Explanation:
\(A=3 N, B=2 N\)
then \(R=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
\(R = \sqrt {9 + 4 + 12\cos \theta } \,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Now \(A=6 N, B=2 N\) then
\(2R = \sqrt {36 + 4 + 24\cos \theta } \,\,\,\,\,\,(2)\)
From (1) and (2) we get \(\cos \theta=-\dfrac{1}{2} \therefore \theta=120^{\circ}\)