Answer
$F=960N$
$\theta=45.2^{\circ}$
Work Step by Step
We are given that in the figure, that the magnitude of the resultant force is 500N. We are asked to determine the magnitude of the force, $F$, and its direction, measured counterclockwise
from the positive x axis.
We can treat the forces like a triangle and use the law of cosines:
$c^2=a^2+b^2−2a∗b∗\cos C$
$c=\sqrt{a^2+b^2−2a∗b∗\cos C}$
$c=\sqrt{(500N)^2+(700N)^2−2∗700N∗500N∗\cos(90^{\circ}+15^{\circ})}$
$c=\sqrt{(500N)^2+(700N)^2−2∗700N∗500N∗\cos(105^{\circ})}$
$c=960N$
$F=960N$
Now we can use the law of sines to find the angle between $F$ and the x-axis.
$\sin\alpha/700N=\sin105^{\circ}/960N$
$\alpha=\arcsin(\sin105^{\circ}/960N∗700N)$
$\alpha=44.8^{\circ}$
This is measured from the y-axis. We need to subtract this from 90 to get the angle from the x-axis, $\theta$.
$\theta=90^{\circ}-44.8^{\circ}=45.2^{\circ}$
