Answer
$F=19.2N$
$\theta= 2.37^{\circ}$ $\searrow$
Work Step by Step
We are asked to determine the magnitude of the resultant force, $F_R$, and its direction, $\theta$
We can apply the law of cosines to find the magnitude of $F'$, which is the result of $F_1$ and $F_2$
$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$
$c=\sqrt{20^2+50^2-2*20*50*\cos(70^{\circ})}$
$c=47.07$
$F'=47.07N$
Using this result, we can apply the law of sines to find $ \theta '$
$\sin70^{\circ} / 47.07 = \sin \theta' / 20$
Solving for $\theta'$, we obtain:
$\theta'=23.53^{\circ}$
We can apply the law of cosines to find the magnitude of $F_R$, which is the result of $F'$ and $F_3$
$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$
$c=\sqrt{47.07^2+30^2-2*47.07*30*\cos(13.34^{\circ})}$
$c=19.18$
$F=19.2N$
Using this result, we can apply the law of sines to find $ \theta $
$\sin13.14^{\circ} / 19.18 = \sin \phi / 30$
Solving for $\phi$, we obtain:
$\phi=21.15^{\circ}$
$\theta= \theta' -\phi = 23.53^{\circ}-21.15^{\circ}=2.37^{\circ}$
