You got pi/6 since you calculated the inside angle from doing tan^-1(y/x) but you didn't consider the region it is in. Its in the 2nd region so it is pi/6 above the negative x axis or 5 pi/6 from the positive axis. You would just do pi - pi/6 to find the angle of inclination in this case. So you would say formally it has an angle of 5pi/6.
You got pi/6 since you calculated the inside angle from doing tan^-1(y/x) but you didn't consider the region it is in. Its in the 2nd region so it is pi/6 above the negative x axis or 5 pi/6 from the positive axis. You would just do pi - pi/6 to find the angle of inclination in this case. So you would say formally it has an angle of 5pi/6.
When doing these question all you need to know is what quadrant it is in and the tan^-1(y/x)
ie.
first quadrant - tan^-1(y/x)
second quadrant - pi - tan^-1(y/x)
third quadrant - pi + tan^-1(y/x)
fourth quadrant - 2pi - tan^-1(y/x)