#### Answer

$F_{BC}=434lb$
$\phi=33.5^{\circ}$

#### Work Step by Step

We are asked to determine the force in member BC and the required angle $\phi$
Given that the force on member AB is 650lb and the force F is 850lb, we can apply the law of cosines to find force in member BC, $F_{BC}$
$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$
$c=\sqrt{850^2+650^2-2*850*650*\cos(30^{\circ})}$
$c=433.64$
$F_{BC}=434lb$
Using this result, we can apply the law of sines to find the$ \phi$
$\sin(45^{\circ}+\phi) / 850 = \sin 30^{\circ} / 433.64$
Solving for $\phi$, we obtain:
$\phi=33.5^{\circ}$