Angle made with the x-axis by a straight line drawn through (1,2) so that it intersects x + y=4 at a distance $\sqrt{6}/3$ from (1,2) is

Let the angle be $\theta$.Then, the equation of the given line is

$\frac{\mathrm{x}-1}{\cos \mathrm{\theta }}=\frac{\mathrm{y}-2}{\sin \mathrm{\theta }}-----\left(1\right)$

The coordinates of a point on (i) at a distance $\sqrt{6}/3$  from (1, 2)
are $(1+\sqrt{6}/3\cos \mathrm{\theta },2+\sqrt{6}/3\sin \mathrm{\theta })$. This point lies on x +y=4. Therefore, 

$\begin{array}{rl} & 1+\frac{\sqrt{6}}{3}\cos \mathrm{\theta }+2+\frac{\sqrt{6}}{3}\sin \mathrm{\theta }=4\\ \mathrm{or} & \cos \mathrm{\theta }+\sin \mathrm{\theta }=\sqrt{\frac{3}{2}}\\ \mathrm{or} & \frac{1}{\sqrt{2}}\cos \mathrm{\theta }+\frac{1}{\sqrt{2}}\sin \mathrm{\theta }=\frac{\sqrt{3}}{2}\\ \mathrm{or} & \cos (\mathrm{\theta }-\frac{\mathrm{\pi }}{4})=\cos (\pm \frac{\mathrm{\pi }}{6})\\ \mathrm{or} & \mathrm{\theta }-\frac{\mathrm{\pi }}{4}=\pm \frac{\mathrm{\pi }}{6}\\ \mathrm{i}.\mathrm{e}.& \mathrm{\theta }={75}^{\circ }\text{or}\mathrm{\theta }={15}^{\circ }\end{array}$