by parallelogram law of vector
Using cosine rule
$$
\begin{aligned}
&R^{2}=20^{2}+30^{2}-2(2)(30) \cos 120^{\circ} \\
&R^{2}=400+900-1200 \times\left(-\frac{1}{2}\right) \\
&R^{2}=1300+600 \\
&R^{2}=1900 \\
&R=\pm \sqrt{1900}=\pm 43.59 N
\end{aligned}
$$
To find the angle \((\theta)\) the resultant makes with the \(30 N\) cable (see figure C)
Using sine rule,
$$
\begin{aligned}
\frac{\sin \theta}{20} &=\frac{\sin 120}{43-59} \\
\theta &=\sin ^{-1}\left[\frac{\sin 120}{43.59} \times 20\right] \\
\theta &=\sin ^{-1}[0.3974] \\
\theta &=23.4^{\circ}
\end{aligned}
$$