In the plane, there are $n$ rays such that any two rays form an obtuse angle. How large can $n$ be?
The rays from the center of a regular tetrahedron to its vertices form obtuse angles with each other, so $n$ can be 4 (and obviously $3, 2, 1, 0$ as well).

We will show that $n$ cannot be greater than 4. Consider the unit vectors in the direction of these rays: according to our assumption, their dot product is negative. Choose the positive $x$-axis of the coordinate system to be one of these rays, then the first coordinate of the others is negative. If $\left(a_{1}, a_{2}, a_{3}\right)$ and $\left(b_{1}, b_{2}, b_{3}\right)$ are two of the others, their dot product is $a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}$, which can only be negative if $a_{2} b_{2}+a_{3} b_{3}<0$, i.e., the dot product of the vectors $\left(a_{2}, a_{3}\right),\left(b_{2}, b_{3}\right)$ is also negative. Therefore, there are also obtuse angles among the projections of the others onto the $(y, z)$ plane. Choose the $y$-axis to be one of these projections, then the second coordinate of the others is also negative, so the product of the third coordinates is also negative. However, the set of real numbers in which the product of any two numbers is negative can have at most two elements, since the set cannot contain 0, and if it had more than two elements, there would be two with the same sign. It can be seen that our reasoning is valid in any dimension, so in general, in a $k$-dimensional space, $n \leq k+1$.