The point ( x, the square root of 7/3) in the second quadrant corresponds to angle θ on the unit circle.

If (x, 7√3) is a point on the unit circle, then x2 + (7√3)2 = 1 and -1 ≤ x ≤ 1. So, x2 = 1- (7√3)2 = 1- 147 = -146 Since there is no value of x where -1 ≤ x ≤ 1 and x2 = -146, the given point is NOT on the unit circle.