Answer
$-1.98KN; 184.53^{\circ}$
Work Step by Step
We know that
$F_x=-30sin 30^{\circ}-26(\frac{5}{13})$
$\implies F_x=-25KN$
and $F_y=-30 cos30^{\circ}+26(\frac{12}{13})$
$\implies F_y=-1.98KN$
Now the resultant force can be calculated as
$F_R=\sqrt{(F_x)^2+(F_y)^2}$
We plug in the known values to obtain:
$F_R=\sqrt{(-25KN)^2+(-1.98KN)^2}$
$\implies F_R=25.08KN$
The orientation can be determined as
$\phi=tan^{-1}(\frac{F_y}{F_x})$
$\phi=(\frac{-1.98KN}{-25KN})=4.53^{\circ}$
$\implies \theta=184.53^{\circ}$
