Answer
$103.08lb;30.3^{\circ}$
Work Step by Step
First of all, we fine the sum of the forces in x and y direction as follows:
$\Sigma F_x=30+91(\frac{5}{13})+40(\frac{3}{5})=89lb$
and $\Sigma F_y=40(\frac{4}{5})-91(\frac{12}{13})=-52lb$
Now the resultant forces can be determined as:
$F_R=\sqrt{(\Sigma F_x)^2+(\Sigma F_y)^2}$
We plug in the known values to obtain:
$F_R=\sqrt{(89)^2+(-52)^2}=103.08lb$
The direction of the resultant force is given as
$\theta=tan^{-1}(\frac{F_y}{F_x})$
$\implies \theta+tan^{-1}(\frac{52}{89})$
$\implies \theta=30.3^{\circ}$
