AC is alfo less than CB, (13.) and therefore in this case, viz. when the perpendicular falls within the triangle, and when the fum of the fides is lefs than a femicircle, the least segment is adjacent to the leaft fide.

2. Again, in the triangle FCA the two fides FC and CA are less than a femicircle; for, fince 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 than CF is; therefore in this cafe alfo, viz. when the perpendicular falls without the triangle, and when the fum of the fides is less than a femicircle, the least segment of the base AD is adjacent to the leaft fide.

3. But in the triangle FCE the two fides FC and CE are greater than a femicircle; for, fince FC is greater than CA, FC and CE are greater than AC and CE. And because AC is lefs 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 fide.

4. In the triangle ECB the two fides EC, CB are greater than a femicircle; for, fince by fuppofition CB is greater than CA, EC and CB are greater than EC and CA. Also, EC is greater than CB, wherefore in this cafe, alfo, the leaft fegment of the base EG is adjacent to the greatest fide of the triangle. Therefore, when the fum of the fides is greater than a femicircle, the leaft fegment of the base is adjacent to the greatest fide, whether the perpendicular fall within or without the triangle: and it has been fhewn, that when the fum of the fides is lefs than a femicircle, the least segment of the base is adjacent to the leaft of the fides, whether the perpendicular fall within or without the triangle. Wherefore, &c. Q. E. D.

PROP.

PROP. XVIII.

IN right angled spherical triangles, the fine of either of the fides about the right angle, is to the radius of the sphere, as the tangent of the remaining fide is to the tangent of the angle oppofite to that fide.

E

Let ABC be a triangle, having the right angle at A; and let AB be either of the fides, the fine of the fide AB will be to the radius, as the tangent of the other fide AC to the tangent of the angle ABC, oppofite 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 fine of the arch AB, and from the point F, let there be drawn in the plane BDC the ftraight line FE at right angles to BD, meeting DC in E, and let AE be joined. Since therefore the ftraight line DF 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 paffes through DF is

D

FB

perpendicular to the plane AEF (17. 2. Sup.), and the plane AEF perpendicular to ABD: But the plane ACD or AED, is alfo perpendicular to the fame ABD, because the spherical angle BAC is a right angle: Therefore AE, the common fection of the planes AED, AEF, is at right angles to the plane ABD, (18. 2. Sup.) and EAF, EAD are right angles. Therefore AE is the tangent of the arch AC; and in the rectilineal triangle AEF, having a right angle at À, AF is to the radius as AE to the tangent of the angle AFE, (1. Pl. Tr.); but AF is the fine of the arch AB, and AE the

tangent

tangent of the arch 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 fine of the arch AB is to the radius as the tangent of the arch AC to the tangent of the oppofite angle ABC. Q. E. D.

COR. And fince by this propofition, fin AB: R:: tan AC: tan ABC; and because R: cot ABC :: tan ABC : R (1. Cor. def. 9. Pl. Tr.) by equality inversely, fin AB : cot ABC:: tan AC: R.

PRO P. XIX.

N right angled fpherical triangles the fine of the fide is to the fine of the angle oppofite to that fide.

Let the triangle ABC be right angled at A, and let AC be either of the fides; the fine of the hypotenuse BC will be to the radius as the fine of the arch AC is to the fine 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 fine of the hypotenuse BC; and from

the point E let there be drawn in the 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 fhown of EA in the preceding propofition, it is the common fection of two planes, DCF, ECF, each perpendicular to,

E

F

the plane ADB. Wherefore CFD, CFE are right angles, and CF is the fine of the arch AC; and in the triangle CFE having the right angle CFE, CE is to the radius, as CF to the

fine of the angle CEF, (1. Pl. Tr.). But, fince CE, FE are at right angles to DEB, which is the common fection 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 fine of the hypotenuse CB, is to the radius, as the fine of the fide AC to the fine of the oppofite angle ABC. Q. E. D.

I

N right angled fpherical triangles, the co-fine of the hypotenuse is to the radius as the co-tangent 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 co-fine of the hypotenufe BC to the radius as the cotangent of the angle ABC to the tangent of the angle ACB. Defcribe the circle DE,

of which B is the pole, and let it meet AC in F, and the circle BC in E; and fince the circle BD paffes through the pole B, of the circle DF, DF will pafs through the pole of BD, (4.). And fince AC is perpendicular to BD, the plane of the circle AC is perpendicular to the plane of the circle BAD, and therefore AC will alfo (4.) pass through the pole of BAD; wherefore, the pole of the circle BAD is in

B.

F

A

the point F, where the circles AG, DE interfect. The arches

FA,

FA, FD are therefore quadrants, and likewise the arches 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 arch ED, the measure of the angle ABC, and FC, the hypotenuse of the triangle CEF, is the complement of AC, and the arch AD, which is the measure of the angle CFE, is the complement of AB.

But (18.) in the triangle CEF, fin CE: R:: tan EF: tan ECF, that is, in the triangle ACB, cof BC: R:: cọt ABC ; tan ACB. Q. E. D.

COR. Because cof BC: R:: cot ABC: tan ACB, and (Cor. 1. def. 9. Pl. Tr.) cot ACB: R:: R: tan ACB, ex æquo,

cot ACB

cof BC: R cot ABC.

PROP.