#### Answer

The magnitude of $F$ is 90.2 N.
The direction of the force is $75.6^{\circ}$ south of west.

#### Work Step by Step

We can find the horizontal component $F_x$ of the fourth force.
$F_x = 100.0~cos(30.0^{\circ}) - 80.0~sin(30.0^{\circ}) - 40.0~cos(53^{\circ}) = 0$
$F_x = -86.6~N + 40.0~N + 24.1~N = -22.5~N$
We can find the vertical component $F_y$ of the fourth force.
$F_y = 100.0~sin(30.0^{\circ}) + 80.0~cos(30.0^{\circ}) - 40.0~sin(53^{\circ}) = 0$
$F_y = -50.0~N -69.3~N + 31.9~N = -87.4~N$
We can use $F_x$ and $F_y$ to find the magnitude of $F$.
$F = \sqrt{(-22.5~N)^2+(-87.4~N)^2} = 90.2~N$
We can find the angle $\theta$ south of west.
$tan(\theta) = \frac{F_y}{F_x}$
$\theta = tan^{-1}(\frac{87.4}{22.5}) = 75.6^{\circ}$