Among the lines $a, b, c$ passing through a point, $a$ and $b$ are perpendicular to each other, $c$ forms a $45^{\circ}$ angle with $a$, and a $60^{\circ}$ angle with $b$. What angle does the line $c$ form with the plane defined by the lines $a$ and $b$?
Let the common point of lines $a, b, c$ be denoted by $O$ and let $C$ be the point on line $c$ such that $O C=1$.

Since $c$ is not perpendicular to the plane $S=[a, b]$, the angle it forms with the plane is given by the angle it forms with its orthogonal projection onto the plane. Let the projection of $C$ onto $S$ be denoted by $C^{\prime}$. We want to determine the angle $\angle C O C^{\prime} = \varphi$. Drop perpendiculars from $C^{\prime}$ to lines $a$ and $b$, and denote the feet of these perpendiculars by $A$ and $B$, respectively.
![](https://cdn.mathpix.com/cropped/2024_05_02_73ef7c6ecad3a9ae6abag-1.jpg?height=764&width=1682&top_left_y=344&top_left_x=210)

Since $C^{\prime} A \perp a$ and $b \perp a$, it follows that line $C^{\prime} A$ is parallel to $b$. Similarly, since $C^{\prime} B \perp b$ and $a \perp b$, $C^{\prime} B$ is parallel to $a$. Thus, quadrilateral $O A C^{\prime} B$ is a rectangle and $O A = C^{\prime} B$, $O B = C^{\prime} A$. To determine the lengths of segments $O A$ and $O B$, we need to show that triangles $O A C$ and $O B C$ are right triangles. It is known that if a line is perpendicular to two (non-parallel) lines in a plane, then it is perpendicular to every line in the plane. Since $C C^{\prime} \perp S$, it follows that $C C^{\prime} \perp a$, but $C^{\prime} A \perp a$. Therefore, line $a$ is perpendicular to two lines in the plane $C C^{\prime} A$, and thus it is perpendicular to line $A C$. Similarly, we can prove that $O B \perp C B$.

Since $\angle C O A = 45^{\circ}$, $\angle C O B = 60^{\circ}$, and $\overline{O C} = 1$, we have $\overline{O A} = \frac{\sqrt{2}}{2}$, $\overline{O B} = \frac{1}{2}$.

Thus, the diagonal of rectangle $O A C^{\prime} B$ is $O C^{\prime} = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^{2} + \left(\frac{1}{2}\right)^{2}} = \frac{\sqrt{3}}{2}$, and from the right triangle $O C C^{\prime}$,

$$
\cos \varphi = \frac{\sqrt{3}}{2} \quad \text{or} \quad \varphi = 30^{\circ}
$$

Gábor Molnár (Zalaegerszeg, Zrínyi M. Gymnasium, 1st year)