The question is in the attachment!!

Explanation:

|F|= B/2

B/2 = √(A²+B²+2ABcosθ) (1)

Since the resultant of A and B is perpendicular to Vector A

tan90°= BSinθ/(A+BCosθ)

(A+BCosθ)=0

Cosθ=-A/B (2)

Using equation (1)

B/2 = √(A²+B²+2ABcosθ)

B/2 = √(A²+B²+2AB×-A/B)

B/2=√(A²+B²-2A²)

B/2=√(B²-A²)

B²/4=B²-A²

A²=B²-B²/4

A²=3B²/4

A=√3B/2

Using equation (2)

Cosθ=-A/B

Cosθ=-[√3B/2]/B

Cosθ=-√3/2

θ= cos^-1 (-√3/2)

θ= 150°

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