Answer
$F_1\approx473.5\;N$
$\phi\approx10.92^\circ$
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. Thus we have
$F_{1x}=F_1\sin\phi\;N$
$F_{1y}=F_1\cos\phi\;N$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=200\;N$. Thus, we have
$F_{2x}=200\cos0^\circ\;N=200\;N$
$F_{2y}=200\sin0^\circ\;N=0\;N$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=260\;N$. Using proportional parts of similar triangles and considering the directions we have
$\frac{F_{3x}}{260\;lb}=\frac{5}{13}$
or, $F_{3x}=260\;N\Big(\frac{5}{13}\Big)=100\;N$
Similarly,
$\frac{F_{3y}}{260\;N}=-\frac{12}{13}$
or, $F_{3y}=-260\;N\Big(\frac{12}{13}\Big)=-240\;N$
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, $(F_R)_x=(F_1\sin\phi+200+100+)\;N=(F_1\sin\phi+300)\;N$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(F_1\cos\phi+0-240)\;N=(F_1\cos\phi-240)\;N$
The magnitude of the resultant force $(\vec F_R)$ acting on the bracket is to be $F_R=450\;N$ 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
$(F_R)_x= 450\cos30^\circ\;N$
or, $F_1\sin\phi+300=389.7$
or, $F_1\sin\phi=89.7$ ...................$(1)$
$(F_R)_y= 450\sin30^\circ\;N$
or, $F_1\cos\phi-240=225$
or, $F_1\cos\phi=465$ ...................$(2)$
$\therefore \tan\phi=\frac{89.7}{465}$
$\implies\;\phi=\tan^{-1}\Big(\frac{89.7}{465}\Big)\approx10.92^\circ$
Substituting $\phi=10.92^\circ$ in eq. $1$, we get $F_1\approx473.5\;N$
