Two forces 3N and 2 N are at an angle q such that the resultant is R. The first force is now increased to 6N and the resultant become 2R. The value of q is
120∘
A=3N,B=2NthenR=√A2+B2+2ABcosθR=√9+4+12cosθ
Now A =6 N, B =2 N then
2R=√36+4+24cosθ
From (i) and (ii) we get cosθ=−12∴θ=120∘