Answer
$82.98 \ N$ and $99.57^{\circ}$
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:
$F_{1x}=30 \cos 45^{\circ}=15 \sqrt 2 \approx 21.21 \\F_{1y}=30 \sin 45^{\circ}=15 \sqrt 2 \approx 21.21 \\ F_{2x}=70 \cos (120)^{\circ}=-35 \\F_{2y}=70 \sin (120)^{\circ}= 35 \sqrt 3 \approx 60.62$
Now, $F_1=F_{1x}+F_{1y} \approx -13.79 $ and $F_2=F_{2x}+F_{2y} \approx 81.83$
So, the resultant force $F$ magnitude is:
$|F|=\sqrt {(-13.79)^2+(81.83)^2}=82.98 \ N$
and $F_y=\tan^{-1} \dfrac{F_y}{F_x} \approx 99.57^{\circ}$
