If A-B perpendicular to B and |A-B|=A/2 angle between A,B is

                                                B

                                               /

                                             / x degree

                                           /___________  A

                                         / 90\

                                       /         \  A/2

                                     / x           \

                                   /____________\              

                                -B        A         A-B

let the angle between A and B vector be x.

then we know A-B is perpendicular to B so its also perpendicular to -B.

hence  a triangle forms by -B , A-B , and A.in which angle between -B and A is x

and angle between A-B and -B is 90.

applying sine rule.

   |A|/sin 90= |A-B|/sin x

solving we get x=30 degree (ANS).

hope i helped you.

Approved

|A-B|=A/2 i.e. A²+B² - 2ABCos@ = A²/4------------>(1)

also, (A-B).B = 0 as angle bwtween is 90 (dot product |a.b|=|a||b|Cos@ )

ABCos@ = B²---------->(2)

on substituting in (1) from (2) we get A = 2/(3)^1/2B

again on substituing in (2) we get Cos@= (3)^1/2/2

hence @ = 30

Approved

A-B PERPENDICULAR TO B

LET ANGLE BETWEEN A AND B BE θ

THEN BSinθ/(A-BCosθ)=tan 90degree=1/0

that gives A-BCosθ=0

A=BCosθ

A/B=COSθ

and also

√A²+B²-2ABCOSθ=lA-Bl=A/2

√A²+B²-2AA=A/2                  (BCOSθ=A)

B²-A²=A²/4

4B²=5A²

A²/B²=COS² θ=4/5

COSθ=2/√5