Assuming the observer is stationary, in 532 minutes (9h2m-10m) the star's GHA will have changed by 133 degrees. The star's GP will have changed by 11.8 degrees. A spherical triangle with vertices at the pole and the two GPs of the star is solvable because you know it's isosceles. Leaving its calculation as an exercise for you, the answer is 83 34.7 North or South.
At this point (2 days later) OP must have heard a complete answer from whomever asked. Analysis in last paragraph is from spherical law of cosines. $\cos(a) = \cos(b)\cos(c) + \sin(b)\sin(c)\cos(A)$. Here, upper case is for angles, and lower case is for sides opposite angles as arcangles. For this question the only known angle is the angle at the pole, ($A$), and it's 133 degrees. The only known side, ($a$) is the one opposite: 11.8 degrees. The star's declination and the observer's latitude are the same, since the star passed through (presumably exact) zenith. Neither OP's latitude or star's will change. So law of cosines for this case becomes $\cos^2(a) = \cos^2(b) + \sin^2(b)\cos(A)$$\cos(a) = \cos^2(b) + \sin^2(b)\cos(A)$. Remember $\sin^2(x) + \cos^2(x) = 1$. If there's any interest, I can show the steps to the answer.
Triangle is small enough that solving as if it were planar will give good approx results.
