In one of the faces of a dihedral angle equal to $\phi$, a point $A$ is taken at a distance $a$ from the edge. Find the distance from point $A$ to the plane of the other face.
Let $\phi=90^{\circ}$ (Fig.1). Through point $A$, lying in the face $\alpha$ of the given dihedral angle, draw a line perpendicular to the edge $l$ of the dihedral angle. Let the drawn line intersect the edge $l$ at point $B$. In the plane $\beta$ of the other face of the dihedral angle, draw the line $BC$, perpendicular to $l$. Then $ABC$ is the linear angle of the given dihedral angle. In the considered case, $\angle ABC=90^{\circ}$, so the line $AB$ is perpendicular to two intersecting lines $l$ and $BC$ in the plane $\beta$. Therefore, $AB$ is perpendicular to this plane. Then the distance from point $A$ to the plane $\beta$ is $a$. If $\phi90^{\circ}$, then similarly we get,

$$
AA1=AB \cdot \sin \left(180^{\circ}-\phi\right)=a \sin \phi
$$

This formula is also valid in the case when $\phi=90^{\circ}: AA1=a \sin 90^{\circ}=a$.

## Answer

$a \sin \phi$.