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In like manner,

(5)

Also, (6) cos C

Introducing R, applying logarithms, and reducing, (4) becomes

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1. Given

2. Given

cos B

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=

log cos 4[log p+log (p-a)+a.c. log b+a.c. log c].

In like manner introduce R and apply logarithms to (5) and (6).

a == 125.

b 150.

C= 100.

By subtracting both members of (1) from 1 and reducing we find

(7) sin A=

(8) sin B

(9) sin C

(7) ÷ (4) = (10) tan A==

(8) (5) (11) tan B=

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V

V

864.

1308.

1086.

Z p ( p

ab

(p-b) (p-c)
Z
pp-a)
(pa) (pc)
1 pp-b)

(9)÷(6)=(12) tan C-p-a) (≥ p—b)

1 pp-c)

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99. Examples.

(p-b) (p-c).

b)

—c)

bc

( p − a) (≥ p—c)

ac

(pa) (≥ p—b)
ab

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100. Problem.

To find the area of a triangle when two sides and their included angle are given.

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Introducing R, and applying logarithms, we have

log (2 k) = log blog clog sin A-10.

bc

C

101. Examples.

1. Two sides of a triangle are 345.6 and 485, respectively, and their included angle is 38° 45' 40"; what is the area? Ans. 52468.

2. Two sides of a triangle are 784.25 and 1095.8, respectively, and their included angle is 85° 40′ 20′′; what is the area. Ans. 428470.

102. Problem.

To find the area of a triangle when the three sides are given.

By the last problem we find

(1) kbe sin A,

(2) sin A = 2 sin A cos A. Article 95, (5).
(3) sin Ap-b) p-c). Article 98, (7).

(p

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103. Examples.

1. The sides of a triangle are 40, 45, 55, required the area. Ans. 887.412. 2. The sides of a triangle are 467, 845, 756, required the area. Ans. 175508.

104. Problem.

Given the perimeter and angles of a triangle, required the sides.

(1)

(3)

bc

k = √≥p(≥p—a) (≥ p—b) (≥ p—c).

b

sin B

a sin A

p(p-a) Article 98, (4).

bc

2 √pp-a) (p—b) (p-c)

(2)

Adding and reducing by Articles 96, (5) and 95, (5), we have

(5)

.. (6)

=

(3)

sin (B+C)= cos

b+c sin (BC) cos

a

sinA cos

and sin

(4) a

p

a

b + c cos

a

a

A,

Adding 1 to both members, we have

a+b+c

(B-C)

cos (B+C)

C

a

cos (B+C) + cos 1⁄2 (B−C) ̧ cos (B+C)

Let p=a+b+c, and reduce by 96, (7), we have

2 cos B cos C

sin A

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p sin A

cos B cos C

sin C

sin A

(B-C)

A

A= cos(B+ C).

Introducing R and applying logarithms, we have

log a log p+ log sin A+

=

a. c. log cos B+ a. c. log cos C — 10.

Similar formulas can be found for b and c. But, after a is found, b and c can be more readily found by article 69.

105. Examples.

=

=

1. Given p 150, A = 70°, B = 60°, C-50°, required a, b, c.

=

Ans. a

50.51, c = 44.68. 2. Given p = 31234.36, A = 35° 45', B = 45° 28', C98° 47', required a, b, c.

Ans. a = 7985, b 9742.5, c = 13506.86.

54.81, b

=

3. Given p 375, A= 55° 46′ 18′′, B 82° 49′ 08′′, C41° 24′ 34′′, required a, b, c.

Ans. a = 125, b

=

150, c = 100.

106. Problem.

Given the three sides of a triangle, to find the radius of

the inscribed circle.

A

B

(1) BOC AOC AOB = ABC.

(2) BOCar.

(3) AOC

br.

(4) AOB cr.

.. (5) BOC+AOC+AOB=4 (a + b + c) r = 1 pr.

a

But (6) ABC= √ ≥ p (≥ p − a) (≥p—b) (ž p—c).

S. N. 9.

C

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107. Examples.

1. The three sides of a triangle are 20, 30, 40, respectively, required the radius of the inscribed circle. Ans. 6.455.

Let O be the center of the circle, and R the radius.

2. The three sides of a triangle are 100, 150, 200, respectively, required the radius of the inscribed circle. Ans. 32.275.

Let OD be perpendicular to b, then

b

AD

2

108. Problem.

Given the three sides of a triangle to find the radius of the circumscribed circle.

b

2

.. (2) R

sin B2 sin B cos B .. (3) R

=

b

2 sin B

AO sin O

A

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=

B

R

The angle the angle B, since each is measured O by one-half the arc AC.

(1) AD

ac

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a

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D b

R sin B.

C

21 ≤p(≥p—a) (p—b)({p−c) ̧

abc

4 k

abc
4vp (pa) (≥p—b) (≥p— c)

Prove that the formula will be the same if the center is without the triangle.

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