The problem you're dealing with seems to involve finding the angle $\theta$ required to make the resultant force act along the positive $u$-axis, with a given magnitude of 3.8 kN. You already have the magnitude of $F_B$ as 1.92 kN from Part A. In Part B, you're attempting to determine the angle $\theta$.

Approach:

1. Force Components:

   • $F_A = 3 \, \text{kN}$ acts horizontally along the $x$-axis.
   • $F_B = 1.92 \, \text{kN}$ acts at an angle $\theta$ to the horizontal (from the figure).

2. Resultant Force along the $u$-axis: The resultant force is the vector sum of $F_A$ and $F_B$, and it's required to lie along the positive $u$-axis with a magnitude of 3.8 kN.

3. Resolve the Forces:

   • Decompose both forces into their components along the $x$-axis and $y$-axis:
     • $F_A$ is purely horizontal: $F_A = (3 \, \text{kN}, 0)$.
     • $F_B$ will have both $x$- and $y$-components depending on $\theta$: $F_{Bx} = F_B \cos(\theta) = 1.92 \cos(\theta)$ $F_{By} = F_B \sin(\theta) = 1.92 \sin(\theta)$

4. Resultant Force: Since the resultant force is along the $u$-axis, its components along the $x$- and $y$-axes should satisfy the following:

   • The total force in the $x$-direction equals the component along $u$-axis: $F_{A} + F_{Bx} = 3.8 \cos(30^\circ)$
   • The total force in the $y$-direction should equal the component along $u$-axis: $F_{By} = 3.8 \sin(30^\circ)$

5. Solve for $\theta$: Using these two conditions, solve for $\theta$. If you attempt the equation for the resultant components and solve algebraically, you'll arrive at the required angle.

Would you like me to provide the full solution for $\theta$ based on this approach?