Answer
Component along $AB=$ $869.33\;lb$
and component along $AC=$ $636.40\;lb$
Work Step by Step
The magnitude of force $\vec F$ is given: $F=450\;lb$
Let $F_{AB}$ and $F_{AC}$ are tow components of $\vec F$ along the members $AB$ and $AC$ axes.
Using the law of sines in the triangle formed by $F$, $F_{AB}$, and $F_{AC}$, we get
$\frac{F_{AB}}{\sin105^\circ}=\frac{F_{AC}}{\sin45^\circ}=\frac{F}{\sin30^\circ}$
$\therefore\;F_{AB}=\frac{F}{\sin30^\circ}\times\sin105^\circ$
or, $F_{AB}=\frac{450}{\sin30^\circ}\times\sin105^\circ\;lb\approx 869.33\;lb$
and $F_{AC}=\frac{450}{\sin30^\circ}\times\sin45^\circ$
or, $F_{AC}=\frac{450}{\sin30^\circ}\times\sin45^\circ\;lb\approx 636.40\;lb$
