3.25. In the face of a dihedral angle equal to $\alpha$, a line is drawn that makes an angle $\beta$ with the edge of the dihedral angle. Find the angle between this line and the other face.
3.25. By the condition, $\gamma_{1} \cap \gamma_{2}=l$, the dihedral angle $l$ is equal to $\alpha, A B \in \gamma_{1}$, $(; ; A B)=\beta$; it is required to find $(\gamma_{2}; A B)$ (Fig. 3.27). Draw $A O \perp \gamma_{2}$, $A C \perp$; then $\angle A C O=\alpha$ (as the linear angle of the dihedral angle $A B C$), $\angle A B C=\beta$. Since $O B$ is the projection of $A B$ on $\gamma_{2}$, $\angle A B O$ is the required angle. Let $A O=a$; then

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-514.jpg?height=616&width=683&top_left_y=1057&top_left_x=470)

Fig. 3.27

$$
A C=\frac{a}{\sin \alpha} (\text { from } \triangle A O C)
$$

and in $\triangle A C B$ we have $A B=\frac{A C}{\sin \beta}=\frac{a}{\sin \alpha \sin \beta}$. Finally, from $\triangle A O B$ we find $\sin \angle A B O=\frac{A O}{A B}=\sin \alpha \sin \beta$. Therefore, $\angle A B O=\arcsin (\sin \alpha \sin \beta)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-515.jpg?height=576&width=541&top_left_y=287&top_left_x=45)