the measure of the angle BAC, because A is the pole of GH; Sph. Trig. therefore DE is the fupplement of the measure of the angle

BAC. In the fame manner, it may be proved, that DF is the C 3. S. P. fupplement of the measure of ABC, and EF of ACB.

Again, because A, B are the poles of DE, DF, each of the arches AG, BM is the fourth part of the circumference b; therefore AG, BM, that is, AB, GM together, are half of the circumference: and MG is the measure of the angle EDF c, because D is the pole of MG; therefore the measure of the angle EDF is the fupplement of the fide AB. In the fame manner, it may be proved, that the measures of the angles DEF, DFE are the fupplements of the fides AC, BC, in the triangle ABC.

COR. If the circles DF, EF meet again in P, the triangle DEP is femi-fupplemental to the triangle ABC; that is, the fide DE, and the measure of the oppofite angle DPE, are the fupplements of the measure of BAC, and of BC: but the fides DP, PE are the measures of ABC, ACB; and the measures of PDE, PED are equal to the fides AB, AC.

b

2. S. P.

THE

THE three angles of a fpherical triangle are together greater than two, and lefs than fix right. angles.

Because the measure of each of the angles at A, B, C, together with the fide oppofite to it of the triangle DEF, is equal to half the circumference ; the meafures of the three angles a 13. s. F. A, B, C, together with the fides of the triangle DEF, are three halves of the circumference: and the three fides of the triangle DEF are lefs than the circumference ; therefore the measures b 10. s. P. of the angles at A, B, C are together greater than half of the circumference; and the angles are therefore greater than two right angles.

b

And because all the exterior and interior angles of a triangle are equal to fix right angles, the interior angles are less than fix right angles.

COR. Because DE, EF are greater than DF; twice DF, to- c 9. s. F. gether with the measures of the angles at A, B, C, are less than three halves of the circumference ; and therefore the three angles at A, B, C, together with twice the fupplement of the least of them, are lefs than fix right angles.

PROP.

Sph. Trig.

I'

F arches of great circles be drawn to the circumference of any great circle, from a point in the fuperficies of the fphere, which is not its pole; the greatest of them is that which paffes through the pole, and its fupplement is the leaft; and of the others, that which is nearer to the greatest, is greater than that which is more remote.

Let ABD be a great circle, of which E is the pole, and let F be any other point in the fuperficies of the fphere, and let the great circle AFD pass through the points E, F, and meet the circle ABD in the diameter AD: Of all the arches of great circles FB, FC, &c. that can be drawn from F to the circumference ABD, FA is the greatest, and FD the leaft; and of the others, FB which is nearer to FA, is greater than FC which is

more remote.

Draw FG perpendicular to AD, and join GB, GC, FA, FB, FC, FD: And because the circle AFD is perpendicular to the ar. Cor.4 circle ABD 2, for it paffes

S. P.

through its pole E, and that

FG is at right angles to their
common fection AD, FG is

b

b4.Def.11. perpendicular to the plane
ABD; therefore FGB, FGC
are right angles. And because
G is a point in the diameter A
AD, which is not the centre,
for a fraight line from E to

S. P.

the centre is perpendicular to

c1.Cor. 3. AD ; GA which paffes through the centre is the greateft d, and GD the least of all the ftraight lines that can be drawn ₫ 7.3. from G to the circumference ABD; and GB which is nearer

€ 47. 1. f 15. 3.

to GA, is greater than GC which is more remote: and because GA is greater than GB, the fquares of AG, GF are greater than the fquares of BG, GF; that is, the fquare of AF is greater e than that of BF; therefore AF is greater than BF: and they are in equal circles; therefore the arch AF is greater than the arch BF. In the fame manner, it may be demonftrated, that the arch BF is greater than the arch CF, and the arch CF than the arch DF. Wherefore FA is the greateft, and FD the leaft, and FB greater than FC.

PROP.

IN

Na right angled spherical triangle, the fides about the right angle are of the fame affection with their oppofite angles; that is, if one of the fides be greater than the fourth part of the circumference, the angle oppofite to it is greater than a right angle; and if equal, equal; and if less, lefs.

Sph. Trig.

Let ABC be a spherical triangle, having BAC a right angle, the fide AB is of the fame affection with the oppofite angle ACB. If AB be the fourth part of the circumference, B is the pole 2 a 2. Cor. 4. of AC, and ACB is a right angle.

G

S. P.

bi.Cor.4.

If AB be not the fourth part of the circumference, take AE the fourth part of it, and let a great circle pass through

b

S. P.

E and C; therefore E is the pole of AC 2, and ECA is a right a 2. Cor.4. angle But if AB be greater than AE, the angle ACB is greater than ACE; and if AB be less than AE, the angle ACB is less than ACE; that is, than a right angle.

IN

PROP. XVII.

Na right angled fpherical triangle, if the fides about the right angle be of the fame affection, the hypotenufe is lefs than the fourth part of the circumference; and if they be of different affection, the hypotenufe is greater than the fourth part of the circumference: or if one of the other two be the fourth part of the circumference, the hypotenufe is equal to it.

Let

Sph. Trig.

Let ABC be a spherical triangle, having the right angle BAC; if one of the fides AB, AC be the fourth of the circumference, the hypotenufe BC is also the fourth; but if AB, AC be both less, or both greater, the hypotenufe is less than the fourth part of the circumference; and if one of them be less, and the other greater, the hypotenufe is greater.

Let the circumferences AB, AC meet again in D, and bisect a 2. S. P. the arch ABD in E; therefore E is the pole of ACD a; bifect alfo ACD in G, and G is the pole of ABD, and let a great circle pafs through E, C: If the point B coincide with E, that is, if AB be the fourth part of the circumference, BC coincides with EC, and is the fourth part of the circumference. If AB, AC be both lefs than AE or AG; CE is nearer to CGD which paffes through the pole G, than CB is; therefore CB is b 15. S. P. lefs b than CE the fourth part of the circumference. In the fame manner, in the triangle CBD, of which the fides CD, BD are each of them greater than CE, the hypotenufe CB is lefs than CE.

b

But if AB be greater than AE, and AC lefs, the arch CB is nearer to CGD than CE is; therefore CB is greater than CE the fourth part of the circumference.

COR. 1. On the contrary, if the hypotenuse be less than the fourth part of the circumference, the fides are of the fame affection; for if they be of different affection, the hypotenuse is greater than the fourth: and if the hypotenuse be greater than the fourth part of the circumference, the fides are of different affection.

COR. 2. Hence, in a right angled triangle, if the angles be of the fame affection, the hypotenufe is less than the fourth part of the circumference: and if they be of different affection, the hypotenufe is greater than the fourth and converfely. For the angles are of the fame affection with their oppofite fides.

IN

:

PROP. XVIII.

N any fpherical triangle, if the perpendicular drawn from the vertex to the bafe, fall within the triangle, the angles at the base are of the fame affection and if it fall without, they are of different affection.

Let ABC be a fpherical triangle, and let the arch CD of a great circle be drawn from C perpendicular to AB.

a 16. s. T.

First, Let CD fall within the triangle ABC; and becaufe Sph. Trig. ADC is a right angled triangle, the angle CAD is of the fame n affection with the oppofite fide CD 2. For the fame reason, the angle CBA is of the fame affection with CD; therefore the angles CAB, CBA are of the fame affection.

Next, Let the perpendicular CD fall

B

D

C

without the triangle; and because CDA is a right angle, the
angle CAD is of the fame affection with CD *; that is, with
the angle CBD; but CAD, CAB are
of different affection, because they are
together equal to two right angles;
therefore the angles CAB, CBA are of
different affection.

COR. Hence, if the angles at A, B be of the fame affection, the perpendicular falls within the triangle; for, if it fell without, they would be of different

B

A

affection. And if the angles at A, B be of different affection, the perpendicular falls without the triangle; for, if it fall within, they would be of the fame affection.

IN

PROP. XIX.

Na right angled fpherical triangle, the fine of either of the fides about the right angle, is to the radius of the fphere, as the tangent of the other fide to the tangent of the angle opposite to it.

Let ABC be a spherical triangle, having the right angle BAC; as the fine of the fide AB to the radius, fo is the tangent of the fide AC to the

tangent of the angle ABC.

to

Let D be the centre of the fphere, and join DA, DB, DC, and from the point A draw AF perpendicular BD; and from F draw b FE, in the plane BDC, at right angles to BD; and let it meet DC in E, and join AE and D because DF is at right angles, both to AF and FE, it is perpendicular to the plane AFE;

:

Kk therefore