First we draw the free body diagram (FBD).
Since the particle is in equilibrium, the third force must be the equilibrant \(F _{3}\) having the same magnitude as the resultant of the two given forces but in the opposite direction.
$$
\begin{aligned}
R^{2} &=40^{2}+30^{2}, \\
R &=\sqrt{40^{2}+30^{2}}=50 \\
\theta &=\tan ^{-1}\left(\frac{F y}{F x}\right) \\
&=\tan ^{-1}\left(\frac{30}{40}\right)=36.870 \approx 37^{\circ} \\
\theta &=E 37^{\circ} N \text { or } 90-37=N 53^{\circ} E
\end{aligned}
$$