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(53.) Let the preceding substitutions be made in the formula O, as converted in (51); that is, for

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COS.

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write i
+

&c.

2 p2 2.3.4 pt which will give Ige 1 r

a? + B2 at + 6a'B'+84 1 + + &c. 1+

&c. 2 r2 24 pt

27%

24 pt
αβ af (@?+)
+
cos. C-

cos. C + &c. r2

6r4 Take both sides from 1, and multiply by 2ro,

at+62 82 + 34-4040 (a +62) cos. C ge=a? +B2 2aß cos. C

+ &c.

12 12

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a

in which the succeeding terms have r4, rs, &c., in their denominators. By increasing r without limit, the terms which have powers of r in their denominators are diminished without limit; rejecting these terms, there remains

72 = a + B2 - 2aß cos. C,

? a well-known property of a rectilinear triangle, whose sides a and ß contain the angle C.

(54.) The following table contains references to the Articles of the Treatise on Trigonometry, which contain the formulæ, to which the preceding formulæ are reducible, by the process, and under the suppositions, of the last article. It must be remembered, however, that a, b, and c, in the Treatise on Trigonometry, correspond to a, b, and y, in the results which the student will obtain. In some cases the resulting formulæ are given, being reducible to geometrical facts too well known to be specially mentioned in the Trigonometry.

(R) a? + B2 = 72; (Rg, Ro R) A + B = 90°; (R3, R., R;) Tr. 54; (0, 0) Tr. 55; (02 , 09) Tr. 56; (0) Tr. 57 ; (0) Tr. 53. (55.)*Returning to (51), the spherical excess (43) 2S – becomes

īt less and less, as 0 is removed farther from the chordal triangle, and vanishes altogether at the limit, since the three angles of the chordal triangle do not exceed two right angles. But the area of the spherical triangle, or p2 (2 S 7), in which the first factor increases, and the second decreases without limit, approximates continually to the area of the chordal triangle.

By converting E, as in (50), it will appear that the limit of 2 78 sin. E is a ß sin. C. But from E, it will appear that the limit of cos. E is unity, or that E diminishes without limit. Hence since the ratio of the sine to its angle approaches without limit to unity as the angle is decreasedt, we have

sin. Σ limit of 2 72 sin. 3 = limit of 2 r2 E

limit of 2r? 2. Σ

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* The rest of this chapter must be omitted by the elementary student.

+ See the first pages of the Treatise on Elementary Illustrations of the Differential and Integral Calculus.

But the last is the limit of the area of the spherical triangle and the first is } « ß sin. C, a well-known expression for the area of the chordal triangle, a and ß being the sides containing the angle O.

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Let us now suppose r to be very great, so that the spherical triangle is nearly plane, and its spherical excess small, and let a plane triangle be taken whose sides are a, b, y. This must not be confounded with the chordal triangle. Let the angles of the latter triangle be A', B', C'. Then (Tr. 55, 56, developing sin.? C in the latter)

a + B2 - y2 cos. C' ~

2αβ 4 «?ße sin. C' = 2 * 6 + 2 B7 + 2 y al - a - b - * whence the preceding expression for cos. C becomes

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rejecting, as inconsiderable, those terms which contain p* &c. in their denominators. Let C' +z = C, then since z is small, cos. C = cos. C' z sin. C/ nearly

αβ whence

sin. C' nearly 6 r2

z =

But } a 6 sin. C' is the area of the rectilinear triangle, the sides of which are a, ß and y: which being very nearly equal to the area of the

=

spherical triangle, we may write 2 r E for it without sensible error. This
gives
2 Σ spherical excess

nearly
3

3

spherical excess C' = C

nearly

3 By proceeding in this way for the other angles, we find the following very remarkable theorem.

If a spherical triangle of very small curvature be flattened without altering the length of its sides, its angles are diminished by quantities which are very nearly equal to one another, and to the third of the spherical excess in the spherical triangle.

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CHAPTER VII.

ON SPHERICAL Polygons*.

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(57.) If the whole surface of a sphere be covered by spherical polygons,

whose sides are arcs of great circles, and if the chords of all these arcs be drawn, it is plain that the chords of each polygon will enclose one face of a solid inscribed in the sphere; which solid will have as many faces as there are polygons, as many edges as there are arcs in them all, and as many solid angles as there are distinct angular points.

(58.) And if with any point in the interior of a solid as a centre, a sphere be described, and if planes be drawn through the several edges of the solid, and the centre of the sphere, these planes intersect the sphere in as many arcs, the whole of which will divide the surface of the sphere into a number of polygons, the whole number of sides being that of the edges of the solid, and so on. If a solid be inscribed inside the sphere as in (57), we shall have a second solid inscribed in a sphere, having the same number of faces, edges, and solid angles as the first.

(58.) Hence any relation which is found to exist between the number of faces, edges, and solid angles of a solid which can be inserted in a sphere, is equally true of solids which have no circumscribing sphere.

(59.) Let E, F, and S be the number of edges, faces, and solid angles in the solid. Then the sum of all the angles of all the spherical polygons must be 4 S right angles, for every solid angle corresponds to angles of the polygons, which are together equal to four right angles. And the total number of sides in all the polygons is 2 E, since every edge has its arc

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* This chapter may be omitted at the first reading...

area is

common to two different polygons. This is also the total number of angles in all the polygons.

(60.) The area of a spherical polygon of n sides is ihus found. From any angular point draw diagonal arcs as in the figure. These divide the polygon into n

2 triangles, the sum of the angles of all of which is the sum of the angles of the polygon. If

the sums of the angles of the several triangles be A,,A,, &c., the area of the polygons is therefore (43)

pel (A,

- 7 + A, - 1 + &c.) = 1 (B — (n − 2) 7) where B is the sum of the angles of the polygon. (61.) If the polygon be equiangular, and O be one of its angles, the

po? (n 0 - (n − 2) )

2) (62.) Let B, , B,, &c., be the sum of the angles of the several polygons in (57), and n,, Na, &c., the number of sides in each. Then (60) the sum of the areas of all the polygons will be

pol [B, - (1, - 2)* + B, – (nz – 2) ++ &c.] In which are F terins (59) similar to B, – (1, - 2) 7. But (59) B. + Be + &c. = 2ST, n, +ng + &c. = 2 E

B, and the sum of all the areas is the whole sphere, or 4 7 ye. Therefore

4 T r = 2 { 2 S + – 2 ET + 2F.} or

S+F = E + 2 Therefore ;-In every solid, the number of solid angles and faces together exceeds the mumber of edges by 2. (63.) Let the solid have F, triangular faces, F, quadrilateral faces, and

Then
F = F, + F, + &c. 2 E = 3F; + 4F, + &c.

1

1

1

2

T

SO on.

3

and the preceding theorem gives

2 S= 4+ F: + 2F, + 3F, + &c. which gives

F, + F + &c. = 2 (S-2-F,-F,-&c.)

3

5

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or;--the number of faces which have an odd number of sides is always

even.

(64.) The number of all the angles of all the polygons is 2 E; this cannot be less than 35, because there must be at least three plane angles to one

solid angle.

Hence generally 2 E> 3 S except only when all the solid angles are made by three planes, in which case 2 E=3S. And the signs < and > must be understood with this limitation throughout the present article. Applying the theorem in (62) we have s < 2F - 4, and E < 3F - 6.

7

4

5

Since

2E> 3 S we have 3F3+4F.+5 F: +6F6+7F,+&c. >6+3 F: +3F.+F+6F6+"F, &c.

3F3 + 2F + F5 > 12+ F, + &c. Therefore F3, F4, F; cannot all be nothing at once; or there is no solid entirely composed of faces of a higher order than pentagonal: and if there be neither quadrilateral nor pentagonal faces, there must be more than four triangles, except only when all are triangles : if there be neither triangles nor pentagons, there must be more than six quadrilaterals, except where all are quadrilaterals: if there be neither triangles nor quadrilaterals, there must be more than twelve pentagons, except where all are pentagons. (65.) In a similar a

way

it
may

be shown that in solids which have four or more plane angles at every solid angle or 2 E > 4 S, there cannot be less than eight triangles; and that in this case S <F-2 and E< 2F-4. Also that in solids which have five or more plane angles at every solid angle, or in which 2 E > 5 S, there cannot be less than twenty triangles; and 3 S < 2 F 4, 3E < 5F - 10. Also, that a solid all whose solid angles have either five or six plane angles, the number of solid angles having five plane angles is neither more nor less than 12; and that it is impossible to form a solid all whose solid angles shall contain six or more plane angles.

(66.) We now proceed to inquire how many regular solids may be formed, that is, solids, the faces of which are all repetitions of the same equilateral and equiangular polygon. The sphere will then be covered in as many ways by equilateral and equiangular spherical polygons.

Let F be the number of such polygons and n the number of sides in each; which last must not be > 5 (65). Let T be the number of plane angles which meet at each of the solid angles; then since the T angles of the spherical polygons which lie round any point of the sphere and meet at the vertex of a solid angle, make up four right angles, every such angle is 2 T divided by T, and the sum of all the angles of one n-sided equiangular polygon is 2 na divided by T. Therefore (60) the area of such a spherical polygon is

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[ ??" – (n − 2)

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T

T

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But F such polygons cover the whole sphere, or make up 4 nr: which equated to F times the preceding, gives by reduction

4T

2n-(1-2) T We must therefore look for every value of T and n, not greater than 5, or less than 3, which will make the preceding value of F a whole number. This will give the values of F; we have then (59) E= } Fn, and S= E + 2 – F.

(67.) First, we inquire for the regular solids all whose sides are pentagons. Let n = 5; then

4T F

10 - 3T

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