Answer
$\theta=23.6^{\circ}$
Work Step by Step
We can determine the required angle $\theta$ as follows:
$F_s=k(sin\theta-sin0)=6sin\theta$
The virtual displacements are given as
$\delta_{yA}=\frac{d(3sin\theta)}{d\theta}=3cos\theta$
and $\delta_{yD}=\frac{d(1sin\theta)}{d\theta}=1cos\theta$
Now, the virtual-work equation is given as
$\delta U=0$
$\implies P\delta_{yA}+F_x\delta_{yC}=0$
$\implies 0.8(3cos\theta)-6sin\theta cos\theta=0$
This simplifies to:
$\theta=23.6^{\circ}$
