#### Answer

$\theta=9.21^{\circ}$

#### Work Step by Step

The required angle can be determined as follows:
$F_s=k(2sin\theta-2sin0)$
$\implies F_s=50(2sin\theta-0)=100sin\theta$
The virtual displacements are given as
$\delta_{yC}=\frac{d(4sin\theta)}{d\theta}=4cos\theta$
and $\delta_{yD/2}=\frac{d(2sin\theta)}{d\theta}=2cos\theta$
Now, according to the virtual-work equation
$\delta U=0$
$\implies P\delta_{yC}F_s\delta_{yD/2}=0$
We plug in the known values to obtain:
$8(4)cos\theta-100sin\theta(2)cos\theta=0$
This simplifies to:
$\theta=9.21^{\circ}$