The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B.
Then the angle between A and B is

Explanation

Given : `theta = 90^(@) , r = (b)/(2)`
Formula : `r^(2) = (a^(2) +b^(2) + 2ab cos phi) " ". . . (1)`
`tan theta = (b sin phi)/(a + b cos phi)" " . . . (20`
Since, `theta = 90^(@) , tan 90^(@) = infty`
i.e., `a + b cos phi = 0 (or) a = - b cos phi " ". . . (3)`
(3) in (1) ` r^(2) = (( - b cos phi)^(2) + b^(2) + 2 (-b cos phi) b cos phi )`
`= b^(2) - b^(2) cos^(2) phi`
`= b^(2) (1 - cos^(2) phi)`
`r^(2) = b^(2) sin^(2) phi`
`sin phi = pm (r)/(b) = pm (b)/(2b) = pm (1)/(2)`
`phi = 30^(@) (or) 150^(@)`
`phi` cannot be `30^(@)` as, resultant makes an angle `90^(@) ` with a .
`:. a and b is 150^(@)`
`:. Ans: (c) 150^(@)`