2EOX OF From the second of these equations EF2=EO+OF22EOX OF X COS EOF which substituted in the first equation gives COS EDF rad rad (ED2+DF2 — Eo2 — of2) + 2eo × OF X COS EOF 2DEX DF But OE-EDOD2 (Euclid I. and 47.) and or2-DF2=OD2, con(EO X OF X COS EOF) — (OD2 × rad) sequently cos EDF = DE X DF (C) Since the preceding conclusion does not depend on any peculiar relation which the c has to the other angles, a similar equation will be equally true for the angles A and B. Hence (rad. cos a)-(rad. cos b. (rad .cos b)-(rad. cos a. cos c) Cos 4A sine b. sine c Cos 4B sine a sine c (rad2, cos c)-(rad. cos b. cos a) Cos 2c= sine b. sine a * These are the formulæ from which Lagrange and Legendre begin their investigations, and of the four quantities involved, any three being given the fourth may be found. They are applicable to every species of spherical triangles, whether rightangled, quadrantal, or oblique-angled, but the formule for right-angled and quadrantal triangles have already been given. (O. 164, and P. 165) (D) If A, B, C represent the three angles of a spherical triangle, the opposite sides may be represented by 180°-4, 180°-B, 180°-c; and if a, b, c represent the three sides, • The former in the Journal de L'Ecole Polytechnique, and the latter in his Élé wents de Géométrie. their opposite angles may be represented by 180° — a, 180°—b, 180°c (X. 134.). Hence Cos (180°-a)= rad2, cos (180° — ▲)—rad . cos (180° — B). cos (180)°— c) sine (180°-B). sine(180° —c) But cos (180°-a)=-cos a; cos (180°-A) =—cos a, &c. (N. 95.) consequently COS a (rad2. cos A)+(rad. cos B. cos c) sin B sine c In the same manner the cosines of the other sides determined. may be (E) It has been shewn (S. 194.) that the sines of the sides of any spherical triangle have the same ratio to each other as the sines of their opposite angles; hence by using the notation of Legendre, we shall have (F) The general expressions for the cosines, which have been obtained by this proposition, may be arranged thus: III. Cos A= Cos B Cos C (rad'. cos a)-(rad. cos b. cos c) And by reducing these last equations. IV. Cosa= Cos c= (cos A. sine b. sine c) + (rad. cos b. cos c) rad2 (Cos B. sine a. sine c) + (rad • cos a. cos €) (Cos c. sine b. sine a)-(rad. cos b. cos a) rad2 rad V. Cos And by reducing these equations, we shall have (cos a. sine B. sine c)-(rad. cos B. Cos c) VI. Cos A= Cos B= Cos c: (cos b. sine a. sine c)-(rad. cos A. cos c) rad2 rad2 rad2 (cos c. sine a. sine B)-(rad.cos A. COS B) (G) The six preceding articles afford solutions to all the different cases of oblique-angled spherical triangles. The Ist. Finds the angles, when two sides and an angle opposite to one of them are given. The IId. Finds a side, when two angles and a side opposite to one of them are given. The IIId. Finds the angles, from the three sides being given. The IVth. Finds the third side, when two sides and their contained angle are given. The Vth. Finds the sides, from the three angles being given. The Vith. Finds the third angle, when two angles and the side adjacent to both of them are given. (H) But none of the foregoing formulæ are conveniently adapted to logarithmical calculation. Let the value of the cosine of c (F. 213.) be substituted in the formula rad2-rad. cos c=2 sinec (2d equation I. 111.) we COS C 2 sine2 c (rad2.cose)-(rad.cosa.coso) shall have I sine a. sine b. rad ; but (sine a. sine b) 2 sine2 + c rad +(cos a. cos b)=rad, cos (a−b) (D. 109.), hence rad. cos (a-b)—(rad . cos e) sine a. sine b From the 4th equation (F. 110.) rad. cos q-rad. cos P= 2 sine (PQ). sine (P-Q), which, by putting Q=(a−b) and PC, becomes rad.cos (a-b) - rad⋅ cos c= 2 sine (ca-b). sine (c-a+b), hence we obtain 2. 2 sine2 ¿C_2 . sine ÷ (c+a−b). sine (c−a+b) ̧ sine a sine b sine c=rad sine (c+a-b). sine (c−a+b) that is } (c + a−b) = {(a+b+-c)—b, and (c+b−a)=(a+b+c)−a, sine (a+b+c)-b.sine (a+b+c) — a hence sine crad and it is evident that the same formula will be obtained, with only a change of the letters, for the angles A and B. When c is near 90° this will not be a convenient rule for producing an accurate result, because the difference of the logarithmical sines for 1" is then very small (see the note page 49); if c be less than 45°, it will be proper to use this rule. (I) Again, if the value of the cosine of c (F. 213) be substituted in the formula rad2+rad. cos c=2 cos2c (1st Equation I. 111.) we shall have COS C 2 COS 1+ rad 2 I rad2 C (rad. cos c)-(cos b. cos a) (sine a. sine b)+(rad. cos c)—(cos b. cos a); but (cos acos b) sine a. sine b -(sine a. sine b)=rad. cos (a+b), (D. 109.). Hence (sine a. sine b)-(cos a. cos b)=-rad. cos (a+b), therefore 2 cosc (rad. cos c)-rad. cos (a+b) From the 4th equation (F. 110.) (rad. cos q)-rad. cos P= 2 sine (P+Q). sine (P-Q), which, by putting Q=c and P= (a+b), becomes (rad. cos c)-rad. cos(a+b)=2 sine (a+b+c) .sine (a+b-c), hence we obtain 2 cos2+c_2 sine (a+b+c). sine (a+b−c) that is sinę a sine b But sine a. sine b sine (a+b+c). sine (a+b-c) 1⁄2 (a + b − c ) = 1⁄2 (a + b + c) − c, hence obvious that the same formula will be obtained, with only a change of the letters, for the angles A and B. When 4c is very small, this rule should not be used where a very accurate result is wanted, because the logarithmical cosines of very small arcs, in a table carried to seven places of figures, differ but little from each other (see the note page 49); if if 4c be between 45° and 90°, this rule may be used with advantage. rad . sine c (K) Also, because tangc (N. 98), and cos c rad. cos c sine c Tang crad =cotc (O. 98.) we shall obtain, by division, sine (a+b+c) − b . sine ž (a+b+c)− a Cotc rad, sine (a+b+c). sine (a+b+c)—c sine (a+b+c)-b.sine (a+b+c)-—a (L) Any of the three preceding formulæ (H. 214, I. 215, or K. 216.) will determine an angle when the three sides are given, and by continuing Legendre's * mode of investigation, formulæ for determining a side in terms of the three angles may be obtained; thus, let the value of cos a (D. 212.) be substituted in the formula rad2-rad. cos a=2 sine2 La (2d Equation I. 111.) we have cos a 1. rad 2 sine2 a (rad2. cos A)+(rad. cos B. cos c) sine B. sine c. rad (sine B. sine c) - (rad. cos A)-(cos B cos c) sine B . sine c ; but rad.cos (B+c)=(cos B cos c)-(c sine B. sine c), (D. 109.) or, (sine B sine c)-(cos B. cos c)=-rad. cos (B+c), therefore 2 sine a-rad. cos (B+ C)-(rad. cos. A) sine B. sine c rad2 tion (F.110.) rad.cos p+rad.cos Q=2 cos (P+Q) ⋅ cos 1(P—Q), which, by putting PB+C and Q=A, becomes rad. cos (B+c)+ rad.cos A=2 cos (B+C+A). COS (B+C-A), hence we obtain 2 sine2 a -2 cos(B+C+A). Cos (B+C-A), that is * Éléments de Géométrie, 6th edition, page 391. et seq. Though the quantity under the radical sign appears under a negative form, it is always positive; for, since the three angles of every spherical triangle are together greater than two right angles (W. 134), (A+B+C) must be greater than 90°, and the cosine of an arc greater than 90° is negative, (K. 94.) therefore the expression |