Force balance in the horizontal direction (the x-direction) gives
$\begin{align*}\left|\mathbf{F}_T \cdot \mathbf{\hat{x}}\right| = 20 \mbox{ N}\end{align*}$
Force balance in the vertical direction (the y-direction) gives
$\begin{align*}\left|\mathbf{F}_T \cdot \mathbf{\hat{y}}\right| = (2\mbox{ kg}) \cdot g = 20 \mbox{ N}\end{align*}$
Since the tension force must act along the rope, the angle that the rope makes with the vertical is exactly:
$\begin{align*}\arctan \left(\frac{20 \mbox{ N}}{20 \mbox{ N}} \right) = \boxed{\arctan 0.5} \end{align*}$
So, answer (A) is correct.

Solution to 1986 Problem 7