We know:
$$ T_1 = 1.19 \cdot T_2 \cdots 1)$$
We can determine:
$$T_3 \text{(other rope)} = T_2 \cdots 2)$$
$$ T_{1y} = T_1 \text{ (no x-component [horizontal])}\cdots 3)$$
$$ T_{1x} = 0 \cdots 4)$$
$$T_{2y} + T_{3y} = -T_{1y} = -T_1 \text{(box isn't moving up or down....)} \cdots 5)$$
$$T_{2x} = -T_{3x} \text{(box isn't moving right or left)}\cdots 6)$$
$$ T_2 = \sqrt{{T_{2y}}^2 + {T_{2x}}^2)}\cdots 7)$$
There are 7 equations and only 6 unknowns ($T_1$, $T_2$, $T_3$, in $x$ and $y$)
So find $T_{2x}$ and $T_{2y}$ and you will have your angle.
I guess you need to assert that the angle of rope 2 with the vertical is the same as theta, due to the metal ring which allows the rope to slip through it and equalize the tensions on rope 2 and rope 3.