#### Answer

$54.2N; 43.5^{\circ}$

#### Work Step by Step

We know that
$F_1=50(\frac{3}{5})i+50(\frac{4}{5})j$
$F_2=80 cos\space 255i+80 sin\space 255j$
and $F_3=30i+0j$
Now we find the horizontal and vertical resultant forces as
$R_x=50(\frac{3}{5})+80cos 255+30=39.3N$
and $R_y=50(\frac{4}{5})+80sin\space 255+0=-37.3N$
Now the magnitude of the resultant force is
$F_R=\sqrt{R_x^2+R_y^2}$
We plug in the known values to obtain:
$F_R=\sqrt{(39.3)^2+(-37.3)^2}=54.2N$
Now the direction can be determined as
$\theta=tan^{-1}(\frac{R_y}{R_x})$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{37.3}{39.3})=43.5^{\circ}$