PROP. III.

F the pole of a great circle be the fame with the interfection of other two great circles; the arch of the first-mentioned circle intercepted between the other two, is the measure of the fpherical angle which the fame two circles make with one another.

Let the great circles BA, CA on the fuperficies of a sphere, of which the centre is D, interfect one another in A, and let BC be an arch of another great cirele, of which the pole is A; BC is the measure of the spherical angle BAC.

Join AD, DB, DC; fince A is the pole of BC, AB, AC are quadrants, (2.), and the angles ADB, ADC are right angles; therefore (4. def. 2. Sup.), the angle CDB is the inclination of the planes of the circles AB, AC, and is (def. 3.) equal to the fpherical angle BAC. Q. E. D.

*

COR. If two arches of great

circles, AB and AC, which inter

B

A

fect one another in A, be each of them quadrants, A will be the pole of the great circle which paffes through B and C, the extremities of thofe arches. For fince the arches AB and AC are quadrants, the angles ADB, ADC are right angles, and AD is therefore perpendicular to the plane BDC, that is to the plane of the great circle which paffes through B and C. The point A is therefore (Cor. 1. 2.) the pole of the great circle which paffes through B and C.

PROP.

When in any reference no mention is made of a Book, or of the Plane

Trigonometry, the Spherical Trigonometry is meant.

PROP. IV.

F the planes of two great circles of a sphere be at right angles to one another, the circumference of each of the circles paffes through the poles of the other; and if the circumference of one great circle pass through the poles of another, the planes of these circles are at right angles.

Let ACBD, AEBF be two great circles, the planes of which are at right angles to one another, the poles of the circle AEBF are in the circumference ACBD, and the poles of the circle ACBD in the circumference AEBF..

From G the centre of the fphere, draw GC in the plane ACBD perpendicular to AB. Then, because GC, in the plane ACBD at right angles to the plane AEBF, is at right angles to the common fection of the two planes, it is (Def. 2. 2. Sup.) alfo at right angles to the plane AEBF, and therefore (Cor.1. 2.) C is the pole of the circle AEBF; and if CG be produced to D, D is the other of the circle AEBF.

In the fame manner, by drawing GE in the plane AEBF, perpendicular to

A

F

G

B

E

AB, and producing it to F, it is fhewn that E and F are the poles of the circle ACBD. Therefore, the poles of each of thefe circles are in the circumference of the other.

Again, If C be one of the poles of the circle AEBF, the great circle ACBD, which paffes through C, is at right angles to the circle AEBF. For, CG being drawn from the pole to the centre of the circle AEBF is at right angles (Cor. 1. 2.) to the plane of that circle; and therefore, every plane paf

fing through CG (17. 2. Sup.) is at right angles to the plane AEBF; now, the plane ACBD paffes through CG. Therefore, &c. Q. E. D.

COR. 1. If of two great circles, the firft paffes through the poles of the fecond, the fecond alfo paffes through the poles of the firft. For, if the firft paffes through the poles of the fecond, the plane of the first must be at right angles to the plane of the second, by the fecond part of this propofition; and therefore, by the first part of it, the circumference of each paffes through the poles of the other.

COR. 2. All great circles that have a common diameter have their poles in the circumference of a circle, the plane of which is perpendicular to that diameter.

PROP. V.

'N ifofceles spherical triangles the angles at the base are equal.

IN

Let ABC be a spherical triangle, having the fide AB equal to the fide AC; the spherical angles ABC and ACB are equal.

Let D be the centre of the fphere; join DB, DC, DA, and from A on the ftraight lines DB, DC, draw the perpendiculars AE, AF; and from the points E and F draw in the plane DBC the ftraight lines EG, FG perpendicular to DB and DC, meeting one another in G: Join AG.

Because DE is at right angles to each of the straight lines AE, EG, it is at right angles to the

A

ရွာ

F

G

D

E

B

plane AEG, which paffes through AE, EG, (4. 2. Sup.); and therefore, every plane that paffes through DE is at right

angles

angles to the plane AEG (17. 2. Sup.); wherefore, the plane DBC is at right angles to the plane AEG. For the fame reason, the plane DBC is at right angles to the plane AFG, and therefore AG, the common fection of the planes AFG, AEG is at right angles (18. 2. Sup.) to the plane DBC, and the angles AGE, AGF are confequently right angles.

But, fince the arch AB is equal to the arch AC, the angle ADB is equal to the angle ADC. Therefore the triangles ADE, ADF, have the angles EDA, FDA equal, as also the angles AED, AFD, which are right angles; and they have the fide AD common, therefore the other fides are equal, viz. AE to AF, (26. 1.) and DE to DF. Again, because the angles AGE, AGF are right angles, the fquares on AG and GE are equal to the fquare of AE; and the fquares of AG and GF to the fquare of AF. But the fquares of AE and AF are equal, therefore the fquares of AG and GE are equal to the fquares of AG and GF, and taking away the common fquare of AG, the remaining squares of GE and GF are equal, and GE is therefore equal to GF. Wherefore, in the triangles AFG, AEG, the fide GF is equal to the fide GE, and AF has been proved to be equal to ÃE, and the base AG is common, therefore, the angle AFG is equal to the angle AEG (8. 1.). But the angle AFG is the angle which the plane ADC makes with the plane DBC (4. def. 2. Sup.) because FA and FG, which are drawn in these planes, are at right angles to DF, the common fection of the planes. The angle AFG (3. def.) is therefore equal to the fpherical angle ACB; and, for the fame reafon, the angle AEG is equal to the fpherical angle ABC. But the angle AFG, AEG are equal. Therefore the fpherical angles ACB, ABC are alfo equal. Q. E. D.

PROP.

PROP. VI.

F the angles at the base of a spherical triangle be equal, the triangle is ifofceles.

Let ABC be a spherical triangle having the angles ABC, ACB equal to one another; the fides AC and AB are also equal.

Let D be the centre of the sphere; join DB, DC, DA, and from A on the straight lines DB, DC, draw the perpendicular AE, AF; and from the

points E and F, draw in the plane
DBC the straight lines EG, FG A
perpendicular to DB and DC,
meeting one another in G; join
AG.

Then, it may be proved, as was done in the laft propofition, that AG is at right angles to the plane BCD, and that therefore the angles AGF, AGE are right angles, and alfo that the angles D AFG, AEG are equal to the

G

E B

angles which the planes DAC, DAB make with the plane DBC. But because the spherical angles ACB, ABC are equal, the angles which the planes DAC, DAB make with the plane DBC are equal, (3. def.) and therefore the angles AFG, AEG are alfo equal. The triangles AGE, AGF have therefore two angles of the one equal to two angles of the other, and they have alfo the fide AG common, wherefore they are equal, and the fide AF is equal to the fide'AE.

Again, because the triangles ADF, ADE are right angled at F and E, the fquares of DF and FA are equal to the square of DA, that is, to the squares of DE and EA; now, the fquare of AF is equal to the fquare of AE, therefore the fquare of DF is equal to the fquare of DE, and the fide DF to the fide DE. Therefore in the triangles DAF, DAE, because DF is equal to DE, and DA common, and alfo AF equal to AE, the angle ADF is equal to the angle ADE;

therefore