sum of the sides is less than a semicircle, the least segment is adjacent to the least side. 2. Again, in the triangle FCA the two sides FC and CA are less than a semicircle; for since AC is less than CB, AC and CF are less than BC and CF. Also, AC is less than CF, because it is more remote from CHG thau CF is; therefore in this case also, viz. when the perpendicular falls without the triangle, and when the sum of the sides is less than a semicirtle, the least segment of the base AD is adjacent to the least side. 3. But in the triangle FCE the two sides FC and CE are greater than . a semicircle; for, since FC is greater than CA, FC and CE are greater than AC and CE. And because AC is less than CB, EC is greater than CF, and EC is therefore nearer to the perpendicular CHG than CF is, wherefore EG is the least segment of the base, and is adjacent to the greater side. 4. In the triangle ECB the two sides EC, CB are greater than a semi circle; for, since by supposition CB is greater than CA, EC and CB are greater than EC and CA. Also, EC is greater than CB, wherefore in this case, also, the least segment of the base EG is adjacent to the greatest side of the triangle. Therefore, when the sum of the sides is greater than a semicircle, the least segment of the base is adjacent to the greatest side, whether the perpendicular fall within or without the triangle: and it has been shewn, that when the sum of the sides is less than a semicircle, the least segment of the base is adjacent to the least of the sides, whether the perpendicular fall within or without the triangle. PROP. XVIII. ▲n right angled spherical triangles, the sine of either of the sides about the right angle, is to the radius of the sphere, as the tangent of the remaining side is to the tangent of the angle opposite to that side. E Let ABC be a triangle, having the right angle at A; and let AB be either of the sides, the sine of the side AB will be to the radius, as the tangent of the other side AC to the tangent of the angle ABC, opposite to AC. Let D be the centre of the sphere; join AD, BD, CD, and let AF be drawn perpendicular to BD, which therefore will be the sine of the arc AB, and from the point F, let there be drawn in the plane BDC the straight line FE at right angles to BD, meeting DC in E, and let AE be joined. Since therefore the straight line DB is at right angles to both FA and FE, it will also be at right angles to the plane AEF (4. 2. Sup.) ; wherefore the plane ABD, which passes through DF, is perpendicular to the plane AEF (17. 2. Sup.), and the plane AEF perpendicular to ABD: But the plane ACD or AED, is also perpendicular to D the same ABD, because the spherical angle BAC is a right angle. Therefore AE, the common section of the planes AED, FB AEP, is at right angles to the plane ABD (18. 2. Sup.), and EAF, EAD are right angles. Therefore AE is the tangent of the arc AC; and in the rectilineal triangle AEF, having a right angle at A, AF is to the radius as AE to the tangent of the angle AFE (1. Pl. Tr.); but AF is the sine of the arc AB, and AE the tangent of the arc AC; and the angle AFE is the inclination of the planes CBD, ABD (4. def. 2. Sup.), or is equal to the spherical angle ABC: Therefore the sine of the arc AB is to the radius as the tangent of the arc AC to the tangent of the opposite angle ABC. COR. Since by this proposition, sin. AB : R :: tan. AC: tan. ABC; and because R cot. ABC tan. ABC: R (1 Cor. def. 9. Pl. Tr.) by equality, sin. AB: cot. ABC :: tan. AC : R. : PROP. XIX. In right angled spherical triangles the sine of the hypotenuse is to the radius as the sine of either side is to the sine of the angle opposite to that side. Let the triangle ABC be right angled at A, and let AC be either of the sides; the sine of the hypotenuse BC will be to the radius as the sine of the arc AC is to the sine of the angle ABC. Let D be the centre of the sphere, and let CE be drawn perpendicular to DB, which will therefore be the sine of the hypotenuse BC; and from the point E let there be drawn in the D F plane ABD the straight line EF perpendicular to DB, and let CF be joined; then CF will be at right angles to the plane ABD, because as was shewn of EA in the preceding proposition, it is the common section of two planes DCF, ECF, each perpendicular to the plane ADB. Wherefore CFD, CFE are right angles, and CF is the sine of the arc AC; and in the triangle CFE having the right angle CFE, CE is to the radius, as CF to the sine of the angle CEF (1. Pl. Tr.). But, since CE, FE are at right angles to DEB, which is the common section of the planes CBD, ABD, the angle CEF is equal to the inclination of these planes (4. def. 2. Sup.), that is, to the spherical angle ABC. Therefore the sine of the hypotenuse CB, is to the radius, as the sine of the side AC to the sine of the opposite angle ABC PROP. XX. E B In right angled spherical triangles, the cosine of the hypotenuse is to the radius as the cotangent of either of the angles is to the tangent of the remaining angle. Let ABC be a spherical triangle, having a right angle at A, the cosine of the hypotenuse BC is to the radius as the cotangent of the angle ABC to the tangent of the angle ACB Describe the circle DE, of which B is the pole, and let it meer AC in f and the circle BC in E; and since the circle BD pases through the pole B, of the circle DF, DF must pass through the pole of BD (4.). An since AC is perpendicular to BD, the plane of the circle AC is perpendi cular to the plane of the circle BAD, and therefore AC must also (4.) pass through the pole of BAD; wherefore, the pole of the circle BAD is in the point F, where the circles AC, DE, intersect. The arcs .FA, FD are therefore quadrants, and likewise the arcs BD, BE. Therefore, in the triangle CEF, right angled at the point E, CE is the complement of BC, the hypotenuse of the triangle ABC; EF is the complement of the arc ED, the measure of the angle ABC, and FC, the hypotenuse of the triangle CEF, is the complement of AC, and the arc AD, which is the measure of the angle CFE, is the complement of AB. But (18.) in the triangle CEF, sin. CE: R:: tan. EF: tan. ECF, that is, in the triangle ACB, cos. BC: R:: cot. ABC: tan. ACB. COR. Because cos. BC: R:: cot. ABC : tan. ACB, and (Cor. 1. def. 9. Pl. Tr.) cot. ABC: R:: R: tan. ABC, ex æquo, cot. ACB: cos. BC : : R : cot, ABC. PROP. XXI. In right angled spherical triangles, the cosine of an angle is to the radius as the tangent of the side adjacent to that angle is to the tangent of the hypotenuse The same construction remaining; In the triangle CEF, sin. FE: R:. tan. CE: tan. CFE (18.): but sin. EF=cos. ABC; tan. CE=cot. BC, and tan. CFE cot. AB, therefore cos. ABC; R:: cot. BC: cot. AB. Now because (Cor. 1. def. 9. Pl. Tr.) cot. BC: R:: R : tan. BC, and cot. AB : R:: R tan. AB, by equality inversely, cot. BC: cot, AB: tạn. BC; therefore (11. 5.) cos. ABC: R:: tan. AB: tan. BC. AB COR. 1. From the demonstration it is manifest, that the tangents of any two arcs AB, BC are reciprocally proportional to their cotangents. COR. 2 Because cos. ABC: R:: tan. AB: tan. BC, and R : cos. BC :: tan. BC: R, by equality, cos. ABC: cot. BC: : tan. AB: R. That is, the cosine of any of the oblique angles is to the cotangent of the hypotenuse, as the tangent of the side adjacent to the angle is to the radius. PROP. XXII. In right angled spherical triangles, the cosine of either of the sides is to the ra dius, as the cosine of the hypotenuse is to the cosine of the other side. The same construction remaining: In the triangle CEF, sin. CF : R: sin. CE sin. CFE (19.); but sin. CF=cos. CA, sin. CE=cos. BC, and sin. CFE cos. AB; therefore cos. CA: R:: cos. BC: cos. AB. PROP. XXIII. In right angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the angle opposite to that side is to the sine of the other angle. The same construction remaining: In the triangle CEF, sin. CF: R:: sin. EF sin. ECF (19.); but sin. CF=cos. CA, sin. EF=cos. ABC, and sin. ECF=sin. BCA: therefore, cos. CA : R:: cos. ABC : sin. BCA. PROP. XXIV. In spherical triangles, whether right angled or oblique angled, the sines of the sides are proportional to the sines of the angles opposite to them. First, let ABC be a right angled triangle, having a right angle at A; herefore (19.) tne sine of the hypotenuse BC is to the radius, (or the sine of the right angle at A), as the sine of the side AC to the sine of the angle B. And, in like manner, the sine of BC is to the sine of the angle A, as the sine of AB to the sine of the angle C; wherefore (11. 5.) the sine of the side AC is to the sine of the angle B, as the sine of AB to the sine of the angle C. C B Secondly, Let ABC be an oblique angled triangle, the sine of any of the sides BC will be to the sine of any of the other two AC, as the sine of the angle A opposite to BC, is to the sine of the angle B opposite to AC. Through the point C, let there be drawn an arc of a great circle CD perpendicular to AB; and in the right angled triangle BCD, sin. BC: R:. sin. CD: sin. B (19.); and in the triangle ADC, sin. AC : R :: sin. CD: sin. A; wherefore, by equality inversely, sin. BC: sin. AC:: sin. A: sin. B. In the same manner, it may be proved that sin. BC: sin. AB :: sin. A sin. C, &c. PROP. XXV. In oblique angled spherical triangles, a perpendicular arc being drawn from any of the angles upon the opposite side, the cosines of the angles at the base are proportional to the sines of the segments of the vertical angle. Let ABC be a triangle, and the arc CD perpendicular to the base BA, the cosine of the angle B will be to the cosine of the angle A, as the sine of the angle BCD to the sine of the angle ACD. For having drawn CD perpendicular to AB, in the right angled triangle BCD (23.), cos. CD: R:: cos. B: sin. DCB; and in the right angled triangle ACD, cos. CD: R:: cos. A: sin. ACD; therefore (11. 5.) cos. B: sin. DCB:: cos. A: sin. ACD, and alternately, cos. B· cos. A :: sin. BCD sin. ACD. PROP. XXVI. The same things remaining, the cosines of the sides BC, CA, are proportiona to the cosines of BD, DA, the segments of the base. For in the triangle BCD (22.) cos. BC: cos. BD:: cos. DC: R, and in |