All the limitations are either contained in the 16th and 17th Sph. Trig. propofitions, or elfe they are eafily deducible from them. Thus, becaufe C is of the fame affection with AB; if AC and C be of the fame affection, AC and AB are of the fame affection; and therefore BC is lefs than yo°, (Prop. XVII.); this is the limitation in cafe 5. Again, if BC be less than 90°, AC and AB, or B and C, are of the fame affection; and therefore, if AC be lefs than 90°, or of the fame affection with CB, AB is lefs than 90°, and for the fame reafon, C is lefs than 90°. But if AC, BC be of different affection, that is, AC greater than 90°, BA is greater than 90°: and the fame may be proved, in the fame manner, when BC is greater than 90°. Thus, the reafon of the limitation in the 2d, 3d, 10th, and 12th cafes, is manifeft.

PROP. XXI.

IN any spherical triangle, the fines of the fides are proportional to the fines of their oppofite angles.

Let ABC be a spherical triangle, as the fine of AC to the fine of BC, fo is the fine of the angle at B to the fine of the angle at A.

If one of the angles, as ACB be a right angle; the fine of AC is to the fine of B as the fine of the hypotenufe AB to a 20. 5. T. the radius or fine of the right angle ACB; that is, as the fine of BC to the fine of A.

But if none of them be a right angle, let CD be the arch of a great circle, perpendicular to AB: and because ADC is a right angle, the fine of AC is to the radius, as the fine of CD to the fine

of A For the fame reafon, the

radius is to the fine of CB, as the B

fine of B to the fine of CD; there

D

fore, by perturbate equality, the fine of AC is to the fine of b 23. 5. CB, as the fine of B to the fine of A.

IN oblique angled triangles, if a perpendicular be

drawn from the vertex to the bafe, or the base produced; and the fegments of the bafe, or the fegments of the vertical angle, contained between the perpendicular and the fides, be taken, with which to compare the other parts of the triangle: most of the cafes of oblique angled triangles may be refolved by the following general proportion.

GENE

Sph. Trig.

GENERAL PROPORTION

III.

THE fines of the parts adjacent to the perpendicular, are reciprocally proportional to the tangents of the parts adjacent to them; and directly proportional to the cofines of the parts feparated from

them.

But the cofines of the parts adjacent to the perpendicular are to be used inftead of their fines, when they are compared with the fides of the triangle.

Let ABC be an oblique angled triangle, and let CD be perpendicular to AB, and fall either within or without the triangle. First, Let the fegments of the base AD, DB be taken, with which to compare the other parts of the triangle ABC; then the angles at A, B are the parts adjacent to them: and because ADC is a right angled triangle, the fine of AD is to the radius, as the tangent of CD to the tangent of A 2. For the same reason, the radius is to the fine of BD, as the tangent of B to the tanb 23.5. gent of CD; therefore, by perturbate equality, the fine of AD is to the fine of BD, as the tangent of B to the tangent of A; which is the first part of the proportion.

2 19. S. T.

Alfo, the fides AC, CB are separated from AD and DB: and because ADC is a right angle, the cofine of AD is to the radius, C 2. G. P. as the cofine of AC to the cofine of CD: and for the fame

reafon, the radius is to the cofine of DB, as the cofine of DC to d 22.5. the cofine of CB; therefore, by equality d, as the cofine of AD to the cofine of DB, fo is the cofine of AC to the cofine of CB; which is the fecond part of the general proportion.

Secondly, Let the vertical angles ACD, BCD be taken, with which to compare the other parts of the triangle ABC; then, the fides AC, CB are adjacent to the vertical angles and because ADC is a right angle, the cofine of ACD is to the radius,

as

as the tangent of CD to the tangent of CA : and for the fame Sph. Trig. reason, the radius is to the cofine of BCD, as the tangent of BC to the tangent of CD; therefore, by perturbate equality, e 1. G. P. as the cofine of ACD to the cofine of BCD, fo is the tangent of BC to the tangent of CA; which is the first part of the proportion.

c

b 23. 5.

Alfo, the angles at A, B are separated from the vertical angles ACD, BCD: and because ADC is a right angle, the fine of ACD is to the radius, as the cofine of A to the cofine of CD ; C 2. G. P. and for the fame reason, the radius is to the fine of BCD, as the cofine of GD to the cofine of B; therefore, by equality d, d 22. 5, as the fine of ACD to the fine of BCD, so is the cofine of A to the cofine of B; which is the fecond part of the general proportion.

IN

PROP. XXII.

N any fpherical triangle, if a great circle be drawn from the vertex, perpendicular to the bafe; the tangent of half the fum of the fegments of the base, is to the tangent of half the fum of the fides, as the tangent of half the difference of the fides to the tangent of half the difference of the fegments of the base.

Let ABC be a spherical triangle, and let the arch CD be perpendicular to AB; the tangent of half the fum of AD, DB is to the tangent of half the sum of the fides AC, CB, as the tangent of half the difference of AC, CB, to the tangent of half the difference of AD, DB.

Let AL be the common fection of the planes AC, AB, and O the centre; make CE, CF each equal to CB; therefore AE is the fum of AC, CB, and AF is their difference: and if AC be equal to CL, A is the pole of CD; therefore AD is equal to DL; and the tangent of half the difference of AC, CB is the cotangent of half their fum, and the tangent of half the difference of AD, DB is the cotangent of half their fum. But if AC be not equal to CL, join EF, and let it meet AL in M; and join BM meeting the arch AB again in G; and join CO, OD, and let them meet EF, BG in H, K; and join BH; and through A draw a AN parallel to OC, meeting the circle ACL a 31. 1. again in P; and join EA, EP, PL: and becaufe EC is equal to CF, OC bifects EF at right angles in H; and because CB is bCor.30.3. equal to CE, CH is the verfed fine of CB, and therefore OHB

C

is a right angle : and OHE is a right angle; therefore OH is c 4. Def. perpendicular P. T.

e 18. 11.

f

Sph. Trig. perpendicular to the plane MBH ; the plane MBH is therefore perpendicular to the plane COD paffing through OH: and the d 4. 11. plane ADB is alfo perpendicular to the plane COD, because ADC is a right angle; therefore their common section BG is f 19. 11. perpendicular to the plane COD; the angle OKB is therefore 8 3.3. a right angle, and BK is equal to KG 8, and the arch BD to the h30. 3. arch DG; therefore AB, AG are the fum and difference of AD, DB and because AN is parallel to OC, ANE is a right angle ; therefore, if EN be made the radius, AN is the tangent 1 of AEN, half of the angle at the centre ftanding upon m 20. 3. AE m, and PN is the tangent of PEN, or the cotangent n of EPN, that is, of half the arch AE: but because the angle APL in the femicircle is equal to the right angle ANM; AM is to ML, as AN to NP P; that is, as the tangent of half AF to the cotangent of half AE. In the fame manner, if a straight

k 29. .

1 1. P. T.

n 10. Def.

P. T.

0 31.3.

p 2. 6.

k

q 11. 5.

line be drawn through A, parallel to OD, it may be proved, that AM is to ML, as the tangent of half AG to the cotangent of half AB; therefore, as the tangent of half AF to the cotangent of half AE, fo is the tangent of half AG to the cotangent of r 16. 5. half AB; and, alternately, the tangent of half AG is to the tangent of half AF, as the cotangent of half AB to the cotangent of half AE; that is, as the tangent of half AE to the tangent of half AB, because the tangents of two arches are re$5.Cor.10. ciprocally proportional to their cotangents. Wherefore, &c. Def. P. T.

t

COR. Hence, the tangent of half the bafe is to the tangent of half the fum of the fides, as the tangent of half the difference of the fides to the tangent of half the difference of the fegments of the bafe, when the perpendicular falls within, or to the tangent of half the fum of the fegments when it falls without the triangle.

LEMMA.

IN

LE M M A.

N a plane triangle, the rectangle contained by half of the bafe and the diftance of the perpendicular from the middle of the bafe, is equal to the rectangle contained by the fines of half the fum and half the difference of the arches of any circle, which are fubtended by the fides of the triangle.

Let ABC be a triangle, and bifect the base BC in D, and draw AE perpendicular to BC: and let FGH be any circle in which are placed FG equal to AB, and GH to AC: the rectangle BD, DE is equal to the rectangle contained by the fines of half the fum and half the difference of the arches FG, GH.

Spb. Trig.

2.10. 1.

Join FH, and bifect it in K; and draw GM, KN perpendicular to KH, and GOP parallel to it; therefore the angle b 11. 12.1 FGP is equal to GFH; and the arch FP equal to GH: and C 31. 1. because KN bifects FH at right angles, it paffes through the centre, and therefore it bifects the ftraight line GP and the arches FGH, PNG". Wherefore FK is the fine of FN halff

the fum of the arches FG, GH, and KM or GO is the fine k of half PG the dif ference of the arches FG, GH: but four

k

d 29. 1.

e 27.3.

Cor. 1. 3.

g 3.3. h 30. 3. 3. Def.

P. Te

times the rectangle B

FK, KM is equal to

twice the rectangle

FH, KM; that is, to

the difference of the

fquares of FG, GH, or of BA, AC; and four times the 1 C... rectangle BD, DE is also equal to the difference of the squares of BA, AC '; therefore the rectangle BD, DE is equal to the rectangle FK, KM, contained by the fine of half the fum of the arches FG, GH and the fine of half their difference.