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 $F_2$
$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$
$c=\sqrt{20^2+30^2-2*20*30*\cos(73.13^{\circ})}$
$c=30.85$
$F=30.85N$
Using this result, we can apply the law of sines to find $ \theta '$
$\sin73.13^{\circ} / 30.85 = \sin (70^{\circ}-\theta') / 30$
Solving for $\theta'$, we obtain:
$\theta'=1.47^{\circ}$
We can apply the law of cosines to find the magnitude of $F_R$, which is the result of $F'$ and $F_3$
$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$
$c=\sqrt{30.85^2+50^2-2*30.85*50*\cos(1.47^{\circ})}$
$c=19.18$
$F=19.2N$
Using this result, we can apply the law of sines to find $ \theta $
$\sin1.47^{\circ} / 19.18 = \sin \theta / 30.85$
Solving for $\theta$, we obtain:
$\theta=2.37^{\circ}$
