given that a straight line passes through the point A(1,2) and makes an angle #theta# with the positive direction of x axes, we can write its equation as
#y-2=(x-1)tantheta ......[1]#
Solving this equation with equation
#x+y=4......[2]#
Subtracting [1] from[2]
#x+2=4-(x-1)tantheta#
#=>x+2=4-xtantheta+tantheta#
#=>x+xtantheta=2+tantheta#
#=>x=(2+tantheta)/(1+tantheta)#
So #y=4-x=4-(2+tantheta)/(1+tantheta)=(2+3tantheta)/(1+tantheta)#
So point of intersection of [1] and [2] is
#((2+tantheta)/(1+tantheta),(2+3tantheta)/(1+tantheta))#
Given that the distance of A from the point of intersection of this line with the line x+y=4 is #sqrt5/3# unit, we can write
#((2+tantheta)/(1+tantheta)-1)^2+((2+3tantheta)/(1+tantheta)-2)^2=(sqrt5/3)^2#
#=>1/(1+tantheta)^2+tan^2theta/(1+tantheta)^2=5/9#
#=>(1+tan^2theta)/(1+tantheta)^2=5/9#
#=>sec^2theta/(1+tantheta)^2=5/9#
#=>sectheta/(1+tantheta)=sqrt5/3#
#=>1/(costheta+sintheta)=sqrt5/3#
#=>costheta+sintheta=3/sqrt5#
#=>1/sqrt2costheta+1/sqrt2sintheta=3/sqrt10#
#=>cos(theta-45^@)=3/sqrt10=cos18.4^@#
#=>theta= 63.4^@#