Consider the ten lines formed by taking the points 2 at a time. Translate if necessary so that all the lines are concurrent. Since translation preserves angles, the sum of the angles between the lines initially must be the same as after this translation, so the sum of the ten angles is 360. Hence the maximum value of the smallest angle is 36 degrees.

In general given $N$ points in the plane, no $3$ collinear, the max of the min of the angles formed by them is $\pi*(N-2)/N$. Let the convex hull have $k$ points, with $N-k$ points in the interior. Consider a point $P$ on the hull, with $Q,R$ adjacent on the hull. Suppose the points in clockwise order about $P$ are $Q,X_1,...X_{N-k},R$. Then $QPX_1,X_1PX_2,...,X_{N-k}PR$ are $N-k+1$ angles which sum to $PQR$. Letting $P$ range over all $k$ points on the hull, we get $k(N-k+1)$ angles which sum to $\pi(N-2)$. Thus the min of all the angles is at most $\pi(N-2)/(k(N-k+1))$. Since $3 \le k \le N$, the denominator is minimized when $k=N$, so the min of all the angles is at most $\pi(N-2)/N$. Equality is realized by a regular $N$-gon.

WLOG, let the minimum angle with measure $m$ be formed at the vertex $A_1$, and draw rays $R_2$, $R_3$, $R_4$, and $R_5$ from $A_1$ to points $A_2$, $A_3$, $A_4$, and $A_5$ respectively. We consider two cases: 1. It is not possible to draw a line through $A_1$ such that all the rays lie to one side of the line. In this case, one of the $4$ angles at $A_1$ which are formed by adjacent rays will have measure $m$. One of the of the other $3$ angles will have measure atleast $(360-m)/3$. WLOG let this angle be $A_5 A_1 A_2$. Then the angle sum in triangle $A_5 A_1 A_2$ will be $\ge 120 - m/3 + m + m$, which gives $m \le 36$. 2. If the condition of case 1 doesn't hold, then we may let rays $R_2$, $R_3$, $R_4$, $R_5$ lie in a clockwise order, with angle $A_2 A_1 A_5$ being acute. Then one of the three acute angles formed by adjacent rays at $A_1$ has measure $m$. Also, the angle sum in triangle $A_2 A_1 A_5$ will be $\ge 3m + m + m$, which gives $m \le 36$. Equality is achieved for a regular pentagon.

Case 1: Convex hull is a pentagon. In the pentagon, consider the vertex around which angle is least ($\implies \text{angle} \leq 108^{\circ}$). Join this vertex to the other 2 vertices as well, creating 2 edges and 2 diagonals of the pentagon. As there are 4 "consecutive" lines with the angle between the furthest 2 at most $108^{\circ}$, we can constrain one of the angles here to be at most $\frac{180}3=36^{\circ}$. (Note: this maxima is achieved in a regular pentagon) Case 2: Hull is a quadrilateral. Again, considering the vertex with the least angle ($\implies \text{angle} \leq 90^{\circ}$), and joining the vertex to the opposite vertex and to the point inside the quadrilateral, we similarly get one of the angles $\leq \frac{90}3=30^{\circ}$. Case 3: Triangle hull Considering vertex with least angle ($\leq 60^{\circ}$ now), join this with 2 points inside the hull to get one of the angles $\leq 20^{\circ}$. Hence we can conclude the maximum minima is $36^{\circ}$, achieved in a regular pentagon. Not nearly as elegant as #4, but should work just as well...

We claim in any configuration, one of the angles is at most 36. The idea is to make cases on the convex hull. Case 1: The convex hull is a pentagon. The sum of the internal angles is 540, so one of the angles is at most 108, WLOG $\angle A_5A_1A_2 \le 108$. Now, $A_3A_1$ and $A_4A_1$ cut this angle into 3 pieces, so one of these is at most 36. Case 2: The convex hull is a quadrilateral. WLOG $A_5$ is inside $A_1A_2A_3A_4$. WLOG $A_5$ is inside $A_2A_3A_4$. The internal angles of $A_2A_3A_4$ sum to 180, so one of them is at most 60, WLOG $\angle A_2A_3A_4\le 60$. Now, $A_5A_3$ cuts this into two smaller angles, so one of them is at most 30. Case 3: The convex hull is a triangle, WLOG $A_1A_2A_3$. One of the angles of this triangle is at most 60, WLOG $\angle A_1A_2A_3\le 60$. Now, $A_4A_2$ and $A_5A_2$ cut this into 3 pieces, so one of them is at most 20. In all cases, the minimum angle created is at most 36. We achieve equality by taking a regular pentagon. Therefore, the answer is 36.