Suppose the 1 kg standard body accelerates at 4.00 m/s2
at 160(degree) from the positive direction of an x axis due to two forces; one is F1 - (2.50N)i + (4.60N)i What is the other
force (a) in unit-vector notation and as (b) a magnitude and
(c) an angle?

accleration = 4

it makes 160^o with +ve x axis so 20^o from -ve x axis ...

now resolve accleration into two rectangular components ,

a = 4cos20^o (-i) + 4sin20^o (j)

net force = ma = 4cos20(-i) + 4sin20(j)                           

(sin20^o similar to sin18^o = 0.3  & 

 cos20^o is similar to cos18^o = 0.67 )

F_net = -2.7(i) + 1.2 (j)           ...........1

force on particle (F_net)  = F₁ + F₂ 

 F₁ = 2.5(i) + 4.6 (j)  so

 F₂ =  -5.2(i) - 3.4 (j)          ............2             ans a)

magnitude = (sum of squares of cofficient of x comp & y comp)^1/2 = 6.2          (ansc)

unit vector = vector/magnitude

  unit vector = -(5.2i + 3.4j)/6.2 = -0.83(i) - 0.55(j)           (ans b)

approve if u like my ans

reso;ve the forces into components and separately apply the newton's laws of motion...