Two vectors a and b have precisely equal magnitudes. in order for the magnitude of a+ b to be 110 times larger than the magnitude of a-b , what must be the angle between them? ,

Due to the magnitudes, as mentioned in the problem, being equal, B has components Acos(θ)x along the x-axis and Asin(θ) along the y-axis. 

The solution is as follows:

A+B has: 
total x-component = A + Acos(θ) = A(1+cos(θ)) 
total y-component = 0 + Asin(θ) = Asin(θ) 
Resultant = √[( A(1+cos(θ)))² + (Asin(θ))²] 
= A√[1 + 2cos(θ) + cos²(θ) + sin²(θ)] 
=A√[2 + 2cos(θ)] 

A-B has: 
total x-component = A - Acos(θ) = A(1-cos(θ)) 
total y-component = 0 - Asin(θ) = -Asin(θ) 
Resultant = √[( A(1-cos(θ)))² + (-Asin(θ))²] 
= A√[1 - 2cos(θ) + cos²(θ) + sin²(θ)] 
=A√[2 - 2cos(θ)] 

A+B/A-B = [A√[2 + 2cos(θ)]] / A√[2 - 2cos(θ)] 
= [√[2 + 2cos(θ)]] / [√[2 - 2cos(θ)]] 

110 = [√[1 + cos(θ)]] / [√[1 - cos(θ)]] 
12100 = [1 + cos(θ)] / [1 - cos(θ)] 

12100 - 12100cos(θ) = 1 + cos(θ)
12101cos(θ)] = 12099 
cos(θ)= 0.999834724
θ = 1.04º

i think the answer is a and b

i wish you luck