Defect of a hyperbolic triangle
Here, we have a triangle ABC in the Poincaré disk, with A = (0,0), B = (0,1/2), and C = (1/2,0). Since the hyperbolic line through B and C lies on a Euclidean circle (centered at X = (5/4,5/4) in this case), we can use Calculus to compute the angle ABC and thus the defect of the triangle.