A flexible chain of mass m hangs between two fixed points at the...

A chain of mass $m$ hangs between two fixed points at the same level with inclination $30°$ at support points.
Consider equilibrium of one half of the chain (mass $m/2$):
Horizontal equilibrium: $T_0 = T_1\cos(30°) = T_1 \cdot \frac{\sqrt{3}}{2}$
Vertical equilibrium: $T_1\sin(30°) = \frac{m}{2}g$
From vertical equation: $T_1 \cdot \frac{1}{2} = \frac{mg}{2}$, giving $T_1 = mg$
Substituting into horizontal equation: $T_0 = mg \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}mg$