0) sine o SOLUTION. Find the other two sides by Case IX, and then find the other angle by Case VI. sine (B~ COS a. tang C Or, I. Cotos ; then, cos A=cos c. sine (CR) II. Coto ; then, cos B=cos A. rad sine (A~0) COS c . tang B III. Cot o= ; then, cos c=cos B. rad Or, the angles A, B, or c, respectively, may be found by the sixth set of equations, page 184. sine sine o CASE XI. Given the three sides of an oblique-angled spherical triangle, to find the angles. SOLUTION. Let a +b+c=s, then 1. Sine f a=rad sine (} s—b). sine (fs—c) sine b. sinec II. Cos į a=rady/sine šs. sine (Hs–a) sine b. sinec 111. Tang} a=rad sine (Is-6). sine (is -c) sine s.sine (i s-a) IV. CotA=rad sine s. sine (is-a) sine (is-6). sine (1 sec) I. Sine | B=rad sine (is-a). sine (Is-c) sine a .sinec II. Cos į B=rad sine į s. sine (is-b) sinec sine, s.sine (fs-) sine s.sine (I s-6) sine (1s-a). sine (I -c) 1. Sine | c=radV sine (is-6), sine (i sama sine a. sine b sine s.sine (i s -c) II. Cos } c=rad sine a'. sine 6 B=radi sine a . a) III. Tang } c=rad sine (is-b). sine (is -a) -c) IV. Cot ț c=rad V sine s.sine (Is -c) sine (i s-6). sine (is-c) Or, the angles A, B, or c, respectively, may be found by the formulæ A. 182. Case XII. Given the three angles of an oblique-angled sine B. sine c sine B. sine c III. Tang | c=rad cos s. cos (s-c) cos (1 S-A A). cos (įs -c) IV. Cot { c = rad Vcos (1 s–A). cos (4 s-B) cos įs cos (1 s-c) Or, the sides a, b, or c, respectively, may be found by the formula in the fifth set of equations, page 184. OF THE AMBIGUITY OF THE DIFFERENT CASES. (F) When two sides, and an angle opposite to one of them, are given, to find the rest (see Case ist, uid, and id), the values of these required parts are sometimes ambiguous. (G) Practical Rules for determining whether the quantities sought are acute, obtuse, or ambiguous, are given in the solutions of the different cases. The two following tables are the same as those given by Legendre at pages 400 and 401 of the 6th edition of his Geometry, and are deduced from Prop. XVII, page 149, and Prop. XVIII, page 150 of this treatise. TABLE I. Let a, a and 6 be the given parts. Then, 1. AC 90°, bc 90° sab one solution. lacb two solutions. a+b=180° two solutions. Sa+b3180° two solutions. 3. A 90°, bc90° Ta+b=180° one solution. ab two solutions. 4. A 90°, 6-90° Lac bone solution. If A=90°, a=b, or a+b=180, the cases will not be ambiguous, but if b=90°, there will be two solutions. When two angles, and a side opposite to one of them, are given, to find the rest (see Case ivth, vth, and vith), the values of the required parts are subject to ambiguity; this triangle being supplemental to that wherein two sides and an angle opposite to one of them are given. TABLE II. Let A, B, and a be the given parts. Then, 1. a 90°, B390% ŞacB one solution. TAB two solutions. A+B180° two solutions. SA+B= 180° two solutions. A+B]180° one solution. ACB two solutions. If a=90°, A=B, or A+B=180°, there will be but one solution, but if B=90°, there will be two solutions. (H) We may likewise remark, that since any side or angle of a spherical triangle is less than 180°, the half of any angle, or half the difference between any two sides, or half the difference between any two angles, must be acute. Hence in the equation, where cotc.cos } (a-6)=tang (A+B), cos } (a+) (M. 188) it is plain that cotic and cos } (a-b) are both positive (K. 100.), and therefore tang 1 (A+B) and cos : (a + 6) must be both positive; consequently, half the sum of any two sides of a spherical triangle is of the same species as half the sum of their opposite angles. This rule is applied in the practical solutions of the different cases, and will frequently remove the ambiguity which would otherwise arise, where a quantity sought is to be determined by means of a sine. CHAP. VI. I. PRACTICAL RULES FOR THE SOLUTIONS OF ALL THE DIFFERENT CASES OF RIGHT-ANGLED SPHERICAL TRIANGLES, WITH THEIR APPLICATION BY LOGARITHMS. Every spherical triangle consists of six parts, three sides, and three angles; any three of which, being given, the rest may be found. In a right-angled spherical triangle, two given parts, besides the right angle, are sufficient to determine the rest. The questions arising from a variation of the given and required parts are 16, but if distinguished by the data, the number of cases is 6. THE GIVEN QUANTITIES ARE, EITHER : B (I) RULE 1. Draw a rough figure as in the margin ; and let AG, AH; FB, FH; CI, CD; EI, EG, be considered as quadrants, or 90° each ; then you have eight right-angled spherical triangles, every two of which will have equal angles at their bases: And the triangles CGF and EDF will have their respective sides and H angles either equal to those of ABC, the triangle under consideration, or they will be the complements thereof. (L. 151.) Then, In triangles having equal angles at their bases. The sines of their bases have the same ratio to each other, as the tangents of their perpendiculars. And, The sines of the hypothenuses have the same ratio to each other, as the sines of the perpendiculars. (M. 167.) (K) Illustration. 1st. In the triangles ABC and' AHG. 1. Sine AG:sine Ac::sine Hg: sine Bc. and these are propor2. Sine Ah:sine AB::tanghy:tang Bc. Štional by inversion. 2d. In the triangles FGC and FHB. - 3. Sine BH : sine cr::sine FB : sine FC. and inversely, &c. 4. Sine FH : sine FG:: tang BH : tang co. 3d. In the triangles CGF and cid. 5. Sine DC : sine fc::sine id : sine FG. 6. Sine cg : sinecr::tang FG : tang iD. and inversely, &c. 4th. In the triangles EDF and EIG. 7. Sine El : sine ED::tang ig : tang DF, and inversely, &c. The student must remember that ABC is the proper triangle in the preceding proportions, and that AG, AH, &c. are each 90°, consequently sine' AG, sine AH, &c. are each equal to radius. BH is the complement of the base AB. ? In any of these CG is the complement of the hypoth. Ac. cases, for sine, or Fc is the complement of the perp. BC. tangent,write coFG is the complement of the angle A. sine, or co-tanED is the complement of the angle c. gent. HG=EF is the measure of the angle A. ID is the measure of the angle c. IG=Ac the hypothenuse, and dr=BC the perpendicular, Since AB and sc are perpendicular to each other, either of them may be considered as the base, and to avoid a number of } }an |