for the plane angle formed by every two contiguous faces of the tetraedron, 70° 31′ 42"; of the hexaedron, 90°; of the octaëdron, 109° 28" 18"; of the dodecaëdron, 116° 33′ 54"; of the icosaedron, 138° 11′ 23′′. But, in these polyedræ, the number of faces meeting about each solid angle, are 3, 3, 4, 3, 5, respectively. Consequently the several solid angles will be determined by the subjoined proportions:

360°:3.90°

Solid Angle. 360°:3.70°31′42′′-180°::1000: 87-73611 tetraëdron. -180°:: 1000:250 hexaëdron. 360°:4.109°28′18′′-360°:: 1000:216-35185 octaëdron. 360°:3.116°33′54′′-180°::1000:471.395 dodecaedron. 360°:5.138°11′23′′ — 540°::1000:419-30169 icosaëdron. 3. The solid angles at the vertices of cones, will be determined by means of the spheric segments cut off at the bases of those cones; that is, if right cones, instead of having plane bases, had bases formed of the segments of equal spheres, whose centres were the vertices of the cones, the surfaces of those segments would be measures of the solid angles at the respective vertices. Now, the surfaces of spheric segments, are to the surface of the hemisphere, as their altitudes, to the radius of the sphere; and, therefore, the solid angles at the vertices of right cones will be to the maximum solid angle, as the excess of the slant side above the axis of the cone, to the slant side of the cone. Thus, if we wish to ascertain the solid angles at the vertices of the equilateral and the right angled cones; the axis of the former is 3, of the latter, 2, the slant side of each being unity. Hence,

Angle at vertex.

1:1-3: 1000: 133-97464, equilateral cone, 1:12::1000: 292-89322, right angled cone.

4. From what has been said, the mode of determining the solid angles at the vertices of pyramids will be sufficiently obvious. If the pyramids be regular ones, if be the number of faces meeting about the vertical angle in one, and a the angle of inclination of each two of its plane faces; if n be the number of planes meeting about the vertex of the other, and a the angle of inclination of each two of its faces: then will the vertical angle of the former, be to the vertical angle of the latter pyramid, as NA(N2) 180°, to na · (n − 2) 180°.

If a cube be cut by diagonal planes, into six equal pyramids with square bases, their vertices all meeting at the centre of the circumscribing sphere; then each of the solid angles, made by the four planes meeting at each vertex, will be of the maximum solid angle; and each of the solid angles at the bases of the pyramids, will be of the maximum solid angle. Therefore, each solid angle at the base of such pyramid, is onefourth of the solid angle at its vertex: and, if the angle at the vertex be bisected, as described below, either of the solid angles arising from the bisection, will be double of either solid angle at the base. Hence also, and from the first subdivision of this inquiry, each solid angle of a prism, with equilateral triangular base, will be half each vertical angle of these pyramids, and double each solid angle at their bases.

The angles made by one plane with another, must be ascertained, either by measurement or by computation, according to circumstances. But, the general theory being thus explained and illustrated, the further application of it is left to the skill and ingenuity of geometers; the following simple examples, merely, being added here.

Ex. Let the solid angle at the vertex of a square pyramid be bisected.

1st. Let a plane be drawn through the vertex and any two opposite angles of the base, that plane will bisect the solid angle at the vertex; forming two trila7

teral angles, each equal to half the original quadrilateral angle.

2dly. Bisect either diagonal of the base, and draw any plane to pass through the point of bisection and the vertex of the pyramid; such plane, if it do not coincide with the former, will divide the quadrilateral solid angle into two equal quadrilateral solid angles. For this plane, produced, will bisect the great circle diagonal of the spherical parallelogram cut off by the base of the pyramid; and any great circle bisecting such diagonal is known to bisect the spherical parallelogram, or square; the plane, therefore, bisects the solid angle.

Cor. Hence an indefinite number of planes may be drawn, each to bisect a given quadrilateral solid angle. Ex. 2. Determine the solid angles of a regular pyramid with hexagonal base, the altitude of the pyramid being to each side of the base, as 2 to 1.

Ans. Plane angle between each two la

teral faces...... ..125° 22′ 35′′ between the base and

Logarithmic Computation of Spherical Triangles. 1. FOR the purposes of exemplifying the rules and formulæ in the preceding chapter, and of assisting the student in deducing the logarithmic computations from the analytical expressions, a few problems are here

added.

Example I.

In the right angled spherical triangle ABC, right angled at a, given the hypothenuse a 64° 40', and one leg b = 42° 12′. Required the rest.

This example falls under case 2, section 2, of the preceding chapter, where

cos c, a req. angle = tan given side x cot hypoth. Hence the following logarithmic operation.

From log sin 42° 12'..9-8271887 From log cos 64° 40'..9.6313258 Take log sin 64° 40′. .9·9560886 Take log cos 42° 12′..9·8697037

Rem, sin B.

48° 0'..9-8711001 Rem. cos c..54° 43'..9.7616221

To log tan 42° 12′.... 9·9574850

Add log cot 64° 40′

....9.6752372

The sum is log cos c 64° 35′.... 9.6327222

Here 10 are added to the index of the remainder, and taken from the index of the sum; conformably with note 2, art. 31 of the preceding chapter.

2dly. To compute the same by means of Napier's circular parts.

Here, if the leg b be assumed as the middle term, (90° — a), and (90° — B) are the opposite parts.

rect cos op. becomes,

and sin mid.

this agrees with the foregoing process.

Again, if (90°-c) be the middle part, then (90°-a)

and b are adjacent parts, and

sin mid.rect tan adja. becomes

cos ccot a tan b = cot hypoth. x tan given side; this also agrees with the preceding.

3dly. If (90° - a) be made the middle part, then and c are the opposite parts, and

sin mid. = rect cos oppos. becomes

cos a = cos b cos c; whence cos c =

cos a

that is, cos c =

cos hypoth. cos given leg

:; which also

cos b

preceding. Hence the logarithmic computation need not be repeated.

Example II.

Given in a spherical triangle, the quadrantal side CA = 90°, an adjacent angle c = 42° 12′, and the opposite angle B = 115° 20′; to find the other angle and sides.

C

Suppose the side CB produced until CD = CA = 90°: then, it is evident from chap. vi. art. 33, that both the angles, CAD, and D, are = 90°, and consequently that AD is the measure of the angle c, and therefore = 42° 12'. It is also evident that ABD, as well as ACD, is a right angled triangle, and that ABD 180° -ABC

A

D

64° 40'. Hence, to find AB, &c. we make use of the triangle ABD, of which we determine the hypothenuse and oblique angles by case 4, of right angled triangles. Thus,

sin hypoth. =

sin given side
sin op. angle

sin req. an. =

cos given angle cos given side From log sin 42° 12′..9.8271887 From log cos 64° 40′..96313258 Take log sin 64° 40'..9.9560886 Take log cos 42o 12'..9-8691037

Rem. sin AB

480..9.8711001 Rem. sin BAD 35°17′ ́
or cos CAB 54°43'

9.761622

sin side req. tan given side x cot op. angle,

16221