alpha$, $\angle D_1FO = \beta$.

Let $D_1O = d$, then from

$\triangle D_1OF (\angle D_1OF = 90^\circ): D_1F = \frac{D_1O}{\sin \angle D_1FO} = \frac{d}{\sin \beta}$.

In $\triangle D_1OE (\angle D_1OE = 90^\circ): EO + D_1O \operatorname{ctg} \angle D_1EO = d \operatorname{ctg} \alpha$.

$EOFD$ is a rectangle and $DF = EO = d \operatorname{ctg} \alpha$.

In $\triangle D_1FD (\angle D_1FD = 90^\circ): D_1F^2 + DF^2 = DD_1^2 \Rightarrow \frac{d^2}{\sin^2 \beta} + d^2 \operatorname{ctg}^2 \alpha = c^2$; $d^2 \left(\frac{1}{\sin^2 \beta} + \operatorname{ctg}^2 \alpha\right) = c^2$

$d = \frac{c}{\sqrt{1 + \operatorname{ctg}^2 \alpha + \operatorname{ctg}^2 \beta}}$.

The volume of the prism $V = S_{ABCD} \cdot d = \frac{abc}{\sqrt{1 + \operatorname{ctg}^2 \alpha + \operatorname{ctg}^2 \beta}}$.

Answer: $\frac{abc}{\sqrt{1 + \operatorname{ctg}^2 \alpha + \operatorname{ctg}^2 \beta}}$.