Two planes: $\mathbf{S}$ and $\mathbf{S}^{*}$ form an angle of $45^{\circ}$ with each other, and their line of intersection is $m$. The line $e$ in $\mathbf{S}$ forms a $45^{\circ}$ angle with $m$. What angle does the line $e$ form with the plane $\mathbf{S}^{*}$?
Let's generalize the question. Let the angle between the two planes be $\alpha$, and the angle between $e$ and $m$ be $\beta$; what is the angle $\varphi$ that the line $e$ makes with the plane $\mathbf{S}^{*}$?

By the angle between a line and a plane, we mean the angle that the line makes with its orthogonal projection on the plane. The projection of the line $e$ on the plane $\mathbf{S}^{*}$ is $e^{\prime}$ (see the figure), which is obtained by connecting the projection of some point $E$ on $e$ to the intersection point of $e$ and $m$, denoted by $M$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4998a54895b18487fbb1g-1.jpg?height=548&width=656&top_left_y=390&top_left_x=732)

Construct a plane through the line $E E^{\prime}$ perpendicular to $m$ (such a plane exists because $E E^{\prime} \perp \mathbf{S}^{*}$): denote the intersection point of the plane $e$ and $m$ by $N$. $E^{\prime} N \perp m$ and $\angle E N E^{\prime} = \alpha$, the angle between the two planes.

Let $E N = 1$, then

\[
\begin{aligned}
& \text{from the right triangle } E E^{\prime} N, \sin \alpha = E E^{\prime}, \\
& \text{from the right triangle } E N M, \sin \beta = \frac{1}{E M}, \\
& \text{from the right triangle } E E^{\prime} M, \sin \varphi = \frac{E E^{\prime}}{E M}
\end{aligned}
\]

Substituting the values of $E E^{\prime}$ and $\frac{1}{E M}$ from (1) and (2) into (3), we get

\[
\sin \varphi = \sin \alpha \sin \beta
\]

In our problem, $\alpha = \beta = 45^{\circ}$, so $\sin \varphi = \left(\frac{\sqrt{2}}{2}\right)^{2} = \frac{1}{2}$, from which $\varphi = 30^{\circ}$.

Sándor Frivaldszky (Bp., II., Rákóczi g. III. o. t.)