Chapter 2 - Force Vectors - Section 2.3 - Vector Addition of Forces - Problems - Page 32: 29

$F_{B} = 1.61 KN$ $\theta = 38.3°$

By applying the cosines law $F_{B} = \sqrt (2^{2} + 3^{2} - 2 \times 2 \times 3 \times \cos(30°)) = 1.615 KN = 1.61 KN$ Now, applying the sines law, we obtain $\frac{\sin(\theta)}{2} = \frac{\sin(30°)}{1.615}$ $\theta = 38.3°$