manner, each solid angle measured by 90°+90°+90°-180° =90°, which is + of the maximum angle. And thus it may be found, that each solid angle of a right prism, with an equilateral triangular base is max. angle = .1000. square base is pentagonal base is hexagonal is heptagonal is octagonal nonagonal is decagonal is Hence it may be deduced, that each solid angle of a reguJar prism, with triangular base, is half each solid angle of a prism with a regular hexagonal base. Each with regular square base = pentagonal = hexagonal = m gonal = decagonal, m gonal base. Hence again we may infer, that the sum of all the solid angles of any prism of triangular base, whether that base be regular or irregular, is half the sum of the solid angles of a prism of quadrangular base, regular or irregular. And, the sum of the solid angles of any prism of tetragonal base is = sum of angles in prism of pentag. base, 2. Let us compare the solid angles of the five regular bodies. In these bodies, if m be the number of sides of each face; n the number of planes which meet at each solid angle; = half the circumference or 180°; and a the plane angle 1 COS 2n made by two adjacent faces: then we have sin A=- 1 sin This theorem gives, for the plane angle formed by every two contiguous faces of the tetraedron, 70°31′42"; of the hexaedron, 90°; of the octaedron, 109°28′18′′; of the dodecaedron, 116°33′54′′; of the icosaedron, 138°11′23′′. But, in these polyedræ, the number of faces meeting about each solid angle, 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 Tetraëdron. 360°: 3.90° Hexaëdron. -180°:: 1000: 250 360°: 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 coness will be to the maximum solid angle, as the excess of the slånt 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 rightangled 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, 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 n 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 u 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 6 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 one-fourth 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 subdivision of this scholium, 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 example, 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 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. SECTION II. Resolution af Spherical Triangles. THE different cases of spherical trigonometry, like those in plane trigonometry, may be solved either geometrically or algebraically. We shall here adopt the analytical method, as well on account of its being more compatible with brevity, as because of its correspondence and connexion with the substance of the preceding chapter*. The whole doctrine may be comprehended in the subsequent problems and theorems. For the geometrical method, the reader may consult Simson's or Playfair's Euclid, or Bishop Horsley's Elementary Treatises on Practical Mathematics. PROBLEM PROBLEM I. To Find Equations, from which may be deduced the Solution of all the Cases of Spherical Triangles. Let ABC be a spherical triangle; AD the tangent, and GD the secant, of the are AB; AE the tangent, and GE the se cant, of the arc Ac; let A tions in art. 6 PI. Trig. applied to the two triangles ADE, GDE, give, for the former, DE = tan b + tan2 c- tan b. tan c. cos A; for the latter, DE2 = sec2 b + sec2 c - sec b secc.cos a. Substracting the first of these equations from the second, and observing that sec2 b - tan2 b = R = 1, we shall have, after a little reduction, 1 + 2 sin b. cos c COS A COS a cos b.cos c 0. Whence the three following symmetrical equations are obtained: COS B } (I.) THEOREM VII. In Every Spherical Triangle, the Sines of the Angles are Proportional to the Sines of their Opposite Sides. If, from the first of the equations marked 1, the value of cos a be drawn, and substituted for it in the equation sin2 A = 1 sin2 A = 1 cos2 a + cos2 b. cos2c-Icosa.cost.cos o sinb sin c Reducing the terms of the second side of this equation to a common denominator, multiplying both numerator and denominator by sin2 a, and extracting the sq. root there will result sin A = sin a. √(1-cos2 a- cos2b- cos2c +2cosa.cosb.cos c) sin a. sin b. sinc Here, if the whole fraction which multiplies sin a, be denoted by K (see art. 8 chap. iii), we may write sin A = K. sin a. And, since the fractional factor, in the above equation, contains terms in which the sides a, b, c, are alike affected, we have similar equations for sin B, and sin c. That is to say, we have In Every Right-Angled Spherical Triangle, the Cosine of the Hypothenuse, is equal to the Product of the Cosines of the Sides Including the right angle. For, if a be measured by O, its cosine becomes nothing, and the first of the equations I becomes cos a = cos b.cosc. Q. E. D. THEOREM IX. In Every Right-Angled Spherical Triangle, the Cosine of either Oblique Angle, is equal to the Quotient of the Tangent of the Adjacent Side divided by the Tangent of the Hypothenuse. If, in the second of the equations 1, the preceding value of cos a be substituted for it, and for sin a its value tan a. cos a= cosa.cosb.cos c; then, recollecting that 1-cos2 c = sin c, there will result, tan a. cOS C. COS B = sinc: whence it follows that, tan a. cos B = tan c, or cos B THEOREM X. tane In Any Right-Angled Spherical Triangle, the Cosine of one of the Sides about the right angle, is equal to the Quotient of the Cosine of the Opposite angle divided by the Sine of the Adjacent angle. |