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for the plane angle formed by every two contiguous faces of the tetračdron, 70° 31' 42”; of the hexačdron, 90°; of the octaëdron, 109°28' 18"; of the dodecaedron, 116° 33' 54"; of the icosačdron, 138° 11’ 23”. But, in these polyedrae, 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: - Solid Angle.

360°:3.70°31'42" — 180°::1000: 87-73611 tetrašdron. 360°:3.90° – 180°::1000:250. hexaedron. 360°:4.109°28'18"–360°:1000:216-351850ctaèdron. 360°:3.116°33'54"–180°::1000:471-895 dodecaedron. 360°:5.138°11'23"-540°::1000:419'30169 icosačdron.

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 comes; 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 comes; the axis of the }. is 34/3, of the latter, 3 v2, the slant side of each being unity. Hence, Angle at vertex. 1 : 1 v3: 1000: 133.97464, equilateral cone, 1 : 1 – v2::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 §: regular ones, if x 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 — (N 2) 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 .# 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 †. will be for 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 theo being thus explaimed and illustrated, the further application of it is left to the skill and ingenuity of geometers; the following simple examples, merely, being added here. Ez. 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 trilateral 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. Er. 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 each face . . . . . . . . . . . . . . . . . 66° 35' 12".

Solid angle at the vertex 89-60648 U. The max. angle Each ditto at the base .. 218-19367 being 1000.

Logarithmic Computation of Spherical Triangles.

1. For the purposes of exemplifying the rules and formulae in the preceding chapter, and of assisting the student in deducing the logarithmic computations from the analytical expressions, a few problems are here added.

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In the right angled spherical triangle ABC, right angled at A, given the hypothemuse 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 sin given side. sin hypoth.

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

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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. and sin mid. = rect cos op. becomes, sin b = sin a sin B. - sin b sin given leg. Hence sin E = i = # > 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 c = cot 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 b and c are the opposite parts, and sin mid. = rect cos oppos. becomes cos a

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cos hypoth.

cos given leg preceding. Hence the logarithmic computation need not be repeated.

that is, cos c = ; which also agrees with the

Example II.

Given in a spherical triangle, the quadrantal side CA = 90°, an j angle c = 42°12', and the opo angle B = 115°20'; to find the other angle and Sicies.

Suppose the side ch produced until co = cA = 90°: then, it is evident from chap. vi. art. 33, c 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 Z ABD = 180°– ABC = 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,

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sin given side cosgiven angle -- -sin op. angle cos given side

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

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