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30 ft., =

b

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24 ft., B = 65°.

2. Solve ABC, given: a = 29 ft., b = 34 ft., A = 30° 20′.
3. Solve ABC when a
4. Solve ABC when a =
5. Solve ABC when a =

30 in., b = 24 in., A = 65°.

15 ft., b 8 ft., B = 23° 25'.

=

57. Case III. Given two sides and their included angle. triangle ABC, a, b, C are known, and it is required to B, c. In this case, c can be determined from the

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relation c2 a2 + b2 - 2 ab cos C, Art. 54; angle A can be determined from the relation

sin A sin C.

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Checks: a2= b2 + c2 - 2 bc cos A, b2 = a2 + c2 — 2 ac cos A, the result in Ex. 1, Art. 54 a; other checks will be found later.

EXAMPLES.

1. In triangle PQR, p = 8 ft., r = 10 ft., Q = 47°. Find q, P, R. q2p2r22 pr cos Q

= 64 + 100 — 2 x 8 x 10 x .6820 = 54.88.

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R

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=

=

8 x .7314
7.408
10 x .7314
7.408

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=.9873.

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FIG. 56.

59° 17'.

B = 35° 25'.

2. Solve ABC, given: a = 34 ft.,

3. Solve ABC, given: a = 33 ft.,

4. Solve RST, given: r = 30 ft.,

5. Solve PQR, given: p = 10 in.,

58. Case IV. Three sides given. If the sides a, b, c are known

in the triangle ABC, then the angles A, B, C can be found by means of the relations (3), Art. 54.

Checks: Relations (1), Art. 54: A+B+C=180°.

checks will be shown later.

Other

EXAMPLES.

1. In ABC, a = 4, b = 7, c = 10; find A, B, C.

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A

b-7

2 ca

a2 + b2

C- 10

2 ab

Ca-4

FIG. 57.

B

c2

2 x 10 x 4

1649. 100

2 × 4 × 7

Angle C is in the second quadrant since its cosine is negative.

Check: 18° 12'+33°7'30"+128°40′52′′-180°0'22". The discrepancy is due to the fact that four-place tables were used in the computation. Had five-place tables been used, the discrepancy would have been less. Find P, Q, R.

2. In PQR, p = 9, q = 24, r = 27.
3. In RST, r = 21, s = 24, t = 27.
4. In ABC, a =
12, b =
5. In ABC, a = 80, b = 26, c = 74.

Find R, S, T.

20, c 28.

Find A, B, C.

Find A, B, C.

6. Solve Ex. 1, using five-place tables.

59. The aid of logarithms in the solution of triangles. It was pointed out in Art. 6 that an expression is adapted for logarithmic computation when, and only when, it is decomposed into factors. In Cases I., II., Arts. 55, 56, the expressions used in solving the triangle can be computed with the help of logarithms. On the other hand, the side opposite to the given angle in Case III., Art. 57, and the angles in Case IV., Art. 58, are found by evaluating expressions which are not adapted to the use of logarithms. Other relations between the sides and angles of a triangle will be found in Arts. 61, 62. By these relations the computations in Cases III., IV., can be made both without and with logarithms. These relations are useful not merely for purposes of computation; they are important in themselves, and valuable because many im portant properties of triangles can be deduced from them.

The explanations given in Arts. 55-57 are presupposed in Arts. 60-62. The general directions to be observed in working the problems are as follows:

1. Write down all the formulas which will be used in the computation.

2. Express these formulas in the logarithmic form.

[As soon as the student perceives that this step does not afford any additional assistance, it may be omitted. See Art. 27, Ex. 1, Note 6.]

3. Make a skeleton scheme, and arrange the arithmetical work neatly and clearly.

The skeleton schemes in the worked examples that follow, are apparent when the numbers are omitted.*

Checks: The various formulas can serve as checks on the results of one another. The relations derived in Exs. 1, 2, Art.

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Since ab, there may be two solutions. Construction shows there are two solutions.

C

Formulas:

-576

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ACB=180°-(A+ABC). ACB1=180°- (A+AB1C).

Ɑ=447

α= 447

47

A

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.. log sin ABC = log b + log sin A · log a = log sin AB1C;

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* Cologarithms are not used in the solutions in the text. In extensive computations the use of cologarithms is favoured by many computers; but it seems best for beginners in trigonometry first to become accustomed to the obvious and direct method of working with logarithms.

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In obtaining log AB, for instance, log sin ACB may be written on the margin of a slip of paper, placed under log a, the addition made, log sin A placed beneath, and the subtraction made.

Solve the triangle ABC, when the following elements are given :

2. A 63° 48', B = 49° 25', a = 825 ft.

=

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61. Relation between the sum and difference of any two sides of a triangle. The Law of Tangents. Use of logarithms in Case III. In any triangle ABC, for any two sides, say a, b,

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That is, the difference of any two sides of a triangle is to their sum as the tangent of half the difference of their opposite angles is

to the tangent of half their sum. This is sometimes called the

law of tangents.

Now A+B=180° – C, and, consequently, (A+B)= 90°

Hence, tan (A+B)=cot

be written

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and, accordingly, relation (1) may

α b tan (4-B) =

a+b cot c.

(2)

Formulas for b, c, and c, a, similar to the formulas for a, b in (1), (2), can be derived in the same way as (1), (2), have been derived. These formulas can also be written down immediately, on noticing the symmetry in formulas (1), (2).

Ex. Write the formulas for sides b, c and c, a. Derive these formulas.

Case III. In a triangle ABC, a, b, C, are known, and c, B, A, are required. Here, } (A+B)= 90° – † C'; also, † (A — B) can be found by (2). Hence, A and B can be found; for

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A = {(A+B) +1(A – B), and B={(A+B)—† (A — B). The side c can then be found by (1), Art. 54. (In using (1), (2), write the greater side and the greater angle first, in order that the difference may be positive.) Formulas (1), (2), can also be used as a check in the cases discussed in the preceding articles. Other checks will be shown in the next article.

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Checks: A+ B + C = 180°, formulas in preceding articles, and formulas shown in the next article.

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