Answer
$\theta_1=16.6^{\circ}$ and $\theta_2=35.8^{\circ}$
Work Step by Step
We can find the required angle $\theta$ as follows:
$F_s=k(4cos45-4cos\theta)=200(4cos45-4cos\theta)$
The virtual displacements are given as
$\delta _{yC}=\frac{d(4sin\theta)}{d\theta}=4cos\theta$
$\delta_{yE}=\frac{d(4sin\theta)}{d\theta}=4cos\theta$
$\delta_{xC}=\frac{d(4cos\theta)}{d\theta}=-4sin\theta$
Now, according to the virtual-work function equation
$\delta U=0$
$\implies P\delta_{yC}+P\delta_{yE}+F_s\delta_{xE}=0$
We plug in the known values to obtain:
$30\times 4cos\theta+30(4)cos\theta-800(cos45-cos\theta)\cdot 4sin\theta=0$
This simplifies to:
$\theta_1=16.6^{\circ}$ and $\theta_2=35.8^{\circ}$
