We are given forces ${F_1} = 4\;{\rm{kN}}$, and ${F_2} = 6\;{\rm{kN}}$.

We are required to determine magnitude of forces in $u$ and $v$ direction.

The forces along with all the angles are shown below.

By the geometry, the angle $\alpha $ can be calculated as,

\begin{array}{c} \alpha = 180^\circ – 75^\circ – 30^\circ \\ = 75^\circ \end{array}

The angle $\beta $ can be calculated as,

\begin{array}{c} \beta = 180^\circ – 75^\circ – 30^\circ – 30^\circ \\ = 45^\circ \end{array}

Now, we shall make the triangle of forces ${F_2}$, ${F_v}$ and ${F_u}$.

Apply the sine rule for the forces ${F_2}$ and ${F_u}$.

\begin{array}{c} \frac{{{F_u}}}{{\sin \alpha }} = \frac{{{F_2}}}{{\sin \left( {180^\circ – \alpha – 30^\circ } \right)}}\\ {F_u} = \frac{{{F_2} \times \sin \alpha }}{{\sin \left( {180^\circ – \alpha – 30^\circ } \right)}} \end{array}

Substitute all the values in the above equation.

\begin{array}{c} {F_u} = \frac{{\left( {6\;{\rm{kN}}} \right) \times \sin 75^\circ }}{{\sin \left( {180^\circ – 75^\circ – 30^\circ } \right)}}\\ = 6\;{\rm{kN}} \end{array}

Apply the sine rule for the forces ${F_2}$ and ${F_v}$ .

\begin{array}{c} \frac{{{F_v}}}{{\sin 30^\circ }} = \frac{{{F_2}}}{{\sin \left( {180^\circ – \alpha – 30^\circ } \right)}}\\ {F_v} = \frac{{{F_2} \times \sin 30^\circ }}{{\sin \left( {180^\circ – \alpha } \right)}} \end{array}

Substitute all the values in the above equation.

\begin{array}{c} {F_v} = \frac{{\left( {6\;{\rm{kN}}} \right) \times \sin 30^\circ }}{{\sin \left( {180^\circ – 75^\circ – 30^\circ } \right)}}\\ = 3.11\;{\rm{kN}} \end{array}