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(13.) Formule relating to three arcs.
sin (a +b+c)=sin a.cos b.cos c+ cos a.sin b.cos c

+ cos a.cos b.sin c- sin a.sin b. sin c.
cos (a +b+c)=cos a.cos b.cosc cos a.sin b.sin c
- sin a.cos b.sin c sin a.sin b.cos C.

tan a + tan b + tan c - tan a. tan b. tan c tan (a+b+c)=

1- (tan a.tan b + tan a tan c + tan b.tanc)

1

(14.) Relations between the sides and angles of plane triangles. Let A, B, C, be the angles,

a, b, c, the sides subtending them ;

s={(a +b+c). sin A sin B sin C

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s (s-a) (cos, 4)*=

bc

(sb)(s -c)
(tan, 4)

8 (8-a)
sin 2 A + sin 2 B + sin 2C=sin A. sin B. sin C.
cos 2 A + cos 2 B + cos2 C= 4 cos A.cos B. cos C-1.
tan A + tan B + tan C=tan A. tan B. tan C.

(W. Ch. ii; L. 365.) The area = 8(8–a) (s—b) (8—c))';

2abc

cos? A.cos į B.cos įC. a+b+c

a

a

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Let R, be the radius of the inscribed, and R, that of the circumscribing circle; then R, = {(8—a)(s—b)(8-c))"; = s.tanĮ A,tanı B.tanĮC,

R,= ] abc.8(8-a)(sb)(8-c)

(Legendre, Geom. Note v; Hind, Trig. 153–64.) (15.)

Values of the side c. sin C

(a - b)*+4 ab (sin _C)*)*. sin A

b cos A + a. 1– (sin B)?) a cos B + b cos A. a’ + bo 2 ab cos C).

b cos A+ao – (b sin A)) (a + b) – 4 ab(cos}C)®)?

cos B + sin B.cot C: (16.)

Values of sin C. sin (A + B).

16+ a-C

1. sin B.

2 ab b

sin B csin B.ao+c2–2ac cos BH

{a cos B+b– (a sin B):]},

6 (17.)

Values of cos C. cos (A + B).

a(sin B)? Fcos B.6-(asin B) ca 1 (sin B) 72

28 (s—c)

-1, ac cos B

ab (a* + c - 2 ac cos B)** (18.)

Values of tan C. - tan (A + B).

2 ab c sin B.6 (e sin B)?7! a+ 62 - 6 c sin B

a cos B +b(a sin B)* a-c cos B'

a sin B-cot B.be-(a sin B)? |

°7

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2

b

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* Analogous formulæ may be obtained by substituting A, a, for B, b, and vice versa, in this, and several of the following formulæ.

(19.)

Right angled triangles : C, the right angle.

a tan A=

[1] Given a, b:

zi
log Tan A* = log r + log a-log b.

c=(a® +62)*; or

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log c=log r + log b-log Cos A. [2] Given a, c:

b=(co – a?)},
log b={log (c + a) + {log (c-a),

b
COS A =

с

log Cos A=log b-log c + log r.

sin B=cos A. [3] Given A, c: a=c sin A;

log a=log c + log Sin A - log r.

B=90° A.

b=C COS A ;

or 6

log b=log c + log Cos A - log r. may

be determined as above.
[4] Given A, b: B=90° - A.

a=b tan A;
log a=log b + log Tan A - log r.

b

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log c= log r + log b-log Cos A; (W. Ch. v; L. 67–71; Leg. 48–52; C. 529-58.)

* The sine, cosine, &c. to the tabular radius r will, for the sake of distinction, be denoted by capital letters. See (Enc. Met. Art. Trigonom.)

(20.) Solution of oblique angled plane triangles.

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log Sin A=log a-log b + log Sin B.
This result is ambiguous if a >b, and B is an acute angle.
C=180° — (A + B).

sin C
c=6

;

sin B log c=log b + log Sin C-log Sin B :

or c=a cos B+b? (a sin B)*7. [3] Given C, a, b;

a-b
tan (A - B) tan { (A + B);

a+b log TanĮ (A-B)=log (ab) – log (a+b) +log Tan Ž (A+B),

A+B=180°C: or thus: suppose a > b, and let tane 7 log Tan (=log r + log a-log b: then

tan Ž (A -- B)=tan { (A + B). tan (0 — 45°), log Tan{(A,B)= log Tan{(A+B)+log Tan (0— 45°) – logr. knowing A + B, and A- B, A and B are determined.

a

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sin C C = a

; sin A

log c=log a + log Sin C – log Sin A.
or c=(a+ b3 - 2 ab cos C)::

2 ab
assume (tan 0)2 =

vers C;

(a - b) log Tan 0=}{log 2+log a+log b+log Vers C+log r}-log (ab);

- 6

then c=

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log c=log (a - b) + log r – log Cos 0.

2 ab
or assume (sin O)? ·(1 + cos C),

(a+b) log Sin0={{log2+log a+log b+log(r+Cos C)+logr}-log(a+b);

then c=(a + b) cos 0; log c=log (a + b) + log Cos 0 – log r.

[4] Given a, b, c.

2 First Method: sin A= .8

(-a)(s—b)(s—c)]; bc

be

log Sin A=log r + log 2-log b – log c

+ {{log 8 + log (8 – a) + log (s – b) + log (s—c)}. Second Method : (sin } 4) = (s—b)(8—c), log Sin A={{log (s-b) + log (s—c) – log b -- log c} + log r. Third Method : (cos į A)= $($ – a); log Cos LA={{log 8 + log (s— a) – log b- log c} + logr.

bc

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