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 But, because cos B' COS A' 2 sin (A+B). sin (A'B') (art. 25 ch. iii), and consequently, a+b. cos (b − c) — cos a = 2 sin 4+6-c. sin a+c-b: we have, obviously, sin2 = sin (a+b−c) . sin † (a+c—b) sin b. sin c 2 Whence, making s = a+b+c, there results The expressions for the tangents of the half angles, might have been deduced with equal facility; and we should have obtained, for example, sin (s—b). sin (sc), (iii.) sins.sin (s-a) Thus again, the expressions for the cosine and cotangent of half one of the angles, are sins. sin (s- a) sin b. sin c sin s sin (s-a) sin (sb). sin (s—c)* The three latter flowing naturally from the former, by means of the values tan == sin cos' Cor. 1. When two of the sides, as band c, become equal, then the expression for sin 4 becomes When all the three sides are equal, or a = b=c, Cor. 3. In this case, if a = bc90°; then sin 44 = =√2 sin 45°: and A = B = c = 90°. Cor. 4. If a b c = 60°: then sin A= sin 35°15′51′′: and A = B = c=70° 31′42′′, the same as the angle between two contiguous planes of a tetraedron. Cor, 5. If a = bc were assumed = 120°: then sin 4= sin 60° sin 120 3 = 1; and A=B = C = 180°; which shows √3 that no such triangle can be constructed (conformably to th. 2); but that the three sides would, in such case, form three Continued arcs completing a great circle of the sphere. PROBLEM PROBLEM III. Given the Three Angles of a Spherical Triangle, to find If from the first and third of the equations marked L (prob. 1), cos c be exterminated, there will result, COS A. sin c + cos c. sin a. cos b = cos a. sin b. But, it follows from th. 7, that sin c = sin 7. sin c COS A sin A SIO A cos a Substituttheir equi sin a ing for sin c this value of it, and for (th. 7). So that the preceding equation at length becomes, COS A. sin c = cos a. sin B In like manner, we have, - sin A. cos c. cos b. . COS C. cos a. COS B. sin C = cos b. sin A - sin B Exterminating cos b from these, there results COS A So like-cos B = wise } cos c = cos c. sin A. sin B This system of equations is manifestly analogous to equation I; and if they be reduced in the manner adopted in the last problem, they will give The expression for the tangent of half a side is (V.) The values of the cosines and cotangents are omitted, to save room; but are easily deduced by the student. Cor. 1. When two of the angles, as B and C, become equal, then the value of cos a becomes cos α = COS sin B COS A sin A b = c = 90°. Cor. 2. When A B c; then cos a = 1. Cor. |
