are right angles; therefore D is the pole of BAEa, and G the a 2. cor. pole of IFCB; AE the complement of the arch AB, and FE equal GD, the measure of the angle B, and DF their complement; and likewife BC equal IF, the measure of the angle G, and CF their complement; and CA, one of the fides about the right angle, equal to HD, and DC their complement.

2d, In the triangles HIC, DCF, right angled at I and F, having the fame acute angle C. Since BA is lefs than a quadrant, DF is to DC as IH is to HC; but HC is a quadrant; there-b 4. 6. fore, the fine of DF is to the fine of DC, as the fine of HI is to the fine of HC, that is, to radius. Wherefore, &c.

3d, For the fame reafon, in the triangles AED, CFD, rightangled at E, F, and having the fame acute angle D, as AE is lefs than a quadrant, the fine of EA is to the fine of CF, as the fine of DA is fo the fine of DC; that is, the cofine of the base is to the cofine of the hypothenufe, fo is radius to the cofine of the perpendicular.

4th, Again, in the triangles BAC, BEF, right angled at A, E, and having the fame acute angle B; as S. BA is to the S. BE, fo is the T. AC to the T. EF; that is, the fine of the bafe is to the rad, as the tangent of the perpendicular to the tan gent of the angle at the bafe.

5th. In the triangle GIF, GHD, right angled at I and H, and having the fame acute angle G; S. GH is to the S. GI, as the T. HD to the TIF; that is, the cofine of the vertical angle C is to rad. as the tangent of the perpendicular is to the tangent of the hypothenuse.

6th. Again, in the fame triangles S. IF is to S. GF, as S. HD to fine GD; that is, the S. of the hypothenufe is to rad, as the fine of the perpendicular is to the fine of the angle at the bafe.

7th. In the triangles HIC, DFC, right-angled at I, F, and the fame acute angle C, as S. CI is to S. CF, fa is T. HI to T. DF, that is, Rad. is to Cof. BC, fo is T. C to Cot. B and S. CH: S. HI. :: S. DC; S. DF; that is, Rad. is to S. ACB ;, Cof. AC: Cof. B.

c 16,

PRO P. XVIII,

THE cofines of the angles at the base are proportional to the fines of the vertical angles.

Let BCD be the triangle, either obtufe or acute angled; let fall a perpendicular CA; then, as the cofine of B is to S. BCA,

fo

fo is the cof. CA to radius; and as cof. CA is to radius, fo is cof. D to the S. DCA: therefore, by equality b, cof. B is to 25. 3. S. BCA fo is cof. D to S. DCA.

b 17:

PRO P. XIX.

Nevery Spherical triangle, the cofines of the fides are proportional to the cofines of the bafes.

Let BCD be the triangle, the fame things fuppofed as in the laft, the cof. BC is to the cof. BA as cof. CA is to rad.; and cof CA is to rad. as cof. DC is to cof. DA; wherefore the cof. BC is to cof. DC as cof. BA is to the cof. of DA".

Noblique Spherical triangles, the fines of the bafes are in the reciprocal proportion of the tangents of the angle at the bafe.

The fame things fuppofed as before, S. BA is to rad. as T. AC is to the tangent of the angle at B; and, inverfely, radius is to S. BA as T. B is to T. AC; and S. DA is to rad. as T.D is to T. AC; wherefore, S. BA is to S. DA as T. D is to T. Bb.

a 17.

b 16.6. cax. I. I. d 14. 6.

IN
N oblique fpherical triangles, the tangents of the fides are in the
reciprocal proportion of the cofines of the vertical angles.

The fame things fuppofed as before, R. is to the cof. ACB as T. BC to T. AC; and R. is to cof. ACD as T. DC to T. AC2; and cof. ACD x T. DC is equal to R. x T. AC; therefore cof. ACB × T. BC is equal to R. × T. AC, therefore cof. ACB x T. BC is equal to cof. ACD x T. DC; therefore T. BC is to T. DC, as cof. ACD to cof. ACB 4.

PRO P. XXII.

IN every oblique Spherical triangle, the fines of the fides are proportional to the fines of their oppofite angles.

For, the fame things fuppofed as before, S. BC is to rad, as a 17. S. CA is to S. B2; and S. DC is to rad. as S. AC is to S. D; b 23. 5. and, inverfely, R. is to S. BC as S. D is to S. AC; wherefore, S. BC is to S. DC as S. D is to S. B.

PRO P. XXIII.

I'
N every Spherical triangle, the cotangent of half the fum of the
angles at the bafe, is to the tangent of half their difference, as
the tangent of half the vertical angle to the tangent of the angle
that the perpendicular makes with the line bifecting the vertical
angle.

a a 17t

Let BDC be the triangle, and let CF bifect the vertical angle. C; then, as rad. : S. ACB :: cof. AC: cof. of the angle Ba, and rad.: S. ACD :: cof. AC: cof. D; therefore, by eq. and permutation, cof. B: cof. D:: S. ACB: S. ACD; therefore, cof. B + cof. D: cof. B cof. D:: S. ACB + S. ACD: S. ACB S. ACD:: cot. B+D:T. B-D::T.BCF: T. ACF=T.B-D.

PROP. XXIV.

IN any Spherical triangle, the rectangle contained under the fines of two fides, is to the fquare of the radius, as the difference of the verfed fines of the bafe, and difference of the fides, to the verfed fine of the angle oppofite to the bafe.

Let ABC be the triangle, and CF, AE, the fines of the fides AB and CB, or MBCB; then CF = MFX AE: AO X ON :: IL, the difference of the verfed fines of the base AC, and the difference of the fides BC and BA, to the verfed fine of the angle B.

For, defcribe a great circle PN about the pole B ; let BP, BN, be quadrants; then the arch PN is the measure of the angle B. From the fame pole B defcribe a leffer circle CFM through C; the plains of thefe circles will be perpendicular to the plain BON; and, let PG, CH, be perpendicular to the fame plain, they will fall on the common fections ON, FM *, a 38. II. fuppofe in G and H ; and through H draw HI perpendicular to AO; then the plain paffing through CH, HI, will be perpendicular to the plain AOB; and AI, which is perpendicular to HI, is likewise perpendicular to CI; and AI is the verfed fineb 4. II.

of

£ 16. 11.

d 23, 6.

of the arch AC, and AL the verfed fine of the arch AM, which
is equal to BM-BABC-BA; and, becaufe MF, NO,
are parallel to CF, PO, the isofceles triangles are equiangu
lar; therefore, if perpendiculars CH, PG, be drawn to the
fides FM, ON, the triangles will be divided fimilarly; there
fore FM ON :: MH: GN; and, becaufe the triangles AOE,
DIH, DLM, are equiangular AE: AO: IL: MH; for IL
=ID, DL, and MH MD, DH; bat FM : ON :: MH :
GN; therefore, AE x FM, or CFA× ON: IL× MH;
MH x GN; the two laft divide by
will be AE x FM :
AO × ON:: IL: GN; that is, the recte gle under the fines of
the legs is to the fquare of the rad. as the difference of the ver-
fed fine of the bafe, and the difference of the legs, is to the
verfed fine of the angle B. Wherefore, &c.

=

IN any Spherical triangle, the difference of the verfed fines of two arches, multiplied into half the radius, is equal to the rectangle contained by the fine of half the fum, and the fine of half the difference of thefe arches.

Let there be two arches AC, BA, whofe difference BC let be bifected in D; then AD is half the fum, and DB half the difference of these arches; and BT = GH the difference of their verfed fines, and CE the fine of half their difference. Then, because the triangles ODV, CBT, are equiangular, as DV; BT:: OD: CBDO: CB; wherefore DV x BC; or, DV × CE = BT × DO = HG × ÷ DO, Wherefore, &c.

T

HE verfed fine of any arch, multiplied by half the radius, is equal to the fquare of the fine of half of that arch.

Let the arch be DB, and C its center; draw the right lines DB, BC, and DE, perpendicular to BC; then DE is the fine of that arch; which let be bifected by the right line CM; then are the triangles CMB, DEB, equiangular; for the angles at M and E are right ones; and the angle at B is common; there fore EB BD:: BM: BC; therefore EB x BC= BMX BD, and EB × ÷ BC = BM × ; BD = BM fquare. Wherefore, &c.

:

IN any fpherical triangle, the rectangle contained under the fines of any two fides, is to the fquare of the radius, as the rectangle contained by the fines of thefe arches, which are the bafe and the dif ference of the fides and the fine of that arch, which is half the dif ference of the fame, to the fquare of the fine of half the angle oppofite to the bafe.

Let the triangle be ABC, and BC, BA, the fides containing the angle B, and AC the bafe fubtending that angle; and let the arch AM be taken equal to the difference of the legs; then,.. the rectangle under the fines of the fides BC, BA, will be to the fquare of the radius, as the rectangle under the fine of the arch AC+ AM; and the fine of the arch AC - AM is to the fine

For, because the rectangle under the fines of the legs AB, BC, is to the fquare of the radius, as IL is to the verfed fine of the angle B ; and, fince R x IL equal to the rectangle under the a 24 fines of the arches AC + AM, and AC AM; and halfь 25.

2

2

radius multiplied by the verfed fine of the angle B, is equal to the fquare of the fine of one half the angle B ; therefore the c 26. rectangle under the fine of the fides, is to the fquare of the rad. as the rectangle under the fines of the arches AC + AM and

2

ACAM is to the fquare of the fine of one half the angle B.

2

Wherefore, &c.