Finding circumcenter In a triangle A(1,2) B(2,3) C(3,1) and $\angle A = cos^{-1}(4/5)$, $\angle B = \angle C = cos^{-1}(\frac{1}{\sqrt{10} }) $ Ordinate of circumcentre of the $\triangle$ is ? I have tried solving by finding the equation for line perpendicular to AC and BC and then finding the intersection point. The ordinate I found was 11/3, but the answer provided is 0. Can anyone pls solve this using the angles provided, and also tell me whether my method is correct or not? Thanks.

Your method is correct but the calculation may be off because it is $\frac{11}{6}$ . There is no way it can be $0$ though.

The angles were probably given because the circumcentre's Y co-ordinate can also be expressed as:

$$Y_c=\frac{y_1sin2A+y_2sin2B+y_3sin2C}{sin2A+sin2B+sin2C}$$

Where $y_1,y_2,y_3$ are the ordinates of the vertices and A,B,C are the angles at those vertices.