shall be equal, each to each, viz. those to Prop. 13) be two spherical triangles havwhich the equal angles are opposite, and ing the two sides of the one equal to the the third angle of the one shall be equal two sides of the other, each to each, and to the third angle of the other. likewise the base B C equal to the base The case in which the equal angles EF. The angle BAC shall be equal lie in the same direction from one ano- to the angle EDF. ther in the two triangles may be demon- Let there be described, as in Prop. 13, strated by superposition, in the same upon the other side of the base B C thé manner as Book I. Prop. 5, to which triangle A' B C, which has its sides and the reader is referred. The other case, angles equal to those of the triangle in which they lie in different directions, ABC, each to each, but in a reverse may be demonstrated as follows:- order. Then, because A' B and D E are Let ABC, DEF (see the figure of each of them equal to AB, they are Prop. 13.) be two spherical triangles, equal to one another: and, for the like which have the two angles A B C, reason, A'C is equal to DF: also BC ACB of the one equal to the two is equal to E F; and these equal parts angles D E F, DFE of the other, each lie in the same direction from one anoto each, but in a reverse order, and like ther in the two triangles A' B C D E F. .wise the side B C equal to the side E F: Therefore, by the first case, the angle their other sides shall be equal each to BAC is equal to the angle EDF. each, and the third angle B AC shall Therefore, because the angle B A C is be equal to the third angle EDF. equal to the angle B A Č, the angle Let there be described, as in the last BAC is likewise equal to the angle proposition, upon the other side of the EDF. base BC the triangle A' BC, which has Therefore, &c. its sides and angles equal to those of the triangle A BC, each to each, but in a PROP. 16. reverse order. Then, because the angles A' B C and D E F are equal, each If two spherical triangles have the of them, to the angle A BC, they are three angles of the one equal to the three equal to one another; and, for the like angles of the other, each to each, they reason the angle A' C B is equal to the shall likewise have the three sides of the angle DFE. Also, BC is equal to E F; one equal to the three sides of the other, and these equal parts lie in the same each to each, viz. those which are oppodirection from one another in the two site to the equal angles.* triangles A'BC, DEF. Therefore, by Let the spherical triangles A B C, the first case, the angle B A' C is equal D E F have the three angles A, B, C of to the angle E D F, and the sides A' B, the one equal to the three angles D, E, F A'C are equal to the sides D E, D F respectively. And, because the angle D' BAC is equal to the angle B A'C, and the sides A B, AC to the sides A'B, A'C, each to each, the angle B A C is equal to the angle E DF, and the sides B E F AB, AC to the sides DE,DF respectively. Therefore, &c. B' E of the other, each to each. The sides AB, If two spherical triangles have two sides of the one equal to two sides of the sides DE, DF and EF, each to AC and B C shall likewise be equal to the other, each to each, and have like each. wise their bases equal, the angle con. tained by the two sides of the one shall D'E' É' be described, they will have the For, if the polar triangles A'B'C', be equal to the angle contained by the three sides of the one equal to the three two sides equal to them of the other. in the same direction from one another, and D' E' are (7.) supplements of the The case in which the equal sides lie sides of the other, each to each, because every two corresponding sides, as A' B' may be demonstrated in the same man measures of equal angles C and F (I. ner as Book I. Prop. 7: the other case, in which they lie in different directions, tion, these polar triangles have likewise ax. 3.). Therefore, by the last proposias follows: Let ABC, DEF (see the figure of # See the Scholium at the end of this Book, D the three angles A', B', C' of the one DPB is greater than D C. Again, equal to the three angles D', E', F' of because the triangles PCD, PED have the other, each to each. And hence the the two sides PC, PD of the one equal triangles ABC, DEF have the three to the two sides PE, PD of the other, sides of the one equal to the three sides each to each, but the angle CPD greater of the other, each to each, because every than the ang!e EPD, the base DC is two corresponding sides, as A B and greater than the base DE (17.). Again, DE are supplements of the measures of because the side ED of the triangle equal angles C' and F' (7.) PDE is greater than the difference of Therefore, &c. PD, PE the other two sides (9.), and that P A is equal to PE, the arc E D PROP. 17. is greater than the difference of PD and If two spherical triangles have two PA, that is, than AD. Therefore, DPB sides of the one equal to two sides of the is the greatest and D A the least of all other, each to each, but the angle which circumference, and D C, which is nearer arcs which can be drawn from D to the is contained by the two sides of the one greater than the angle which is contained to D P B, is greater than D Е, which is by the two sides which are equal to them more remote. of the other ; the base of that which has Also, there can be drawn only two the greater angle shall be greater than equal arcs DE, D F from the point D the base of the other : and conversely. to the circumference, and these make equal angles with the shortest arc D A See the demonstration of Book I. upon either side of it. For, if PF be Prop. 11. drawn, making the angle A PF equal Prop. 18. to the angle A PE, and D F be joined ; then, because the triangles PDF, PDE If any point be taken within a circle have two sides of the one equal to two of the sphere which is not its pole; of all sides of the other, each to each, and the arcs which can be drawn from that the included angles D PF, DPE equal point to the circumference, the greatest is to one another, the bases D F and D E that in which the pole is, and the other are equal to one another, and the angles part of that arc produced is the least; and PDF, PD E which are adjacent to of any others, that which is nearer to the ADF, ADE, likewise equal (13.). greatest is always greater than one more But, besides D F, there cannot be drawn remote : and from the same point to the any other arc from D to the circumfercircumference there can be drawn only ence equal to D E; for if it were postwo arcs that are equal to one another, sible, and D G were such an arc, then which two make equal angles, upon either DG, which is nearer to (or farther from) side, with the shortest arc. DP B than D F is, would be equal toDF, Let A B C be any which is contrary to what was shown in circle (either a great E. the former part of the proposition. or a small circle) Therefore, &c. of the sphere, P its Cor. 1. From a point to a spherical B pole (if a great cir arc, the perpendicular is either the least cle, either of its two or the greatest distance, the least when poles; if a small it is less than a quadrant, the greatest circle, the nearer when it is greater than a quadrant. pole), and D a point on the surface of For the perpendicular arc either passes, the sphere, which is upon the same or may be produced to pass through side of the circle A B C D with the pole the pole of the spherical arc (5. Cor. 1.): P. Of all the arcs D A, D PB, DC, &c. in the former case it is, as DP B, which can be drawn from D to the cir- greater than any other arc which can be cumference, the arc D PB which passes drawn from the point E to the circumthrough P shall be the greatest, and ference of which the spherical arc is a DA the other part of that arc produced portion, and it is, at the same time, the least: and of the others, DC, which greater than (5.) the quadrant PB; in the is nearer to D P B, shall be greater than latter case it is, as D A, less than any DE, which is more remote. other, and it is, at the same time, less Join PE, PC. Then, because the two than (5.) the quadrant P A. sides DP, P C of the triangle D PC are Cor. 2. If there be taken within a together greater than the third D C (9.), circle (that is, anywhere on the surface and that P B is equal to PC, the arc of the sphere, if it be a great circle, or 0 с G a on the same side with the nearer pole, then, A B being (18. if a small circle) a point from which Cor. 1.) the least arc there fall more than two equal arcs to which can be drawn the circumference, that point shall be from A to B C, AC a pole of the circle. which is nearer to it Cor. 3. Hence, also, if there be taken is less than AD, B without a small circle a point from which is more rewhich there fall more than two equal mote (18.); or, searcs to the circumference, that point condly, if AB and BC shall be the more distant pole of the be both greater than small circle. For if the opposite point quadrants, then A B through which the equal arcs pass being (18.Cor. 1.) the (4 Cor. 2.) be taken, three distances greatest arc which B of the latter from the circumference can be drawn from (1. ax. 3.) will be equal, and therefore, by A to BC, AC which is further from it is Cor. 2., it will be the nearer pole, less than AD, which is not so remote (18.); therefore, in both cases, A C is less than PROP. 19. a quadrant. But, if either of the sides, as AB, be greater than a quadrant, and In a spherical triangle, which has the other, B C, less, AB being the one of its angles a right angle, either greatest arc which can be drawn from A of the other angles is greater or less to B C, A C which is nearer to it is than a right angle, according as the greater than A D, which is more remote opposite side is greater or less than a (18.); therefore, A C is greater than a quadrant; and if the sides, which con quadrant. tain the right angle, are both greater or Therefore, &c. both less than quadrants, the third side, Cor. 1. It is evident from the demonor hypotenuse, shall be less than a quad- stration, that if one of the two sides rant'; but if one of these sides be greater which contain the right angle be equal than a quadrant, and the other less, the to a quadrant, the opposite angle will hypotenuse shall be greater than a quad- be a right angle, and the hypotenuse rant. likewise a quadrant. Let A B C be a spheri Cor. 2. Hence, in a right-angled cal triangle, having the sphericai triangle, either of the two sides right angle A B C, and let is of the same affection with the opposite ACB be one of the other angle; and the hypotenuse is, 1°, equal two angles: the angle to, or, 2°, less than, or, 3o, greater than, ACB shall be greater or a quadrant, according as, 1°, one of the less than a right angle, adjoining angles is a right angle, or B according as the opposite these two angles are, 2°, of the same or, side A B is greater or less than a quad- 3°, of different affections. See note, rant. For, if from BA, or from BA produced, BP be cut off equal to a quadrant, and SECTION 3. — Of Equal Portions of PC joined, the angle PC B will be a right Spherical Surface, and the Measure angle, because P is the pole of B C (5.); of Solid Angles. and it is evident that the angle ACB is greater or less than the angle PCB, PROP. 20. according as A B is greater or less than PB. Spherical triangles which have the Also, if the sides A B, B C are both three sides of the one equal to the three greater or both less than quadrants, the sides of the other, each to each, contain hypotenuse A C shall be less than a also equal portions of spherical surface, quadrant; but if one of them be greater whether the triangles are symmetrical than a quadrant, and the other less, or otherwise. A C shall be greater than a quadrant. Let A B C D E F be two spherical For, if from B C, or from B C pro- triangles, which have the three sides of duced, B D be cut off equal to a quad- the one equal to the three sides of the rant, and AD be joined, D will be the other, each to each, viz. AB to DE, pole of A B (5.), and, therefore, A D will AC to D F, and B C to EF: the surbe equal to a quadrant. And, first, if AB face of the triangle ABC shall be equal and B C be both less than quadrants, to the surface of the triangle D EF. page 61. the former three, if the point P falls within the triangle ABC; or to the difference between the sum of two of them and the third, if it fall without the triangle; and the triangle D E F is equal to the sum of the latter three, in the first case, or to the difference between the sum of the corresponding two of them and the third, in the other*. Therefore, the triangle ABC is equal to the triangle DEF (I. ax. 2. 3.). Therefore, &c. F B Let a plane be drawn through the PROP. 21. points A, B, C, to cut the sphere in a circle passing through those points (1.): half the difference between the hemi Every spherical triangle is equal to from the centre of the sphere, draw a perpendicular to the plane A'BC, and spherical surface, and the sum of three produce it to meet the surface of the lunes which ħave their angles equal to the sphere in P: then P is the pole of the three angles of the triangle respectively. circle A B C (def. 3.); and, consequent Let A B C be any spherical triangle; ly, (2.) if the spherical arcs PA, PB, its surface shall be equal to half the RC are drawn, joining the point P with difference between the hemispherical the points A, B, C, respectively, they will be equal to one another. Make the spherical angle DEQ equal to the spherical angle A BP: make also the arc EQ equal to the arc B P, and join QD, Q F. Then, because the angles DEF (15.) and D E Q are equal to the angles A B C and A B P, each to each, the angle Q EF is equal to the angle PBC (I. ax. 3.). And, because in the surface, and the sum of three lunes triangles Q E F, PB C, two sides of the having their angles respectively equal one are equal to two sides of the other, to the angles A, B, C. each to each, and the included angles Let the circles be completed, of which QEF, PB C likewise equal to one ano the sides A B, A C, and B C are parts, ther (13.) the arc Q F is equal to the and let A a, Bb, and C c be the diameters arc PC, and the angle EQ F to the passing through the points A, B, and C. angle B PC. In the same manner, it Then, because C Ac and A ca are semimay be shown that the arc QD is circumferences (4.), they are equal to one equal to the arc P A, and the angle another, and A c being taken from each, D QE to the angle A P B. There- the remainders AC, ac are likewise fore, because the angles DQ E and equal. In the same manner it may be EQF are respectively equal to the shown that A B is equal to ab, and BC angles A P B and B PC, the angle to bc. And, because the triangles DQF is equal to the angle A PC: A B C, abc have the three sides of the and because the arcs PA, PB, PC one equal to the three sides of the are equal to one another, the arcs Q D, other, each to each, their surfaces are Q E, Q F, which are severally equal to equal to one another (20.) Now, the lune them, are likewise equal to one another A cab A has its angle equal to the angle (I. ax. 1.). A of the triangle A B C (8. Cor. 1.), and Now, it is evident that isosceles tri- the lunes B ABC B, C BCAC have for angles, as P B C and Q E F, which have their angles the angles B, C of the same equal vertical ingles, and equal sides triangle. And these three lunes are containing them, may be made to coin * If the arcs BP, E Q, or these arcs produced, cut cide, and, therefore, are equal to one the arcs A C, D F in the points R, S, respectively, it another. Therefore, the isosceles trian- may be shewn by prop. 14, that B R is equal to ES: gles P AB, PBC, and PAC are re therefore, since B P is equal to EQ, if ihe point P falls within the triangle A BC, the point Q must spectively equal to the isosceles triangles likewise fall within the triangle DEF; or if P fall Q DE, QEF, and Q DF. But the without the triangle A BC, and within the angle A BC, Q must likewise fall without the triangle triangle A B C is equal to the sum of DE F, and within the corresponding angle DE F. с together equal to the surface of the the sum of its angles minus two right hemisphere A B C b c together with the angles : therefore, the sum of all the tri. triangles A B C, abc; that is, to the angles, that is, the polygon, is measured surface of the hemisphere, together with by the sum of all their angles minus twice the triangle ABC. Therefore, twice as many right angles as there are twice the triangle A B C is equal to the triangles. But the sum of all the andifference between the sum of the three gles of the triangles is equal to the sum lunes and the surface of the hemisphere; of all the angles of the polygon, together and the triangle A B C is equal to half with the angles of the point P, that is, (8. that difference. Cor. 2.) together with four right angles: Therefore, &c. and there are as many triangles as the Cor. 1. Hence, if the angles of a polygon has sides. Therefore, the surspherical triangle are given, we may face of the polygon is measured by the find the proportion which its surface sum of its angles plus four right angles bears to the surface of the sphere. minus twice as many right angles as For, since the surface of any lune is to the polygon has sides. the surface of the sphere, as the angle of Let 3 be the sum of its angles, and the lune to four right angles (V. 21.), or n the number of its sides : then the suras twice its angle to eight right angles; if face of the polygon is measured by the surface of the sphere be represented 3+ 4R-2 n R. by 8 right angles, or 8 R,* the surfaces of the lunes which have the angles Scholium. A, B, and C will be represented by 2 A, It follows from the last corollary, that 2 B, and 2C; and the hemispherical if S be the number of the solid angles of surface will be represented by 4 R: a polyhedron, F the number of its faces, F therefore, the surface of the triangle and E the number of its edges, S+F-E A B C will be represented by half 2 At will be equal to 2. For, if a sphere be 2 B+2 C - 4 R, that is, by A+B+C- described about any point within the 2 R. And hence, it is commonly said, polyhedron as a centre, the lines which that the surface of a spherical triangle are drawn from this point through the is measured by the excess of the sum of solid angles of the polyhedron will cut its angles above two right angles ; it the surface of the sphere in S points, being understood that the surface of and the planes which are drawn from the the sphere is measured by eight right same point through the edges will cut angles. the surface in E spherical arcs; and thus Cor. 2. The surface of a spherical the whole surface of the sphere will be triangle whose angles are A, B, C is divided into F spherical polygons, corequal to the surface of a lune whose responding to the F faces of the polyA+B+C hedron. Now, the surface of each of angle is - R. For both these spherical polygons is measured by surfaces are measured by A + B + C - angles minus twice as many right an the sum of its angles plus four right 2 R. gles as the polygon has sides. ThereCor. 3. The surface of a spherical fore, the surfaces of all the polygons polygon A B C D E is measured by the taken together are measured by the sum excess of the sum of of all their angles taken together plus all its angles, together with four right as many times four right angles as there are polygons minus twice as many right angles, above twice angles as all the polygons together have as many right angles sides. But the sum of all their angles as the polygon has taken together is the sum of all the angles, sides. at the S angular points, and, therefore, For the polygon may be divided into is equal to S times four right angles as many triangles as it has sides, by (8. Cor. 2.): the number of the polygons joining its angles with any point P taken is F; and the number of the sides of the within the polygon; and the surface of polygons is 2 E, because the number of each of these triangles is measured by different sides is E, and each of these is at once a side of two adjoining polygons. * In which case, R represents the surface of the spherical triangle, which has each of its angulae Therefore, the surfaces of all the polypoints the pole of the side opposite to it (7. Scholium.). gons, taken together, are measured by We shall see (28. Scholium.) that the surface of this triangle measures a solid right angle at the centre (S + F - E) times four right angles. of the sphere. Therefore, since the surface of the 2 A . |