Answer
$F=62.5\;lb$
$\theta\approx14.29^\circ$
Work Step by Step
First we resolve each force into its $x$ and $y$ components using the parallelogram law,
Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force in which $F_1=F\;lb$. Thus we have
$F_{1x}=F\cos\theta\;lb$
$F_{1y}=F\sin\theta\;lb$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=50\;lb$. Thus, we have
$F_{2x}=50\cos0^\circ\;lb=50\;lb$
$F_{2y}=50\sin0^\circ\;lb=0\;lb$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=90\;lb$. Using proportional parts of similar triangles and considering the directions we have
$\frac{F_{3x}}{90\;lb}=-\frac{3}{5}$
or, $F_{3x}=-90\;lb\Big(\frac{3}{5}\Big)=-54\;lb$
Similarly,
$\frac{F_{3y}}{90\;lb}=-\frac{4}{5}$
or, $F_{3y}=-90\;lb\Big(\frac{4}{5}\Big)=-72\;lb$
Second, using scalar notation and indicating the positive directions of components along the $x$ and $y$ axes with symbolic arrows, we sum these components algebraically:
Summing the x components, we have
$\xrightarrow{+} (F_R)_x=\sum F_x$
or, $(F_R)_x=(F\cos\theta+50-54)\;lb=(F\cos\theta-4)\;lb$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(F\sin\theta+0-72)\;lb=(F\sin\theta-72)\;lb$
The magnitude of the resultant force $(\vec F_R)$ acting on the bracket is to be $F_R=80\;lb$ and directed along the $u$ axis. If $(F_R)_x$ and $(F_R)_y$ are two components of $(\vec F_R)$ along positive $x$ and $y$ axes, then
$(F_R)_x= 80\times\cos45^\circ\;lb$
or, $F\cos\theta-4=56.57$
or, $F\cos\theta=60.57$ ...................$(1)$
$(F_R)_y= -80\times\sin45^\circ\;lb$
or, $F\sin\theta-72=-56.57$
or, $F\sin\theta=15.43$ ...................$(2)$
$\therefore \tan\theta=\frac{15.43}{60.57}$
$\implies\;\theta=\tan^{-1}\Big(\frac{15.43}{60.57}\Big)\approx14.29^\circ$
Substituting $\theta=14.29^\circ$ in eq. $1$, we get $F=62.5\;lb$
