#### Answer

The magnitude of the net force exerted on A is 7.3 N and it is directed at an angle of $79.2^{\circ}$ below the negative x-axis.

#### Work Step by Step

We can find the angle $\theta$ below the positive x-axis from A to D.
$tan(\theta) = \frac{100}{141}$
$\theta = arctan(\frac{100}{141})$
$\theta = 35.3^{\circ}$
We then find the horizontal component of the force exerted on A.
$F_x = F_B+F_{Dx}$
$F_x = -3.0~N+(2.0~N)~cos(35.3^{\circ})$
$F_x = -1.37~N$
We then find the vertical component of the force exerted on A.
$F_y = F_C+F_{Dx}$
$F_y = -6.0~N-(2.0~N)~sin(35.3^{\circ})$
$F_y = -7.16~N$
We then find the magnitude of the force F.
$F = \sqrt{(F_x)^2+(F_y)^2}$
$F = \sqrt{(-1.37~N)^2+(-7.16~N)^2}$
$F = 7.3~N$
We find the angle below the negative x-axis.
$tan(\theta) = \frac{7.16}{1.37}$
$\theta = arctan(\frac{7.16}{1.37})$
$\theta = 79.2^{\circ}$
The magnitude of the net force exerted on A is 7.3 N and it is directed at an angle of $79.2^{\circ}$ below the negative x-axis.