The statement is represented diagrammatically as;
Using cosine rule, we have
\(R 2=\overline{A^{2}}+\overline{B^{2}}-2 \overrightarrow{A B} \cos \theta\)
\(18^{2}=15^{2}+B^{2}-2 \times 15 \times B \cos 120\)
\(18^{2}-15^{2}=B^{2}-30 B \times-\frac{1}{2}\)
\(99=B^{2}+15 B\)
\(B^{2}+15 B-99=0\)
Using the formula method
\begin{aligned}
&B=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
&=\frac{-15 \pm \sqrt{15^{2}-4 \times(-99)}}{2 \times 1}
\end{aligned}
\begin{aligned}
&=\frac{-15 \pm \sqrt{225+396}}{2} \\
&B=\frac{-15 \pm \sqrt{621}}{2} \\
&B=\frac{-15 \pm 24.92}{2}= \\
&B=\frac{-15+24.92}{2}=\frac{9.92}{2} \\
&\therefore B=4.96 N
\end{aligned}