Answer
$\theta = 29^{\circ}$
Work Step by Step
In part (c), we found that:
The horizontal component $T_{3x} = 28~N$
The vertical component $T_{3y} = 50~N$
We can find $\theta$:
$tan~\theta = \frac{T_{3x}}{T_{3y}}$
$\theta = tan^{-1}~\frac{T_{3x}}{T_{3y}}$
$\theta = tan^{-1}~\frac{28~N}{50~N}$
$\theta = 29^{\circ}$
