#### Answer

$\theta=54.9^{\circ}$
$F=10.4kN$

#### Work Step by Step

We can apply the law of sines to find the angle between the y-axis and $F_A$, ($\theta$)
$\sin(90^{\circ}-\theta) / 6 = \sin 50^{\circ} / 8$
Solving for $\theta$, we obtain:
$\theta=54.9^{\circ}$
By summing the forces and creating a triangle, we can solve for the missing angle in the triangle knowing that the whole triangle is $180^{\circ}$
$\phi=180^{\circ}-(90^{\circ}-54.9^{\circ})=94.93^{\circ}$
Now we can apply the law of cosines to find the magnitude of the resultant force, $F_R$.
$c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$
$c=\sqrt{8^2+6^2-2*8*6*\cos(94.93^{\circ})}$
$c=10.4$
$F=10.4kN$