By N I Lobachevskiĭ; Athanase Papadopoulos

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Y/ D p sin A sin ! cos2 A sin2 ! y/ D sin A p d! tan ! : cos2 A sin2 ! From this, integrating from ! D 0, which corresponds to x D 0, and noting that the area expressed by the integral is also equal to 2 A B, we deduce the following: Z ! d! A B D sin A p 2 0 cos2 A sin2 ! where A is a constant angle and where B is determined by the equation cos B D sin A : cos ! If ! D 2 , the hypotenuse becomes parallel to the edge y and the angle B becomes equal to zero. Thus we have in this case Z 2 A d! tan !

In this way, the aforementioned equations become sin A sin c D sin a; cos b sin A D cos B; cos a cos b D cos c: (5) Equations (5) concern a right spherical triangle that can be deduced from a right rectilinear triangle, and whose edges therefore cannot exceed 2 . Let us add that if we draw an arc of a great circle from the vertex of the angle A perpendicularly to the edge b, this arc will intersect the arc a or its extension in such a way that each of the arcs, from the intersection point until b, is equal to 2 , and the angle made by these arcs is b.

The perpendicular to ˇ drawn from the endpoint of ˇ is parallel to the extension of a. c C ˇ/ and which will thus be parallel to the extension of a. c C ˇ/ Figure 22. The quadrilateral used in the proof of the formula giving the area of a right triangle as angular deﬁcit. c C ˇ/, is the expression of the area of the triangle. ˇ/; that is, the area of a right rectilinear triangle is equal to the difference between two right angles and the sum of the three angles of the triangle. 51 This 51 In hyperbolic geometry, the area of a triangle is usually dened as the angular deﬁcit, that is, as the difference between two right angles and the angle sum of the triangle.