How many obtuse angles can be formed from the 15 rays on a single point on a same plane? Consider 15 rays that originate from a point. What is the maximum number of obtuse angles they can form, assuming that the angles between two rays is less than or equal to 180 degrees?
 A: Call two rays near if they form a non-obtuse angle. If there are $n$ ordered pairs of near rays, there are exactly ${15\choose 2}-\frac n2$ obtuse angles among the rays. Hence we want to minimize $n$.
Claim. $n\ge 60$.
Proof. Suppose one of the rays (wlog the positive $x$ axis) is near  $a\le 2$ other rays (i.e., we have $a$ other rays within the (closed) first or fourth quadrant). Then there are $b$ rays in the second and $c$ rays in the third quadrant (with the negative $x$ axis being counted as either of these quadrants) where $b+c\ge12$.
Then we have (e.g., by Jensen's inequality) $$n\ge b(b-1)+c(c-1)\ge 2\cdot 6\cdot 5=30.$$
Therefore,  we need only consider configurations where each ray is near $\ge 3$ other rays. 
Again, suppose the some ray is near $a=3$ other rays, and define $b$ and $c$ as above, where now $b+c=11$. This time, we  have 
$\ge b(b-1)+c(c-1)=(b-5)^2+(b-6)^2+49\ge50$ near pairs within the left half plane. Additionally, each of the $4$ rays in the right half plane is first component of at least $3$ near pairs. Hence,
$$ n\ge 50+4\cdot 3=62.$$
Remains the case that each ray is near at least $4$ other rays. Then clearly, 
$$n\ge 4\cdot 15=60.$$
$\square$
As we can achieve the lower bound $n=60$ (e.g., with five rays per each of the  directions $0^\circ$,  $120^\circ$, $240^\circ$), the maximal number of obtuse angles is
$$ 75.$$
