Tension? Brilliant

Let the left-side string tension be $T_{1}$ and the right-side string be $T_{2}.$ The Brilliant sphere will remain in static equilibrium according to $\Sigma F_{x} = 0, \Sigma F_{y} = 0$ , or :

$\Sigma F_{x} \Rightarrow T_{1} \cos(\pi/3) - T_{2} \cos(\pi/6) = 0;$ (i)

$\Sigma F_{y} \Rightarrow T_{1} \sin(\pi/3) + T_{2} \sin(\pi/6) - mg = 0$ (ii)

We first find that $T_{1} = \sqrt{3}T_{2}$ from (i). If we substitute this expression into (ii), then we obtain:

$\sqrt{3}T_{2} (\sqrt{3}/2) + (1/2)T_{2} = (3)(10) \Rightarrow 2T_{2} = 30 \Rightarrow T_{2} = 15, T_{1} =15\sqrt{3}$ newtons.

Hence, $T_{1} + T_{2} = 15(\sqrt{3}+1) = \boxed{40.98}$ newtons.