## Engineering Mechanics: Statics & Dynamics (14th Edition)

$\theta=54.9^{\circ}$ $F=10.4kN$
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, $F_R$. $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{8^2+6^2-2*8*6*\cos(94.93^{\circ})}$ $c=10.4$ $F=10.4kN$