Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

HBM, HKM, therefore HB will be equal to BM, and the triangle DBG equiangular to BHK. For the same reason, MCL will be bisected by CK, and MC equal to CL.

Now, because BF, FC, are equal to BH, CL ; AH, AL, are equal to the sum of the sides AB, BC, AC, and AH equal to half the sum of the fides. And, because the triangles DBG, BHK, are equiangular, GD is to DB as BH is to İKf; and f 4. 6. the rectangle under DG, HK, equal to the rectangle DBH8, 8 16. 6. that is, to BFC; but the triangles ADG, AHK, are equiangular ; therefore AD is to DG as AH is to HK, and the rectangle under DG, HK, equal to the rectangle HAD; therefore the rectangle HAD is to the rectangle DBH, or BFC, as the square of AD is to the square of DG", but AD is the excess h 22. 6. of AH above the base BC, and BF, FC the right lines by which AH exceeds the sides AB, AC; and, if AD is taken rad. then DG is the tangent of half the angle BAC. Wherefore, &c.

[blocks in formation]

IV every plain triangle, the base is to the sum of the sides as the

difference of the sides is to the sum or difference of the segments of the base, as the greater or lefser side of the triangle is taken for

the base.

AD the

Let ABC be a triangle ; from the vertex A let fall the perpendicular AD, then the base is to the sum of the sides as the difference of the sides is to the sum or difference of the segments CD, BD, according as the base BC is the lefler or greater fide. From the vertex A, let fall the perpendicular upon

the base BC; with the center A, and distance AC, greater of the other two sides, describe the circle CEF, and produce AB both ways to F and E, and CB to G ; then, because the right lines FE, GC, cut one another in B, the reco tangle FBE is equal to the rectangle GBC"; but CB is to BF asa 35. s. BE is to GB b; that is, when the base BC is the greatest, theb 16. 6. base to the sum of the sides, as the difference of the sides to the difference of the segments of the base ; but, when BC is the least, GB is the sum of the segments of the base. Wherefore, &c.

PRO P. VI.

THE fum and difference of any two quantities being given to

find these quantities.

U

Let

Let AB, BC, be the two quantities ; place them in the same right line, as AC ; and bisect AC in E ; and cut off AD equal to BC; then DB is the difference of the two quantities, and EB half their difference; therefore, if to AE, half their fum, EB, half their difference, be added, the sum is equal to AB, the greater quantity; and if from AE, half the sum, ED, half their difference, be taken, gives AD equal to BC, the lefser quantity. Wherefore, &c,

[merged small][ocr errors]

THE

Ε L Ε Μ Ε Ν Τ S.

OF

S P H E R I C A L

TRIGONOMETRY

中中中中中中中中中中

DEFINITIONS:

I. T! HE poles of a sphere are two points in the fuperficies of the sphere that are the extremes of the axis.'

II. The pole of a circle in a sphere, is a point in the superficies of

the sphere from which all right lines, drawn to the circum. ference of the circle, are equal to one another.

III. A great circle in a sphere is that whose plain passes through the

center of the sphere, and whose center is the same with that of the sphere, or whose plain bisects the sphere.

IV: A spherical triangle, is a figure comprehended under the arches of three great circles of a sphere.

V. A spherical angle is that which is contained under two arches of

greater circles in the superficies of the sphere.

[ocr errors][merged small]

G

RE AT circles in a sphère mutually bisect each other.

a

Let the two great circles be ACB, AFB, they will mutually bisect each other; for their common section AB is the diameter of both circles.

PROP:

[blocks in formation]

I from the pole of any circle, to its

center, a right line be drawn, it will be perpendicular to the plain of that circle.

a def. 26

b 8. I.

Let the circle be AFB, and its pole C; from which draw CD to the center, then CD will be perpendicular to the plain of that circle.

For, in it draw any diameters EF, GH, and join CG, CH, CE, CF; then, in the triangles CDF, CDE, the two fides CD, DE, are equal to the two fides CD, DF, and their bases CF, CE, are equal“; therefore the angle CDF is equal to the angle CDE b; therefore.CD is perpendicular to the plain of the circle AFB. Wherefore, &c.

COR. I. Hence, if this circle be a great circle, the distance upon the superficies of the sphere betwixt the pole and great circle is a quadrant, for the plain of it bisects the sphere.

II. Great circles, that pass through the pole of some other

circle, make right angles with it ; for the right line CD is the d 19. 11. common section of such plains d.

IC 4. II.

P R O P.

LE
F a great circle is described about the pole of a sphere, and

from that pole two right lines be drawn to the circle, the arch of that circle contained by the two.right lines is the measure of the angle at the pole.

at A.

a cor. %.

Let A be the pole of a sphere, and ECF the great circle de. scribed about it, and let the right lines AC, AF, be drawn to the.great circle; then the arch CF is the measure of the angle

For, let D be the center of the sphere, then the angles ADC, ADF, are right angles"; and the angle CDF is the inclination

of the plains ACB, AFB, and equal to the spherical triangle b def. 6. ' CAF, or CBF b.

COR. I. If the arches AC, AF, are quadrants, then A is the pole of the circle passing through the points C, F; for AD is at right angles to the plain FDC.

II. The vertical angles are equal, for each is equal to the inclination of the circles ; also, the adjacent angles are equal to two right angles.

II.

C4. II.

[blocks in formation]

IF F two spherical triangles have two sides of the one, equal to two

sides of the other, and the angle contained by the two sides of the one equal to the correspondent angles of the other, the two trim angles will be equal.

For, if the two arches containing the angles are equal, their chords or subtenses are likewise equal", and contain equal angles; a 29. :3 therefore their bases are equal, and remaining angles of the one, equal to the remaining angles of the other, each to each; and the right lined triangles equal ); but equal right lines cut off equal b 4. 3• circumferences *; wherefore the spherical triangles are equal to © 28. 3. one another.

COR. I. Hence triangles will be equal and congruous, if two angles of the one be equal to two angles of the other, each to cach, and a side of the one equal to a side of the other, either the side that lies betwixt the equal angles, or subtending one of them d.

d 29. and

24. 3. and II. Equilateral triangles are likewise equiangular.o.

III. In isosceles triangles, the angles at the bases are equal ; c 29. and and, if the angles at the bases are equal, the triangles are ifor- 24. 3. and celes f.

f 29 and IV. Any two fides of a triangle are greater than the third; for 24. 3. S. any two of their chords or subtenses, are greater than the third 8. and 6, s.

26. I.

8 20, 1.

[blocks in formation]

A a

NY fide of a spherical triangle is less than a semicircle.

Let AC, AB, the fides of the triangle ABC, be produced till they meet in D, then the femicircle ACD is greater than the arch AC.

[blocks in formation]

TH HE three sides of a Spherical triangle are less than a whole

circle.

For,

« ΠροηγούμενηΣυνέχεια »