The sides of each of the 80 angles pass through the endpoints of a side of the 10-gon $P$. We say that such an angle and such a side are adjacent; each side is adjacent to exactly 8 angles. Call black the 59 angles that are known to be equal and $\alpha$ the measure of a black angle. Let $a$ be a side of $P$. The locus of points $X$ such that $a$ subtends angle $\alpha$ at $X$ is the union of two circular arcs. Denote by $y_a$ the one of them which lies on the same side of $a$ as the polygon $P$. We also name $y_a$ the entire circle containing $y_a$. All black angles adjacent to $a$ have their vertices on $y_a$.

For a side $a$ let $m_a$ be the number of black angles adjacent to $a$. The hypothesis can be stated as $\sum m_a \ge 59$. We show that the inequality implies that $P$ is cyclic. Consider two cases.

Let there be a side $a$ with $m_a \ge 7$. Then $y_a$ contains the vertices of at least 7 black angles; these vertices are different from the endpoints of $a$. Hence at least $2+7=9$ vertices of $P$ lie on $y_a$.

Suppose that there is a vertex $A \in y_a$, and let $AB = b$, $AC = c$ be the sides with common vertex $A$.
Then $y_b \ne y_a$, $y_c \ne y_a$ by $A \in y_a$. So arcs $y_a$ and $y_b$ has (at most) two common points; one such point is $B$. Because all vertices except $A$ are on $y_a$, we see that $y_b$ contains at most one vertex different from $A$ and $B$, implying $m_b \le 1$. On the other hand it is immediate that $y_a = y_a$ for every $s \ne b, c$, hence $A \in y_a$ for $s \ne b, c$. Thus $m_s \le 7$ for each of the 8 sides $s$ different from $b$ and $c$. In conclusion $\sum m_a \le 1+1+8 \cdot 7 = 58$, contradicting the hypothesis.

Suppose now that $m_a \le 6$ for each side $a$. Then a direct computation using $\sum m_a \ge 59$ shows that $m_a = 6$ holds for at least 9 sides $a$: the last side satisfies $m_a = 5$ or $m_a = 6$. We show that $y_b = y_c$ for every two consecutive sides $b = AB, c = AC$; this is enough to imply that $P$ is cyclic. Indeed $y_b$ contains at least $m_b - 1$ vertices different from $A, B$ and $C$. Likewise $y_c$ contains at least $m_c - 1$ vertices different from $A, B$ and $C$. Both arcs combined contain at least $m_b + m_c - 2 \ge 6 + 5 - 2 = 9$ vertices $D \ne A, B, C$. It follows that there are two vertices $D, E \ne A, B, C$ that are common for $y_b$ and $y_c$. One more such vertex is $A$, so $y_b = y_c$, as stated.

Now that $P$ is cyclic, each side $BC = a$ there corresponds an angle $\alpha_a$ such that $\vec{BC} = \alpha_a$ for every vertex $V \ne B, C$. Also $\alpha_a$ occurs among the 80 angles of division exactly $8k$ times where $k$ is the number of sides with length $a$. Because at least 59 angles are known to be equal, $P$ has at least 8 equal sides. The angle $\alpha_a$ corresponding to them occurs 64, 72 or 80 times. It is straightforward

now that the 80 angles of division can assume at most 3 different values. If these are exactly 3 then one of them occurs 64 times, and each of the other two occurs 8 times.