The resultant of the vectors A and B is perpendicular to the vector A and it`s magnitude is equal to half the magnitude of vector B the angle between A and B is

Let the angle between the two vectors A & B be x And angle between A vector And resultant vector be yTherefore ,tany=Bsinx/A+BcosxBut y=90 giventan90=Bsinx/A+BcosxBut tan90= infinity =1/01/0=Bsinx/A+BcosxCross multiply A+Bcosx=0cosx=-A/BBut |A+B|=B/2 given √A×A+B×B+2ABcosx=B/2Thus,√A×A+B×B-2AB(-A/B)=B/2√A×A+B×B-2A×A=B/2Squaring both sides B×B-A×A=B×B/2A×A=3/4B×BTaking square rootA=B√3/2Cosx = -A/B=-B√3/2÷B=-√3/2Cosx =-√3/2Therefore, x=150Thus angle between A&B vectors is 150