Good afternoon. Please provide the solution for the attachment

Let P and Q be the magnitude of vectors.P:Q=1:2 => 2P=QNow let a be the angle made by resultant vector R with vector P. And b be angle between P and Qa= tan–¹(√3/2)tan a = √3/2By vector formula tan a= Qsinb / (P + Qcosb)So √3/2= Qsinb / (P + Qcosb)Put Q = 2P√3/2= 2Psinb/ (P+2Pcosb)= 2sinb/ 1+2cosbSquaring both sides3/4 = 4 sin²b /( 1 + 4 cos²b + 4cosb)Cross multiply & Substitute sin²b=1- cos²b3+12cos²b+12cosb= 16(1-cos²b)Simplify into28 cos²b+ 12cosb -13=0cos b= {-12 ± √[144– 4×28×(-13)] } / 2×28 cos b= (-12 ± 40 )/ 2×28 cos b = -52/56 or 28/56cos b = -13/14 or 1/2b = cos ^-1( -13/14 or 1/2)The negative can be neglectedTherefore b = cos ^-1(1/2)b = 60° or [π/3]