Answer
$80.27 \ N$
and $ -16.2^{\circ}$
Work Step by Step
The resultant vector $F$ is given by: $F=F_1+F_2$
Where: $F_1=F_{1x}+F_{1y}$ and $F_2=F_{2x}+F_{2y}$
The components of the force vectors are:
$F_{1x}=40 \cos 30^{\circ}=20\sqrt 3 \approx 34.64 \\F_{1y}=40 \sin 30^{\circ}=20 \\ F_{2x}=60 \cos (-45)^{\circ}=30\sqrt 2 \approx 42.43 \\F_{2y}=60 \sin (-45)^{\circ}=- 30 \sqrt 2 \approx 42.43$
Now, $F_1=F_{1x}+F_{1y} \approx 77.07 $ and $F_2=F_{2x}+F_{2y} \approx -22.43$
So, the resultant force magnitude is:
$|F|=\sqrt {(77.07)^2+(-22.43)^2}=80.27 \ N$
and $F_y=\tan^{-1} \dfrac{F_y}{F_x} \approx -16.2^{\circ}$
