Let $A=(1,2)$ equation of line is

$\frac{x-1}{\cos \theta }=\frac{y-2}{\sin \theta }=r=\frac{\sqrt{6}}{3}$

$\therefore x=\frac{\sqrt{6}}{3}\cos \theta +1,y=\frac{\sqrt{6}}{3}\sin \theta +2$

This point lies on line $x+y=4$

$\therefore \frac{\sqrt{6}}{3}\cos \theta +\frac{\sqrt{6}}{3}\sin \theta =1$

$\Rightarrow \sin \theta +\cos \theta =\frac{\sqrt{3}}{\sqrt{2}}$

$\Rightarrow \sin \left(\frac{\pi }{4}+\theta \right)=\frac{\sqrt{3}}{2}$

$\Rightarrow \frac{\pi }{4}+\theta =60°$ or $120°$

$\Rightarrow \theta =15°$ or $75°$

$\therefore$ Larger angle $=75°$