Note. The angle B is acute like its opposite side A C, because the side A B is not ambiguous; and that its acute value applied to the side B C is of the same affection with the sum of the angles. PROBLEM III. Given Two Sides of an Oblique-angled Spherical Triangle, and the Angle contained between them; to find the other Two Angles and the Third Side. RULE. 1.-To find the other two angles. As the log. co-sine of half the sum of the two given sides, is to the log. co-sine of half their difference; so is the log. co-tangent of half the contained angle, to the log. tangent of half the sum of the other two angles. Half the sum of the angles thus found, will be of the same affection with half the sum of the sides.-Again: As the log. sine of half the sum of the two given sides, is to the log. sine of half their difference; so is the log. cotangent of half the contained angle, to the log. tangent of half the difference of the other two angles.-Half the difference of the angles, thus found, will always be acute. Now, half the sum of the two angles, added to half their difference, will give the greater angle; and half the difference of the angles subtracted from half their sum will leave the lesser angle. 2.-To find the third side. The angles being known, the third or remaining side is to be computed by Rule 3, Problem I., page 198. To find the Angle B: As the side AC+AB+2=74:20:35" Log. co-sine ar, comp.=10.568834 9.998712 9.702414 To find the Angle C: As the side AC+AB÷2= 74:20:35" Log. sine ar.compt. = 10.016421 Is to the side AC-AB÷2= 4.24. 40 Log. sine = 8.885996 63. 15.10 Log. co-tangent = 9.702414 Note.-The half sum of the angles came out acute, because the half sum of the sides is acute: the half difference of the angles is always acute. 9.942825 To the side B C = 118:45:34" Log. sine = Remark 1.-The side B C may be found directly, independently of the angles B and C, by the following general Rule. To twice the log. sine of half the contained angle, add the log. sines of the two containing sides; from half the sum of these three logs. subtract the log. sine of half the difference of the sides, and the remainder will be the log. tangent of an arch: the log. sine of which being subtracted from the half sum of the three logs. will leave the log. sine of half the required - side. Example. Let the side A C, of a spherical triangle, be 62:10:25", the side A B 50:14 45%, and the included angle A 123:11:40%; required the side BC? Arch. . 81:50:52 Log. tang. = 10.843897 Log. S. 9.9955884 Remark 2.-The side B C may be also computed by the following general rule, viz. To twice the log. sine of half the contained angle, add the log. sines of the two containing sides, and the constant logarithm 6.301030; the sum (rejecting 40 from the index), will be the log. of a natural number :-which being added to the natural versed sine of the difference of the containing sides, the result will be the natural versed sine of the third side. Thus, to find the side B C in the above example. Half included ang. A=61:35:50? twice the log. sine. 62.10.25 Log. sine = Side B C. 94:13:40 N.V.S. 1.073721; the same as above. = Note. This formula will be found exceedingly useful on many Astronomical occasions.-And, when the index of the sum of the four Logarithms is 6, as in the above example, the natural number corresponding thereto may be very readily found in the second part of Table XXVII., between pages 153 and 166. Vol. PROBLEM IV. Given Two Angles of a Spherical Triangle, and the Side comprehended between them; to find the remaining Angle and the other Two Sides. 1.-To find the other two sides. RULE. As the log. co-sine of half the sum of the two given angles, is to the log. co-sine of half their difference; so is the log. tangent of half the comprehended side, to the log. tangent of half the sum of the other two sides. Half the sum of the sides, thus found, will be of the same affection with the half sum of the angles. Again. As the log. sine of half the sum of the two given angles, is to the log. sine of half their difference; so is the log. tangent of half the comprehended side, to the log. tangent of half the difference of the other two sides. Half the difference of the angles, thus found, will always be acute. Now, half the sum of the two sides, added to half their difference, will give the greater side; and half the difference of the two sides, subtracted from half their sum, will leave the lesser side. 2. To find the remaining angle. The sides and two angles being known, the remaining or third angle is to be computed by Rule 3, Problem II., page 200. To half the sum of the sides=63:18:28" Log. tangent = . 10.298628 Half difference of the sides = 2. 49. 10, as in the next operation. To find the Side AB : As the angle A+angleC÷2=61:15:20 L. sine ar. compt. 10.057113 Note. The half sum of the sides came out acute because the half sum of the angles is acute; the half difference of the sides must be always acute. Remark 1.-The angle B may be found directly by the following general rule. To twice the log. co-sine of half the given side, comprehended between the two given angles, add the log. sines of those angles: from half the sum of these three logs. subtract the log. sine of half the difference of the angles, and the remainder will be the log. tangent of an arch.-Now, the log. sine of this arch being subtracted from the half sum of the three logs. will leave the log. sine of half the required angle, Half diff. of do. 2. 35. 5 Log. sine = 8.654144 Arch 85:55:17" Log. tangent 11. 1468844 Lg.S.=9.998899 |