The pole of a circle of the sphere is a point in the superficies of the sphere, from which all straight lines drawn to the circumference of the circle are equal. II. through the centre of the sphere, and whose centre III. a sphere comprehended by three arches of three IV. sphere is contained by two arches of great circles, PROP. I. As they have a common centre, their common section will be a diameter of each which will bisect them. PROP. II. Fig. I. The arch of a great circle betwixt the pole and the circumference of another is a quadrant. Let ABC be a great circle, and D its pole; if a great circle DC pass through D, and meet ABC in C, the arch DC will be a quadrant. Let the great circle CD meet ABC again in A, and let AC be the common section of the great circles, which will pass through E, the centre of the sphere: join DE, DA, DC: by def. 1. DA, DC are equal, and AE, EC are also equal, and DE is common; there fore (8. 1.) the angles DEA, DEC are equal ; wherefore the arches DA, DC are equal, and consequently each of them is a quadrant. Q. E. D. bej 18 Banii iugistia PROP. III. FIG. 2. Krior boobs If a great circle be described meeting two great circles AB, AC passing through its pole A in B, C, the angle at the centre of the sphere upon the circumference BC, is the same with the spherical angle BAC, anch the arch BC is called the measure of the spherical an gle BAC. Q. E. D. Let the planes of the great circles AB, AC, intersect one another in the straight line AD, passing through D their common centre: join DB, DC. Since A is the pole of BC, AB, AC, will be quadrants, and the angles ADB, ADC right angles; therefore (6 def. 11.) the angle CDB is the inclination of the planes of the circles AB, AC; this is (def. 4.), the spherical angle BAC. Cor. If through the point A, two quadrants AB, AC, be drawn, the point A, will be the pole of the great circle BC, passing through their extremities B, C. Join AC, and draw AE, a straight line to any other point E, in BC; join DE: since AC, AB are quadrants, the angles ADB, ADC are right angles, and AD will be perpendicular to the plane of BČ: therefore the angle ADE is a right angle, and AD, DC are equal to AD, DE, each to each ; therefore AE, AC are equal, and A is the pole of BC, by def. 1. In isosceles spherical triangles, the angles at the base are equal. Let ABC be an isosceles triangle, and AC, CB, the equal sides; the angles BAC, ABC, at the base AB, are equal. Let D be the centre of the sphere, and join DA, DB, DC; in DA take any point E, from which draw, in the plane ADC, the straight line EF at right angles to ED, meeting CD in F, and draw in the plane ADB, EG at right angles to the same ED; therefore the rectilineal angle FEG is (6. def. 11.) the inclination of the planes ADC, ADB, and therefore is the same with the aspberical angle BAC: from F draw FH perpendicular to DB, and from H draw, in the plane ADB, the straight line HG at right angles to HD, meeting EG in G, and join GF. Because DE is at right angles to EF and EG, it is perpendicular to the plane FEG (4. 11.), and therefore the plane FEG is perpendicular to the plane ADB, in which DE is (18. 11.): in the same manner, the plane FHG is perpendicular to the plane ADB; and therefore GF, the common section of The planes FEG, FHG, is perpendicular to the plane ADB (19. 11.); and because the angle FHG is the inclination of the planes BDC, BDA, it is the same with the spherical angle ABC; and the sides AC, CB of the spherical triangle being equal, the angles EDF, HDF, which stand upon them at the centre of the sphere, are equal; and in the triangles EDF, HDF, the side DF is common, and the angles DEF, DHF, are right angles; therefore EF, FH are equal : and in the triangles FEG, FHG the side GF is common, and the sides EG, GH, will be equal by the 47. 1. and therefore the angle FEG is equal to FHG (8. 1.); that is, the spherical angle BAC is equal to the spherical angle ABC. PROP. V. Fig. 3. If, in a spherical triangle ABC, two of the angles BAC, ABC be equal, the sides BC, AC opposite to them are equal. Read the construction and demonstration of the preceding proposition, unto the words, 6 and the sides “ AC, CB”, &c. and the rest of the demonstration will be as follows, viz. And the spherical angles BAC, ABC being equal, the rectilineal angles FEG, FHG, which are the same with them, are equal; and in the triangles FGE, FGH, per the angles at G are right angles, and the side FG op posite to two of the equal angles is common; therefore (26. 1.) EF is equal to FH: and in the right angled mu triangles DEF, DHF, the side DF is common; whereefore (47. 1.) ED is equal to DH, and the angles EDF, HDF are therefore equal (4. 1.), and consequently the mis sides AC, BC of the spherical triangle are equal. gesinsg soda 30 storia Sonats a boa : 4 l PROP. VI. Fig. 4. marts at 10 Any two sides of a spherical triangle are greater than the third. em boos toinen Let ABC be a spherical triangle, any two side BC AB, BC will be greater than the other side AOLA 11 Let D be the centre of the sphere: join DA, DB, D : DC. The solid angle at D is contained by three plane 1. and two of them ADB, BDC, are greater than the third ADC, that is, any two sides AB, BC, of the spherical triangle ABC, are greater than the third AC. PROP. VII. FIG. 4. Detta go The three sides of a spherical triangle are less than a circle. Let ABC be a spherical triangle as before, the three sides AB, BC, AC, are less than a circle. Let D be the centre of the sphere: the solid angle at D'is contained by three plane angles BDA, BDC, ADC, which together are less than four right angles (21. 11.); therefore the sides AB, BC, AC, together, will be less than four quadrants, that is, less than a circle. In a spherical triangle, the greater angle is opposite to the greater side, and conversely. Let ABC be a spherical triangle, the greater angle A is opposite to the greater side BC, Let the angle BAD be made equal to the angle B, and then BD, DA will be equal (5. of this), and therefore AD, DC are equal to BC; but AD, DC are greater than AC (6. of this), therefore BC is greater than AC, that is, the greater angle A is opposite to the greater side BC. The converse is demonstrated as't Prop. 19. 1. El. 2. E. D. at een law an PROP. IX. Fig. 6. A lo slog In any spherical triangle ABC, if the sum of the sides AB, BC, be greater, equal, or less than a semicircle, the internal angle at the base AC will be greater, equal, or less than the external and opposite BCD; and therefore the sum of the angles A and ACB will be greater, equal, or less than two right angles. Let AC, AB produced meet in D. 1. If AB, BC be equal to a semicircle, that is, to AD, BC, BD will be equal, that is (4. of this), the angle D, or the angle A, will be equal to the angle BCD. Per 2. If AB, BC together be greater than a semicircle, that is, greater than ABD, BC will be greater than BD; and therefore (8. of this), the angle D, that is, the angle A, is greater than the angle BCD. 3. In the same manner it is shewn that if AB, BC together be less than a semicircle, the angle A is less than the angle BCD. And since the angles BCD, BCA are equal to two right angles, if the angle A be greater than BCD, A and ACB together will be greater than two right angles. If A be equal to BCD, A and ACB together will be equal to two right angles; and if A be less than BCD, A and ACB will be less than two right angles. Q. E. D. PROP. X. Fig. 7. If the angular points, A, B, C, of the spherical triangle ABC be the poles of three great circles, these great circles by their intersections will form another triangle FDE, which is called supplemental to the former ; that is, the sides FD, DE, EF are the supplements of the measures of the opposite angles C, B, A, of the triangle ABC, and the measures of the angles F, D, E, of the triangle FDE, will be the supplements of the sides AC, BC, BA, in the triangle ABC. Let AB produced meet DE, EF, in G, M, and AC meet FD, FE, in K, L, and BC meet FD, DE in N, H. Since A is the pole of FE, and the circle AC passes through A, EF will pass through the pole of AC (2. of this), and since AC passes through C, the pole of FD, FD will pass through the pole of AC; therefore the pole of AC is in the point F, in which the arches DF, EF, intersect each other. In the same manner D is the non of BC and E the not |