Distance is not perpendicular

Use the parametric form of a straight line,
$\frac{x-1}{cos \theta}=\frac{y-2}{sin \theta}=\frac{\sqrt6}{3}$ .
By putting values of $x$ and $y$ in $x+y=4$
we get $\theta=15°$ or
$75°$ . The greater is $75°$ . So, answer is $\boxed{75°}$

x + y = 4 or y = 4 - x (eqn 1)

Consider a circle centered at ( 1 , 2 ) with radius = sqrt(6)/3. It's equation will be (x - 1)^2 + (y -2)^2 = 6/9. (eqn 2) and it touches eqn 1 at 2 points.

Substitute 4 - x for y in eqn 2, then expand.

x^2 - 2x + 1 + 4 - 4x +x^2 = 2 / 3

Solving the quadratic yields x = 1.7875 or 1.2125. Use the second value as that makes the line through (1,2) steeper and thus the bigger angle with x - axis.

y = 4 - 1.2125 or y = 2.7875 which yields a slope of that line (2.7875 -2 ) / ( 1.2125 - 1) = .7875 / .2125 = 3.7059

tan^-1(3.7059) = 74.899 degrees or 75 degrees