Let $S_{1}$ and $S_{2}$ be planes that are perpendicular to each other. The line $e$ makes a $30^{\circ}$ angle with both planes. What angle does the line $e$ form with the intersection line of the two planes?
Let the line $e$ intersect the plane $S_{1}$ at $A$, and the plane $S_{2}$ at $B$. Drop perpendiculars from $A$ and $B$ to the intersection line of the two planes, and let the feet of these perpendiculars be $A_{1}$ and $B_{1}$, respectively.

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The triangles $A B A_{1}$ and $A B B_{1}$ are congruent, as both are right-angled, share the side $A B$, and $\angle A B A_{1} = \angle B A B_{1} = 30^{\circ}$, which are the angles of inclination of $e$ to $S_{2}$ and $e$ to $S_{1}$, respectively; thus, $A_{1} A = B_{1} B = a$. Measure this distance along the perpendicular to $S_{1}$ at $A_{1}$: point $A_{2}$, and along the perpendicular to $S_{2}$ at $B_{1}$: point $B_{2}$. In the isosceles right-angled triangle $A A_{1} A_{2}$, $A A_{2} = a \sqrt{2}$. In the right-angled triangle $A B B_{1}$, one of the acute angles is $30^{\circ}$, so $A B = 2 B B_{1} = 2 a$, and $A B_{1} = a \sqrt{3}$.

From the right-angled triangle $A A_{1} B_{1}$, by the Pythagorean theorem,

$$
A_{2} B = A_{1} B_{1} = \sqrt{A B_{1}^{2} - A A_{1}^{2}} = \sqrt{(a \sqrt{3})^{2} - a^{2}} = a \sqrt{2}
$$

thus, the triangle $A A_{2} B$ is isosceles ($A A_{2} = A_{2} B$) and right-angled; therefore, $\angle A_{2} B A = 45^{\circ}$. Since $A_{1} B_{1} \parallel A_{2} B$, the angle $\angle A_{2} B A$ is equal to the angle of inclination of the line $e$ and the intersection line of the two planes.