Answer
$F_R=8.03kN$
$\phi=1.22^{\circ}$
Work Step by Step
We are asked to determine the magnitude of the force, $F_R$, and its direction, measured clockwise
from the positive u axis.
We can treat the forces like a triangle and use the law of cosines:
$c^2=a^2+b^2−2a∗b∗\cos C$
$c=\sqrt{a^2+b^2−2a∗b∗\cos C}$
$c=\sqrt{(4kN)^2+(6kN)^2−2∗4kN∗6kN∗\cos(105^{\circ})}$
$c=8.03kN$
$F_R=8.03kN$
Now we can use the law of sines to find the angle between $F$ and the x-axis.
$\sin\alpha/6kN=\sin105^{\circ}/8.03kN$
$\alpha=\arcsin(\sin105^{\circ}/8.03kN∗6kN)$
$\alpha=46.22^{\circ}$
We need to subtract 45 from our angle to get the angle as measure from the u-axis, $\phi$
$\phi=46.22^{\circ}-45^{\circ}=1.22^{\circ}$
