Question 1036853
So by equilibrium, the sum of the forces in the x and y directions would equal zero.
{{{F[1x]=28*cos(42)=20.81}}}
{{{F[1y]=28*sin(42)=18.74}}}
{{{F[2x]=65*cos(24)=59.38}}}
{{{F[2y]=65*sin(24)=26.44}}}
So then,
{{{F[1x]+F[2x]+F[3x]=0}}}
{{{20.81+59.38+F[3x]=0}}}
{{{F[3x]=-80.19}}}
and
{{{F[1y]+F[2y]+F[3y]=0}}}
{{{18.74+26.44+F[3y]=0}}}
{{{F[3y]=-45.18}}}
So then,
{{{abs(F[3])=sqrt(F[3x]^2+F[3y]^2)}}}
{{{abs(F[3])=sqrt(80.19^2+45.18^2)}}}
{{{abs(F[3])=sqrt(8471.67)}}}
{{{abs(F[3])=94.02}}}
and
{{{tan(theta)=F[3y]/F[3x]}}}
{{{tan(theta)=45.18/80.19}}}
{{{tan(theta)=0.5634}}}
Since both are negative, {{{theta}}} is in Quadrant 3.
{{{theta=209.4}}}