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?

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
$$ 75.$$
