Answer
Component along $u$ axis is $21.96\;lb$
and component along $v$ axis is $15.53\;lb$
Work Step by Step
The magnitude of force $\vec F$ is given: $F=30\;lb$
Let $F_u$ and $F_v$ are tow components of $\vec F$ along the $u$ and $v$ axes.
Using the law of sines in the triangle formed by $F$, $F_u$, and $F_v$, we get
$\frac{F_u}{\sin45^\circ}=\frac{F_v}{\sin30^\circ}=\frac{F}{\sin105^\circ}$
$\therefore\;F_u=\frac{F}{\sin105^\circ}\times\sin45^\circ$
or, $F_u=\frac{30}{\sin105^\circ}\times\sin45^\circ\;lb\approx 21.96\;lb$
and $;F_v=\frac{F}{\sin105^\circ}\times\sin30^\circ$
or, $F_v=\frac{30}{\sin105^\circ}\times\sin30^\circ\;lb\approx 15.53\;lb$
