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.

54 a, are also useful as checks.

60. The use of logarithms in Cases I., II. An example worked out, will give sufficient explanation.

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

Formulas :

sin ABC = - sin Asin AB1 C.

α

ACB=180°-(A+ABC). ACB1=180°-(A+AB1C).

.. log sin ABC = log b + log sin A - log a =

log sin AB1C;

log AB = log a + log sin ACB – log sin A ;

log AB1 = log a + log sin ACB1 — log sin A.

* 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.

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 :

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,

sin A+ sin B

[By composition and division.]

2 cos(A+B) sin (A-B). [Art. 52, Formulas (5), (6).] 2 sin (A+B) cos (A-B)

1

tan (A-B)

1

[Art. 44, A, B.] (1)

a

a + b

2

tan (A + B)

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.

C

2

Now A+B=180° – C, and, consequently, (A+B) = 90° — — Hence, tan (A+B)= cot and, accordingly, relation (1) may be written

2'

tan (4 – B):

a b

=

a+ & 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

A = {(A+B) + }(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.

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

log tan (BC) = log(b − c) + log cot A − log(b + c),

b = 472

c = 324

log a = log blog sin A – log sin B; or =

A = 78° 40'

b- c = 148

b + c = 796

A = 39° 20'

log clog sin A

log b = 2.67394 log sin A = 9.99145 - 10 log sin B = 9.95223 – 10 ... log a = 2.71316 ... a 516.6

Check: A+ B + C = 78° 40′ + 63° 27′ 1′′ + 37° 52′ 59′′ :

= 180°.

NOTE. Formulas (1), (2), are adapted to logarithmic computation; but the computations can be made without the aid of logarithms.

2. Solve ABC, given b = 352, a = 266, C = 73°.

3. Solve PQR, given p = 91.7, q = 31.2, R = 33° 7' 9".

4. Solve ABC, given a = 960, b = 720, C= 25° 40'.

62. Trigonometric ratios of the half angles of a triangle. Use of logarithms in Case IV.

In any triangle ABC,

The substitution of these values in (1) and (2) gives

28.2(s-a).

(2)

2 sin24=

2 bc

2 (s—b). 2 (sc); 2 cos2 4=

2 bc

Since tan2 A = sin2 | A ÷ cos2 A, it follows that

=

(3)

(4)

NOTE. By geometry, b+c>a. Hence, − a +b+c>0, and, accordingly, s a is positive. Similarly, sb, s

c, are also positive. Therefore, the quantities under the radical signs are positive. The positive sign must be given to the radical, for A is less than 180°, and consequently A lies between 0° and 90°.