We have two forces is 3P and 2P, let angle between two forces is θ. ∴ Resultant R=√(3P)2+(2P)2+12P2cosθ =√P2(13+12cosθ) .......(i) Now first force is doubled. ∴ First force =6P and resultant is also double ∴R=2R ⇒2R=√(6P)2+(2P)2+24P2cosθ ⇒2R=√P2(40+24cosθ) .........(ii) From Eqs. (i) and (ii), 2P√13+12cosθ=P√40+24cosθ ⇒4(13+12cosθ)=40+24cosθ ⇒52+48cosθ=40+24cosθ ⇒24cosθ=−12 ⇒cosθ=−