From Mechanics

From the figure the three known forces can be calculated using,

\displaystyle \begin{align} \mathbf{F} =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}

We get

\displaystyle \begin{align} \mathbf{F}1 =131\mathbf{i}+47\textrm{.}9 \mathbf{j} \end{align}

\displaystyle \begin{align} \mathbf{F}2 =150 \mathbf{j} \end{align}

\displaystyle \begin{align} \mathbf{F}3 =161\mathbf{i}-135\mathbf{j} \end{align}

If the unknown force is \displaystyle \mathbf{F}4

the fact that the forces are in equilibrium gives

\displaystyle \mathbf{F}1+\mathbf{F}2+\mathbf{F}3+\mathbf{F}4=0

or

\displaystyle \mathbf{F}4=-\mathbf{F}1-\mathbf{F}2-\mathbf{F}3=-292\mathbf{i}-62\mathbf{j}\ \text{N}