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 of 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 secant, of the arc AC; let the capital letters A, B, C, denote the angles of the triangle, and the small letters a, b, c, the opposite sides BC, AC, AB. Then the first equa tions in art. 6 PI. Trig. applied to the two triangles ADE, GDE, give, for the former, DE' = tan' b tan2 € - tan b. tan c. cos A; for the latter, DE2 = = sec2 b + sec2 c sec b. sec c cos a. Substracting the first of these equations from the second, and observing that sec2 b tan2 b = R2 = 1, we shall have, after a little -- the three following symmetrical equations are obtained: cos a = cos b. cos c + sin b. sin e. cos A 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 = 1 — cos2 A, we shall have - Reducing the terms of the second side of this equation to a common denominator, multiplying both numerator and denominator by sin' a, and extracting the sq. root there will result √(1-cos2 a- cos2 b— cos2 c + 2 cos a. cos b. cos c) sin b. sin c sin A sin a. sin a. 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 sin A = K. sin a . . . sin B = K. sin b sin c K. sin c. 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 40, 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 cosa be substituted for it, and for sin a its value tan a. cos a= cos α.cos b. cos c; then, recollecting that 1-cos c = sin2 c, there will result, tan a COS C. COS B sinc: whence it follows that, 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. THEOREM XI. In Every Right-Angled Spherical Triangle, the Tangent of either of the Oblique Angles, is equal to the Quotient of the Tangent of the Opposite Side, divided by the Sine of the Other Side about the right angle. COS B sin a tan c Whence, because (th. 8) cos a = cos b. cos c, and since sin a = cos a. tan a, we have sin sin b 1 tan & sin c Q. E. D. sin b tan B = In like manner, tan c = THEOREM XII. In Every Right-Angled Spherical Triangle, the Cosine of the Hypothenuse, is equal to the Quotient of the Cotangent of one of the Oblique Angles, divided by the Tangent of the Other Angle. For, multiplying together the resulting equations of the preceding theorem, we have In Every Right-Angled Spherical Triangle, the Sine of the Difference between the Hypothenuse and Base, is equal to the Continued Product of the Sine of the Perpendicular, Cosine of the Base, and Tangent of Half the Angle Opposite to the Perpendicular; or equal to the Continued Product of the Tangent of the Perpendicular, Cosine of the Hypothenuse, and Tangent of Half the Angle Opposite to the Perpendicular*. This theorem is due to. Prony, who pured it without demonstration in the Connaissance des-Temps for the year 08, and made use of it in the construction of a chart of the course of the Po. Here, Here, retaining the same notation, since we have substituted their values in sines and cosines, there will arise, sin c. cos a cos B. cos c. sin a = cos B. cos c. sin b sin B Then substituting for sin a, and sin c. cos a, their values in the known formula (equ. v chap. iii) viz, it will become, sin (a–c) = sin b. cos c. tan B; which is the first part of the theorem: we introduce, instead of cos c, its value and, if in this result cos a (th. 8), it will be transformed into sin (a−c) = tan b. cosa. tan B; which is the second part of the theorem. Q. E. D. Cor.' This theorem leads manifestly to an analogous one with regard to rectilinear triangles, which, if h, b, and p denote the hypothenuse, base, and perpendicular, and B, P, the angles respectively opposite to b, p; may be expressed thus: hbp.tan 1⁄2P`. b.tan B. These theorems may be found useful in reducing inclined lines to the plane of the horizon. h PROBLEM II. Given the Three Sides of a Spherical Triangle; it is required to find Expressions for the Determination of the Angles. Retaining the notation of prob. 1, in all its generality, we soon deduce from the equations marked 1 in that problem, the following; viz, sin b. sin c Cos a- cos b. cos c COS A = COS B = COS C = cos & cos a. cos c sin a. sin c sin a. sin i As these equations, however, are not well suited for logarithmic computation; they must be so transformed, that their second members will resolve into factors. In order to this, substitute in the known equation 1 Cos A 2 sin2 A, the preceding value of cos A, and there will result. cos (bc) 2 sin2 A |