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SPHERICAL TRIGONOMETRY.

T

DEFINITIONS.

I.

in the fuper

HE pole of a circle of the sphere is a point
ficies of the sphere, from which all straight lines drawn

to the circumference of the circle are equal.

II.

A great circle of the sphere is any whose plane passes through the centre of the sphere, and whose centre therefore is the fame with that of the sphere.

III.

A spherical triangle is a figure upon the superficies of a sphere

comprehended by three arches of three great circles, each of which is less than a femicircle.

IV.

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

PROP. I.

GREAT circles bisect one another.

As they have a common centre their common section will be a diameter of each which will bisect them,

PROP. II. FIG. I.

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

Let ABC be a great circle and Dits 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 section of the great circles, which will

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pass through E the centre 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 confe quently each of them is a quadrant. Q. E. D.

PROP. III. FIG. 2.

IF a great circle be described meeting two great circles AB, AC paffing through its pole A in B, C, the angle at the centre of the sphere 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 an other in the straight line AD paffing through D their common centre; 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 AČ, and draw AE a straight line to any other point E in BC; join DE: Since 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.

PROP. IV. FIG. 3.

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

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

Let

Let D be the centre of the sphere, and join DA, DE, DC; in DA take any point E, from which draw, in the plane ADC, the straight line EF at right angles to ED meeting CD in F, and draw, in the plane ADB, EG at right angles to the same 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 straight 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 section 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 stand upon them at the centre 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.

PROP. V. FIG. 3.

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

Read the construction and demonstration of the preceding propofition, unto the words, "and the fides AC, CB," &c. and the rest 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 opposite to two of the equal angles angles is common; therefore (26. 1.) EF is equal to FH; and in the right-angled triangles DEF, DHF the side DF is common; wherefore (47.1.) ED is equal to DH, and the angles EDF, HDF are therefore equal, (4.1.) and consequently 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 centre of the sphere ; join DA, DB, DC.

The solid 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.

PROP. VII. FIG. 4.

THE three fides of a spherical triangle are less than a

circle.

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

Let D be the centre of the sphere: The solid angle at Dis contained by three plane angles BDA, BDC, ADC, which together are less than four right angles, (21. 11.) therefore the fides AB, BC, AC together, will be less than four quadrants; that is, less than a circle.

PROP. VIII. FIG. 5.

IN a spherical triangle the greater angle is opposite to the greater fide; and converfely.

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,

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DC

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 opposite to the greater side BC. The converse is demonftrated as prop. 19. 1. El. Q. E. D.

PROP. IX. FIG. 6.

IN any spherical triangle ABC, if the sum of the fides AB, BC be greater, equal, or less than a femicircle, the internal angle at the base 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 less 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 semicircle, 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 shewn, that if AB, BC together be less than a femicircle, the angle A is less than the angle BCD. And fince the angles BCD, BCA are equal to two right angles, if the angle A be greater than BCD, A and ACB together will 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 less than BCD, A and ACB will be less than two right angles.

E. D.

PROP. X. FIG. 7.

IF the angular points A, B, C of the spherical triangle 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 supplements of the measures

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