Answer
The angle that each wire makes with the vertical is $48.2^{\circ}$.
Work Step by Step
The sum of the vertical component of tension $T$ in each wire is equal in magnitude to the weight of the frame.
$2T~cos(\theta) = mg$
$2(0.75~mg)~cos(\theta) = mg$
$cos(\theta) = \frac{1}{1.5}$
$\theta = arccos(\frac{1}{1.5})$
$\theta = 48.2^{\circ}$
The angle that each wire makes with the vertical is $48.2^{\circ}$.
