Answer
$F=19.6lb$
$F_b=26.4lb$
Work Step by Step
We are given that the component of force, $F$, acting on line $aa$ is $Fa=30lb$.
Using the law of sines with the given angles, we can find the magnitude of F and the component of F on line $bb$.
$\sin80^{\circ} / 30lb = \sin 40^{\circ} / F$
Solving for $F$, we obtain:
$F=19.6lb$
$\sin80^{\circ} / 30lb = \sin 60^{\circ} / F_b$
Solving for $F_b$, we obtain:
$F_b=26.4lb$
