Answer
$F_R=497N$
$\phi=155^{\circ}$
Work Step by Step
We are given that in the figure, $\theta =60^{\circ}$ and $F=450N$. We are asked to determine the magnitude of the resultant force and its direction, measured counterclockwise
from the positive x axis.
We can treat the forces like a triangle and use the law of cosines:
$c^2=a^2+b^2-2a*b*cosC$
$c=\sqrt{a^2+b^2-2a*b*cosC}$
$c=\sqrt{(700N)^2+(450N)^2-2*700N*450N*cos(60^{\circ}-15^{\circ}}$
$c=497N$
$F_R=497N$
Now we can use the law of sines to find the angle between $F$ and $F_r$, $\alpha$
$\frac{\sin \alpha}{700N}=\frac{\sin 45^{\circ} }{497N}$
$\alpha=\arcsin (\frac{\sin 45^{\circ} }{497N}*700N)$
$\alpha=95.19^{\circ}$
We will have to add this to $\theta$ to get the angle from the x-axis
$\phi = \alpha +\theta= 95.19^{\circ} +60^{\circ}=155^{\circ}$
