SPHERICAL TRIGONOMETRY.

THE

A

DEFINITIONS.

I.

HE pole of a circle of the sphere is a point in the superficies of the fphere, from which all straight lines drawn to the circumference of the circle are equal.

II.

great circle of the sphere is any whose plane paffes through the center of the fphere, and whofe center therefore is the fame with that of the fphere.

III.

A spherical triangle is a figure upon the fuperficies of a sphere comprehended by three arches of three great circles, each of which is lefs than a femicircle.

IV.

A spherical angle is that which on the superficies of a sphere is contained by two arches of great circles, and is the same with the inclination of the planes of these great circles.

PROP. I.

GREAT circles bifect one another.

As they have a common center their common section will be a diameter of each which will bifect them.

13

TH

'HE arch of a great circle betwixt the pole and
circumference of another is a quadrant.

Let ABC be a great circle and D its pole; if a great circle DC pass through D, and meet ABC in C, the arch DC will be a quadrant.

Let the great circle CD meet ABC again in A, and let AC be the common fection of the great circles, which will pafs

through E the center of the sphere: join DE, DA, DC: by def. 1. DA, DC are equal, and AE, EC are also equal, and DE is common; therefore (8. 1.) the angles DEA, DEC are equal; wherefore the arches DA, DC are equal, and confequently each of them is a quadrant. Q. E. D.

PROP. III. FIG. 2.

F a great circle be defcribed meeting two great circles AB, AC paffing through its pole A in B, C, the angle at the center of the fphere upon the circumference BC, is the fame with the spherical angle BAC, and the arch BC is called the measure of the spherical angle BAC.

Let the planes of the great circles AB, AC interfect one another in the straight line AD paffing through D their common center; join DB, DC.

Since A is the pole of BC, AB, AC will be quadrants, and the angles ADB, ADC right angles; therefore (6. def. 11.) the angle CDB is the inclination of the planes of the circles AB, AC; that is, (def. 4.) the spherical angle BAC. Q. E. D.

COR. If through the point A, two quadrants AB, AC, be drawn, the point A will be the pole of the great circle BC, paffing through their extremities B, C.

Join AC, and draw AE a straight line to any other point E in BC; join DE: fince AC, AB are quadrants, the angles ADB, ADC are right angles, and AD will be perpendicular to the plane of BC: therefore the angle ADE is a right angle, and AD, DC are equal to AD, DE each to each; therefore AE, AC are equal, and A is the pole of BC, by def. 1. Q. E. D.

IN

PROP. IV. FIG. 3.

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

Let ABC be an isofceles triangle, and AC, CB the equal fides; the angles BAC, ABC, at the bafe AB, are equal.

Let D be the center of the sphere, and join DA, DB, DC; in DA take any point E, from which draw, in the plane ADC,

the ftraight line EF at right angles to ED meeting CD in F, and draw, in the plane ADB, EG at right angles to the fame ED; therefore the rectilineal angle FEG is (6. def. 11.) the inclination of the planes ADC, ADB, and therefore is the fame with the spherical angle BAC: from F draw FH perpendicular to DB, and from H draw, in the plane ADB, the ftraight line HG at right angles to HD meeting EG in G, and join GF. Because DE is at right angles to EF and EG, it is perpendicular to the plane FEG, (4. 11.) and therefore the plane FEG is perpendicular to the plane ADB, in which DE is: (18. 11.) in the fame manner the plane FHG is perpendicular to the plane ADB; and therefore GF the common fection of the planes FEG, FHG is perpendicular to the plane ADB; (19. 11.) and because the angle FHG is the inclination of the planes BDC, BDA, it is the fame with the spherical angle ABC; and the fides AC, CB of the spherical triangle being equal, the angles EDF, HDF, which ftand upon them at the center of the sphere, are equal; and in the triangles EDF, HDF the fide DF is common, and the angles DEF, DHF are right angles; therefore EF, FH are equal; and in the triangles FEG, FHG the fide GF is common and the fides EG, GH will be equal by the 47. 1. and therefore the angle FEG is equal to FHG; (8. 1.) that is, the spherical angle BAC is equal to the spherical angle ABC.

IF, in a spherical triangle ABC, two of the angles BAC, ABC be equal, the fides BC, AC oppofite to them, are equal.

Read the construction and demonstration of the preceding propofition, unto the words, " and the fides AC, CB," &c. and the reft of the demonstration will be as follows, viz.

And the spherical angles BAC, ABC being equal, the rectilineal angles FEG, FHG, which are the fame with them, are equal; and in the triangles FGE, FGH the angles at G are right angles, and the fide FG oppofite to two of the equal angles is common; therefore (26. 1.) EF is equal to FH; and in the right-angled triangles DEF, DHF the fide DF is common; wherefore (47. 1.) ED is equal to DH, and the angles EDF, HDF are therefore equal, (4. 1.) and confequently the fides AC, BC of the spherical triangle are equal.

PROP. VI.

FIG. 4.

ANY two fides of a spherical triangle are greater

than the third.

Let ABC be a spherical triangle, any two fides AB, BC will be greater than the other fide AC.

Let D be the center of the sphere; join DA, DB, DC. The folid angle at D is contained by three plane angles ADB, ADC, BDC;, and by 20. 11. any two of them ADB, BDC are greater than the third ADC; that is, any two fides AB, BC of the spherical triangle ABC, are greater than the third AC.

HE three fides of a spherical triangle are lefs than a circle.

THE

Let ABC be a spherical triangle as before, the three fides AB, BC, AC are lefs than a circle.

Let D be the center of the fphere: the folid angle at D is contained by three plane angles BDA, BDC, ADC, which together are less than four right angles, (21. 11.) therefore the sides AB, BC, AC together, will be less than four quadrants; that is, lefs than a circle.

IN a led

Na spherical triangle the greater angle is oppofite to the greater fide; and conversely.

Let ABC be a spherical triangle, the greater angle A is opposed to the greater fide BC.

Let the angle BAD be made equal to the angle B, and then BD, DA will be equal, (5. of this) and therefore AD, DC are equal to BC; but AD, DC are greater than AC, (6. of this,) therefore BC is greater than AC, that is, the greater angle A is oppofite to the greater fide BC. The converfe is demonstrated as prop. 19. 1. El. Q. E. D.

Ν

PROP. IX. FIG. 6.

IN any spherical triangle ABC, if the fum of the fides

AB, BC be greater, equal, or lefs than a femicircle, the internal angle at the bafe AC will be greater, equal, or less than the external and opposite BCD; and therefore the fum of the angles A and ACB will be greater, equal, or lefs than two right angles.

Let AC, AB produced meet in D.

1. If AB, BC be equal to a femicircle, that is, to AD, BC, BD will be equal, that is, (4. of this) the angle D, or the angle A will be equal to the angle BCD.

2. If AB, BC together be greater than a femicircle, that is, greater than ABD, BC will be greater than BD; and therefore (8. of this) the angle D, that is, the angle A, is greater than the angle BCD.

3. In the fame manner is it fhewn, that if AB, BC together be less than a femicircle, the angle A is lefs than the angle BCD. And fince the angles BCD, BCA are equal to two right angles, A and ACB together will

if the angle A be greater than BCD, be greater than two right angles. If A be equal to BCD, A and ACB together will be equal to two right angles; and if A be lefs than BCD, A and ACB will be lefs than two right angles. Q. E. D.

F the angular points A, B, C of the fpherical triangle

IF

ABC be the poles of three great circles, these great circles by their interfections will form another triangle FDE, which is called fupplemental to the former; that is, the fides FD, DE, EF are the fupplements of the measures of the oppofite angles C, B, A, of the triangle ABC, and the measures of the angles F, D, E of the triangle FDE, will be the fupplements of the fides AC, BC, BA, in the triangle ABC.

Let AB produced meet DE, EF in G, M, and AC meet FD, FE in K, L, and BC meet FD, DE in N, H.

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