Equilateral pentagon of δ can be seen from the figure: Simplifying, δ is obtained as function of α and β: The remaining angles of the pentagon can be found geometrically: The remaining angles of the orange and blue triangles are readily found by noting that two angles of an isosceles triangle are equal while all three angles sum to 180°. Then formula_7 and the two remaining angles of the green triangle can be found from four equations stating that the sum of the angles of the pentagon is 540°, the sum of the angles of the green triangle is 180°, the angle
Schwarz triangle function case the region in the sector between the radii "OA" and "OB" outside the edge "AB" is convex as the intersection of three convex regions. By induction the angles at "A" and "B" are greater than or equal to /3. Thus the geodesics to "C" from "A" and "B" start off in the region; by convexity, the triangle "ABC" lies wholly inside the region. The quadrilateral "OACB" has all its angles less than (since "OAB" is a geodesic triangle), so is convex. Hence the radius "OC" lies inside the angle of the triangle "ABC" near "C". Thus the angles between