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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

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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.

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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.

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3. Again, because (art. 17. Note 3.) cos A2=(1+cosA), by substitution, cosдA2 — 2bc+b2+c2—a2 _ (b+c) 2—a2

=

4bc

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((b+c)+a) ( (b+c)—a) and consequently

Cos A

46c

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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

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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.

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a-b_tan
tant(A-B), which agrees with Prop. 10.
a+b ̃tanž(A+B)’

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.

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8. Divide the terms of art. 6. by a, and tanž(A

tan (A-B)
tan (A+B)

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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
(b+c)

suvers A);

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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

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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
2bc

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,

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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)=

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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.

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therefore a

sin B+sinC

=(art. 5. and

18. Note 3.)

cos (B-C)

b sinA+c sinA_(b+c)sin(B+C)
sinB+sinC

(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.

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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
(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(ab), and b

((c-(a-b)) (c+(a−b))
c.cosA―(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

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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=

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