A line $e$ is inclined at a $40^{\circ}$ angle to a plane $S$, and at a $49^{\circ}$ angle to a line $f$ in $S$ passing through its foot. What is the angle between $f$ and the projection $s^{\prime}$ of $e$ onto $S$? (We also accept an answer from first-year students based on construction and angle measurement.)





Translating the text into English while preserving the original formatting and structure, the result is as above.
I. solution. Let the intersection point of $e$ with $S$ be $D$, a point different from it be $E$, and its (perpendicular) projection on $S$ - or on $e'$ - be the point $E'$, and on $f$ be the point $F$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5d6a28962bc925ad7750g-1.jpg?height=613&width=751&top_left_y=257&top_left_x=665)

Thus, $D E E'$ and $D E F$ are right triangles, $\angle E D E' = 40^{\circ}$, $\angle E D F = 49^{\circ}$, and we need to determine the angle $\angle E' D F = x$.

Below, we show that the angle at $F$ in the triangle $E' D F$ containing $x$ is a right angle. Based on this,

$$
\begin{gathered}
\cos x = \frac{D F}{D E'} = \frac{D E \cos 49^{\circ}}{D E \cos 40^{\circ}} = \frac{\cos 49^{\circ}}{\cos 40^{\circ}} = \frac{0.6561}{0.7660} = 0.8562 \\
x = 31^{\circ} 6'
\end{gathered}
$$

To prove the used statement, consider the reflection of $D$ over $F$, denoted as $D_1$. Since $E F \perp F D$, $E D D_1$ is an isosceles triangle, $E D_1 = E D$. Drawing the line $D_1 E'$, it is perpendicular to $E E'$ because it lies in $S$, as $D_1$ is also in $S$. Therefore, the right triangles $E D_1 E'$ and $E D E'$, which share a leg, are congruent, $E' D_1 = E' D$, and thus $E D D_1$ is an isosceles triangle, and $E' F$ bisects the base $D D_1$, making it perpendicular to it. - We relied on the definition of perpendicularity between a line and a plane: if a line is perpendicular to a plane, it is perpendicular to every line in the plane.

Judit Székely (Jászberény, Lehel Vezér Gymnasium, 2nd grade)

II. solution. In the first solution, the points $D, E, E', F$ define a triangular pyramid, we draw the net of this pyramid and measure the desired angle $x$ in it.

![](https://cdn.mathpix.com/cropped/2024_05_02_5d6a28962bc925ad7750g-1.jpg?height=837&width=1068&top_left_y=1601&top_left_x=515)

Taking the length of the edge $D E$, we draw two right triangles on each side with it as the hypotenuse, with angles of $40^{\circ}$ and $49^{\circ}$ at $D$ (2nd figure, Thales' circle, $D E E' = \mathrm{I}$. and $D E F = \mathrm{II}$. triangles). Then, we construct the right triangle $E E' F = \mathrm{III}$. with $E E'$ as a leg, whose hypotenuse is the $E F$ obtained in the second triangle. Finally, we construct the triangle $D E' F = \mathrm{IV}$. from the $D E'$ in the first, the $D F$ in the second, and the $E' F$ in the third. In this, by measurement,

$$
x = \angle F D E' \approx 31^{\circ}
$$

(On the other hand, measuring the angle $\angle D F E'$, we find it to be $90^{\circ}$.)

János Reviczky (Budapest, Alsóerdősor utcai Secondary School, 8th grade)