was shewn in the preceding propofition of the straight line FA : wherefore CHD, CHG are right angles, and CH is the fine of the arch AC; and in the triangle CHG, having the right angle CHG, CG is to the radius as CH to the fine of the angle CGH: (1. Pl. Tr.) but fince CG, HG are at right angles to DGB, which is the common section of the planes CBD, ABD, the angle CGH will be equal to the inclination of these planes; (6. def. 11.) that is, to the spherical angle ABC. The fine, therefore, of the hypothenufe CB is to the radius as the fine of the fide AC is to the fine of the oppofite angle ABC. Q. E. D.

COR. Of these three, viz. the hypothenuse, a fide, and the angle oppofite to that fide, any two being given, the third is alfo given by prop. 2.

IN

N right-angled fpherical triangles, the co-fine of the hypothenufe 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 hypothenufe BC will be 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 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 alfo pafs through the pole of BD. (13. 18. 1. Theod. fph.) And fince AC is perpendicular to BD, AC will alfo pafs through the pole of BD; wherefore the pole of the circle BD will be found in the point where the circles AC, DE meet, that is, in the point F: the arches FA, FD are therefore quadrants, and likewife the arches BD, BE in the triangle CEF, right-angled at the point E, CE is the complement of the hypothenufe BC of the triangle ABC, EF is the complement of the arch ED, which is the meafure of the angle ABC, and FC the hypothenufe 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 (17. of this) in the triangle CEF, the fine of the fide CE is to the radius, as the tangent of the other fide is to the tangent

of the angle ECF oppofite to it, that is, in the triangle ABC, the co-fine of the hypothenufe BC is to the radius, as the co-tangent of the angle ABC is to the tangent of the angle ACB. Q. E. D. COR. 1. Of thefe three, viz. the hypothenufe and the two angles, any two being given, the third will also be given.

COR. 2. And fince by this propofition the co-fine of the hypothenufe BC is to the radius as the co-tangent of the angle ABC to the tangent of the angle ACB. But as the radius is to the co-tangent of the angle ACB, fo is the tangent of the fame to the radius; (Cor. 2. def. Pl. Tr.) and, ex aequo, the co-fine of the hypothenuse BC is to the co-tangent of the angle ACB, aş the co-tangent of the angle ABC to the radius.

N right-angled fpherical triangles, the co-fine of an angle is to the radius, as the tangent of the fide adjacent to that angle is to the tangent of the hypothenufe.

The fame conftruction remaining; in the triangle CEF. (17. of this) the fine of the fide EF is to the radius, as the tangent of the other fide CE is to the tangent of the angle CFE oppofite to it; that is, in the triangle ABC, the co-fine of the angle ABC is to the radius as (the co-tangent of the hypothenuse BC to the co-tangent of the fide AB, adjacent to ABC, or as) the tangent of the fide AB to the tangent of the hypothenufe, fince the tangents of two arches are reciprocally proportional to their cotangents. (Cor. 1. def. Pl. Tr.)

COR. And fince by this propofition the co-fine of the angle ABC is to the radius, as the tangent of the fide AB is to the tangent of the hypothenufe BC; and as the radius is to the cotangent of BC, so is the tangent of BC to the radius; by equality, the co-fine of the angle ABC will be to the co-tangent of the hypothenufe BC, as the tangent of the fide AB, adjacent to the angle ABC, to the radius.

PROP. XXI. FIG. 14.

N right-angled fpherical triangles, the co-fine of either of the fides is to the radius, as the co-fine of the hypothenufe is to the co-fine of the other fide.

The fame conftruction remaining; in the triangle CEF, the fine of the hypothenuse CF is to the radius, as the fine of the fide CE to the fine of the oppofite angle CFE; (18. of this) that is, in the triangle ABC the co-fine of the fide CA is to the radius as the co-fine of the hypothenufe BC to the co-fine of the other fide BA. Q. E. D.

I

N right-angled fpherical triangles, the co-fine of either of the fides is to the radius, as the co-fine of the angle oppofite to that fide is to the fine of the other angle.

The fame conftruction remaining; in the triangle CEF, the fine of the hypothenuse CF is to the radius as the fine of the fide EF is to the fine of the angle ECF oppofite to it; that is, in the triangle ABC, the co-fine of the fide CA is to the radius, as the co-fine of the angle ABC oppofite to it, is to the fine of the other angle. Q. E. D.

Fig. 15.

OF THE CIRCULAR PARTS.

N any right-angled spherical triangle ABC, the complement

two fides, are called The circular parts of the triangle, as if it
were following each other in a circular order, from whatever
part we begin: thus, if we begin at the complement of the hy-
pothenuse, and proceed towards the fide BA, the parts following
in order will be the complement of the hypothenufe, the com-
plement of the angle B, the fide BA, the fide AC, (for the right
angle at A is not reckoned among the parts,) and, lastly, the
complement of the angle C. And thus at whatever part we begin,
if any three of these five be taken, they either will be all con-
tiguous or adjacent, or one of them will not be contiguous to
either of the other two: in the first case, the part which is be-
tween the other two is called the Middle part, and the other two
are called Adjacent extremes. In the fecond cafe, the part
which is not contiguous to either of the other two is called the
Middle part, and the other two Oppofite extremes.
For ex-
ample, if the three parts be the complement of the hypothenuse
BC, the complement of the angle B, and the fide BA; fince these
three are contiguous to each other, the complement of the angle
B will be the middle part, and the complement of the hypothenuse
BC and the fide BA will be adjacent extremes: but if the com-
plement of the hypothenufe BC, and the fides BA, AC be
taken; fince the complement of the hypothenufe is not adjacent
to either of the fides, viz. on account of the complements of the
two angles B and C interveening between it and the fides, the
complement of the hypothenufe BC will be the middle part, and
the fides, BA, AC oppofite extremes. The most acute and inge-
nious Baron Napier, the inventor of Logarithms, contrived the
two following rules concerning these parts, by means of which
all the cafes of right-angled spherical triangles are resolved with
the greatest ease.

RULE I.

The rectangle contained by the radius and the fine of the middle part, is equal to the rectangle contained by the tangents of the adjacent parts.

RULE II.

The rectangle contained by the radius, and the fine of the middle part is equal to the rectangle contained by the co-fines of the

oppofite parts.

These rules are demonftrated in the following manner.

First, Let either of the fides, as BA, be the middle part, and Fig. 16. therefore the complement of the angle B, and the fide AC will be adjacent extremes. And by Cor. 2. prop. 17. of this, S, BA is to the Co-T, B, as T, AC is to the radius, and therefore RXS, BA Co-T, BXT, AC.

The fame fide BA being the middle part, the complement of the hypothenufe, and the complement of the angle C, are oppofite extremes; and by prop. 18. S, BC is to the radius, as S, BA to S, C; therefore RxS, BA=S, BC XS, C.

Secondly, Let the complement of one of the angles, as B, be the middle part, and the complement of the hypothenuse, and the fide BA will be adjacent extremes: and by Cor. prop. 20. Co-S, B is to Co-T, BC, as T, BA is to the radius, and therefore Rx Co-S, B Co-T, BCXT, BA.

Again, Let the complement of the angle B be the middle part, and the complement of the angle C, and the fide AC will be oppofite extremes: and by prop. 22. Co-S, AC is to the radius, as Co-S, B is to S, C: and therefore Rx Co-S, B=Co-S, ACXS, C.

Thirdly, Let the complement of the hypothenufe be the middle part, and the complements of the angles B, C, will be adjacent extremes; but by cor. 2. prop. 19. Co-S, BC is to Co-T, C as Co-T, B to the radius: therefore Rx Co-S, BC-Co-T, BX Co-T, C.

Again, Let the complement of the hypothenuse be the middle part, and the fides AB, AC will be oppofite extremes: but by prop. 21. Co-S, AC is to the radius, as Co-S, BC to Co-S, BA; therefore Rx Co-S, BC=Co-S, BAX Co-S, AC. Q. E. D.