BY CALCULATION. I. To find the side ac. Because BC + (AC acute), BC + (AC obtuse), and A+B, are each of the same species, with respect to 180°, ac is ambiguous, (B.235.) being either 80°.19′,or 99°, 41'. III. To find the LC. sine (AC~BC) =8°.14′.30′′ 9.15639 : sine (AC+BC)=72°.4.30" 9.97839 :: tang (A~B) = 3°.53′ 8-83174 :cotc=65°.44.50" 9.65374 2 131°.29.40" LC. If the obtuse value of the side Ac had been used, the c, by the same method, would have been = 155°.22.40". The side BC115°.20′ BY CONSTRUCTION. (Plate V. Fig. 18.) 1. With the chord of 60 degrees describe the primitive circle, through the centre P draw BPC, and DPE at right angles to it. 2. Draw the great circle вce making an angle of 48°.30' with the primitive. (P. 160.) 3. Set off the side BC=115°.20′ from в to m, and draw the parallel circle mcm (Z. 162.), cutting the oblique circle вce in c. 4. With the tangent of the complement of the angle a= 53°.23′, and centre P, describe an arc; and with the secant of the same angle and centre c, cross it in o. 5. With the centre o, and radius oc, draw the great circles Acb, acb. Then ABC is the triangle required; and none of the parts are ambiguous. To measure the required parts. 6. AB, measured by a scale of chords, 82°.26'. 7. AC (C. 163.) 8. The 57°.30'. c(G. 164.)=61°.41′. (T) CASE III. Given two sides of an oblique spherical triangle, and the angle contained between them, to find the rest. BY CONSTRUCTION. (Plate V. Fig. 19.) 1. With the chord of 60 degrees describe the primitive circle, through the centre P draw CPe, and apr at right angles to it. 2. Set off the side AC 80°.19' by a scale of chords, and through the point A, draw the great circle AB, making an angle of 51°.30' with the primitive. (P. 160.) 3. Set off the supplement of AB59°.13′ from n to m, and draw the parallel circle mBm (Z. 162.) cutting the oblique circle ABN in B. Through the three points c, B, e, draw a great circle, then ABC is the triangle required.* * A perpendicular may be drawn from the vertical angle c upon the base AB, by finding p the pole of the oblique circle an, (N. 159.) and drawing a great circle CDP, through p and the point C. (W.161.) To measure the required parts. 4. BC (C. 163.)=63°.50′. 5. The angles B and c (G. 164.) 59°.16′ and 131°.30'. Ᏼ Because (AB+ Ac)is obtuse (B+c)must be obtuse=95°.22′.37". Hence 95°.22′.37" + 36°.6.20"-131°.28'.57" Lc, and 95°.22.37"-36°.6'.20"-52°.16'.17" B. II. To find BC. 1. By using the angles B and C. sine (BC)=36°.6.20" : sine 2. Without the angles в and c. :: tang (ACAB)=20°.14′ 9.56654 : tang 9.99377 9.93405 9.79430 |log sine 4 A = 25°.45′ 9.63794 9.63794 3. In the oblique spherical triangle ABC. The side AB= 40°. 0'.10′′ Ans. The ZA=121°.36'.20". Required the other parts. ZB 42°.15.13". Zc=34°.15. 3". BC 76°.35'.36". (U) CASE IV. Given two angles, and the side adjacent to both of them, to find the rest. BY CONSTRUCTION. (Plate V. Fig. 20.) 1. With the chord of 60 degrees describe the primitive circle, through the centre P draw CPe, and apr at right angles to it. 2. Set off the side AC-80°.19' by a scale of chords, and through the point A, draw the great circle ABn, making an angle of 51°.30' with the primitive. (P. 160.) In the same manner draw the great circle cвe through c, making an angle Bcr=48°. 30', the supplement of c. Then ABC is the triangle required.* To measure the required parts. 3. AB and BC (C. 163.)=120°.47', and 63°. 50'. 4. The B (G. 164.) 59°.16'. Hence 92°.19′+28°.29' 120°.48' AB. And = * A perpendicular CD may be drawn, if required, by the note Case III. 2. In the oblique spherical triangle ABC. AB 50°.10.30". c= 42°.15'.13" Ans. BC=76°.35′.36". LA=121°.86'.20" The side AC 40°. 0'.10" Required the other parts. 3. In the oblique spherical triangle ABC. B=34°.15'.3". Required the other parts. The side AC 30° Answer. BC 24.4. ZB 46°.19'. (W) CASE V. Given the three sides to find an angle. The side AC- 80°.19′ Given The side BC= 63°.50 The side AB=120°.47′ Required the angles. BY CONSTRUCTION. (Plate V. Fig. 21.) 1. With the chord of 60 degrees describe the primitive circle, through the centre P draw CPe, and aer at right angles to it. 2. Set off the side BC=63°.50' from the scale of chords, and draw BPE. 3. Set off the side Ac=80°.19', by a scale of chords, from c to m, and draw the parallel circle mam. (Z. 162.) 4. Set off the supplement of AB 59°.13', by a scale of chords, from E to n, and draw the parallel circle nan. (Z. 162.) 5. Through c and A, and в and A, draw the great circles CAE and BAE. Then ABC is the triangle required.* To measure the required parts. 6. The angles A, B, and c (G. 164.) will be 51°.30′; 59°.16′; and 131°.30'. * A perpendicular co may be drawn from the vertical c upon the base AB, by the note to Case III. Or take the sines of AC and AB, add them together and subtract the sum from 20; the remainder will be the same as the sum of these cosecants without the indices. |