15. Given that 15 rays share a common endpoint. Question: What is the maximum number of obtuse angles (considering the angle between any two rays to be the one not greater than $180^{\circ}$) that these 15 rays can form?
15. First, it is explained that constructing 75 obtuse angles is achievable.

The position of the rays is represented by their inclination angles, with 15 rays placed near the positions of $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$. Within each group, the five rays are sufficiently close to each other (as shown in Figure 10). Thus, the total number of obtuse angles formed is
$$
\mathrm{C}_{15}^{2}-3 \mathrm{C}_{5}^{2}=75
$$

Next, we prove that it is impossible to obtain more than 75 obtuse angles.
For any arrangement, if there do not exist three rays such that the angles between them are all obtuse, then the total number of obtuse angles does not exceed $\frac{\mathrm{C}_{15}^{2} \times 2}{3}=70$; if there exist three rays such that the angles between each pair are all obtuse, then any other ray can form at most two obtuse angles with these three rays. Removing these initial three rays, the total number of obtuse angles is reduced by at most
$$
3+2 \times 12=27 \text{.}
$$

In the remaining 12 rays, continue to search for such three rays (if such three rays do not exist, then the initial total number of obtuse angles does not exceed $27+44=71$), and remove these three rays. Thus, the number of obtuse angles is reduced by at most $3+2 \times 9=21$. Continue this process, and the number of obtuse angles is reduced by at most $3+2 \times 6=15$ and $3+2 \times 3=9$. In the end, only three rays remain.
Therefore, the initial number of obtuse angles is at most
$$
27+21+15+9+3=75
$$