since the ratio of an arc to its sine is ultimately that of equa lity, and the sine of A+dA may be considered as the same with the sine of A; it follows, that 13. Since, by NO. 12. d sinA=cosAdA, or the variation of the sine of an arc is proportional to its cosine; it follows that, near the termination of the quadrant, the slightest alteration in the value of a sine would occasion a material change in the arc itself. Again, from the same Note, d tanA= dA COSA2 or the variation of the tangent is inversely as the square of the cosine, and must therefore increase with extreme rapidity as the arc approaches to a quadrant. 14. It is convenient to reduce the solution of triangles to algebraic formula. Let a, b and c denote the sides of any plane triangle, and A, B, and C their opposite angles. The various relations which connect these quantities may all be derived from the application of Prop. 11. 2. But, since (art. 16. NO. 3.) sin‡A2—†(1—cosA), it fol2bc—b2—c2+a2 lows, by substitution, that sinA2=. 4bc a2—(b—c)3 __(a+b—c) (a-b+c), and therefore, s denoting the 4bc 4bc semiperimeter, Sin}A2 = Prop. 14. 3. Again, because (art. 17. Note 3.) cos A2=(1+cosA), by substitution, cosдA2 — 2bc+b2+c2—a2 _ (b+c) 2—a2 = 4bc ((b+c)+a) ( (b+c)—a) and consequently Cos A 46c 4bc 4. The second expression being now divided by the third, gives tan §A1 — (s—b) (s—c), corresponding to Prop. 12. s(sa) These are the formula wanted for the solution of the first case of oblique-angled triangles. To obtain the rest, another transformation is required. 5. It is manifest that sin A*—1——cos A 4b2c2—(b2+c2—a2)3 and consequently, by Note 5. Book VI., sinA = 462c2 2T sin A = bc 2T For the same reason, sinB = and hence ac sin A a ; which corresponds to Prop. 9. sinB-T 6. Again, by composition, by art. 18. Note 7. a-b_tan 7. Lastly, transforming the first expression, there results, a = √(b2+c2—2bc cos A)=√( (b—c)2+2bc versA) = √((b+c)2—2bc(1+cosA)). The preceding formula will solve all the cases in plane trigonometry; but, by certain modifications, they may be sometimes better adapted for logarithmic calculation. 8. Divide the terms of art. 6. by a, and tanž(A tan (A-B) 9. Again, from art. 7. a=√ ( (b—c)2+2bc versA)= 2bc (b−c) √ (1+ (b—c)3• versA); consequently find tan x= ✔2bc ✔ bc T√vers A = 21 b_sin A, and a = (b—c) sec x= b-c COS X 10. But the expression in art. 1., by a different decomposition, gives a=√((b+c2-2bc suvers A))=(b+c)√(1– 2bc suvers A); 11. Other expressions are likewise occasionally used. Thus, by art. 1., 2bc.cos Ab2+c2-a2, or c2-2bc. cosAa2-b2, and, solving this quadratic, we obtain c = b cosA±√ (a2 — b2 + b3cos A3) = b cosA±√(a2—ba sinA2, or c=b cosA±√((a+b sinA)(a–b sinA)). When two sides and an angle opposite to one of them are given, the third side is thus found by a direct process. sinC 12. From art. 5., c=a- ; but C being a supplementary sinA angle, its sine is the same as that of A+B, and consequently sin AcosB+cos AsinB, c = a( c=a sin A sinC ). By a similar transformation, sinC a sin (B+C) = a(sinBcos C+cosBsinC)=cosB+sinBcotC 13. Lastly, from art. 3. of Note 7, cot A+ cotC= sin (A+C) sin AsinC If the angle A be assumed equal to 90°, the preceding formule will become restricted to the solution of right-angled triangles. 14. From art. 1., cos A=0= b2+c2-a2 whence, ab+c2, which expresses the radical property of the right-angled tri angle. sin B b sin A α 15. From art. 5.,- = and consequently sin B = which corresponds with Prop. 7. 16. Again, from the same article, a b sin B sinB = and C sinC cos B' For the convenience of computing with logarithms, other expressions may be produced. 17. Thus, from art. 14., ba2-c2, and hence b=√((a+c)(a—c)). b 18. Since a2=¿1 (1— ———), put — =tan x, and a=b(sec c)= b 19: Lastly, because ¿a = a2 (1+ —), put — = sin x, and a.cos x. Besides the regular cases in the solution of triangles, other combinations of a more intricate kind sometimes occur in practice. It will suffice here to notice the most remarkable of these varieties. 20. Thus, suppose a side, with its opposite angle and the sum or difference of the containing sides, were given, to de. therefore a sin B+sinC =(art. 5. and 18. Note 3.) cos (B-C) b sinA+c sinA_(b+c)sin(B+C) (b+c)2sin(B+C)cos(B+C)_(b+c)cost(B+C) 2sin(B+C)cos(B-C) But cos (B+C)=sin‡A, and hence cosă(B—C)=(b+c)sin‡A a and the difference of the supplementary angles B and C being known, these angles themselves are hence found. In like manner, it will be found that sin (B—C)= (b—c) cos A a 21. Let a side with its adjacent angle and the sum of the other sides be given, to determine the triangle. By art. 4. tan B2 = and tan B2= Sa.SC S.S-b ; whence tan A2 s—a.s—b. (s—c), and consequently tank Atan}B= 2 Sa.sb.s2 s-c_(a+b)-c s = (a+b)+c2 or cotВ tan▲ (a+b)+c Again by art. 1., 2bc cosA = b2 + c2 — a2, or a2 — b2 — c2 — 2bc.cosA, and adding 2ab+ 2b to both sides, a2+ 2ab + b3-c2=2ab+26-2bc.cos A, or (a+b)—c2=2b(a+b—c.cos A); whence ((a+b) + c)((a + b)—c) = 2b(a+b—c.cosA), and b=z ((a+b)+c) ( a+b)—c). (a+b)-c.cosA If the sign of b be changed, and the supplement of its adjacent angle therefore assumed, we shall obtain cot Вtan A c+(a—b) ((c-(a-b)) (c+(a−b)) The relation of the sides and angles of a triangle might also be in some cases conveniently expressed by a converging series. Thus a sinAsin(B+C) sin BcosC+cosBsinC' and consequently b sinB cosC + b cos B sin C = a sin B, or b.sinC sin B a—b cosCcosB=tanB. Wherefore, by actual division, tanB= |