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:
$c^2=a^2+b^2−2a∗b∗\cos C$
$c=\sqrt{a^2+b^2−2a∗b∗\cos C}$
$c=\sqrt{(700N)^2+(450N)^2−2∗700N∗450N∗\cos(60^{\circ}−15^{\circ})}$
$c=497N$
$F_R=497N$
Now we can use the law of sines to find the angle between $F$ and $F_r$, $\alpha$
$\sin\alpha*700N=\sin45^{\circ}497N$
$\alpha=\arcsin(sin45^{\circ}497N∗700N)$
$\alpha=95.19^{\circ}$
We will have to add this to θ to get the angle from the x-axis
$ϕ=\alpha+\theta=95.19^{\circ}+60^{\circ}=155^{\circ}$
