1

Or, since sec = (chap. i. 19 H), we have

COS

These may obviously be extended to other arcs, by means of formulæ (P), (Q), (R), of this chapter, and prop. 3 of chap. ii.

42. Operations of this kind ought not to be carried far without being subjected to checks and proofs. For this purpose, after the sines and cosines are found, the tangents, secants, &c. are easily verified by their mutual relations. The sines and cosines themselves, are examined by means of some of the following "formula of verification."

(18° A) sin (90° - A)

2 sin 54° cos A

(18° + A)

sin

sin A + sin (36° - a) + sin (72° + a) = sin (36° + ^} + sin (72°-A).

Here A may have any assigned value less than the least specified arc. Thus, if A were 9°, we should have sin 9° + sin 27° + sin 81° = sin 45° + sin 63°, and if from any table the sines thus taken make equal sums, it is highly probable they are all correct. Taking these from Dr. Hutton's tables, we have

43. Of the formulæ investigated in this chapter, those which have letters of reference (A), (B), (C), &c. relate to the sums, differences, multiples, &c. of sincs, fan

gents, &c. while those which have figures of reference, (1), (2), (3), &c. will be employed in the solutions of plane triangles. Other kinds will find their application in subsequent parts of this introduction; and the student will do well, after he has gone through their investigaWe tion, to arrange them in separate tables for use. shall terminate the present chapter by subjoining three examples.

Example I.

Given the three sides of a plane triangle 40, 34, and 25, respectively, to find the largest angle, by formula (15).

Here (a + b + c -c

(a + b

15.5..log

b:

24.5..log

1.1903317

1.3891661

Their prod. shown by the sum of the logs.. 2.5794978

Their prod. shown by the sum of the logs.. 2:6723288

The latter sum taken from the former borrowing 20 for radius squared, gives

19.9071690.

The latter log. by 2 for the, gives 9.9535845. This, by the tables, is nearly tan 41° 56', and by the formula it is tan ; therefore a = 41°56′, and the required angle 83° 58′ nearly.

Note. In chap. iii. case 3, the same result is obtained by a different process.

Example II.

What are the angles of that plane triangle whose natural tangents are integers?

It is evident from equa. (4) that the sum of the three tangents must be equal to their continual product. Now,

the only three integers which possess this property, are 1, 2, and 3; which are, therefore, the tangents required. The angles of which they are the tangents, are 45°, 63° 26′ 6′′, and 71° 33′ 54′′; whose sum makes 180°, as it ought to do.

Example III.

There is a plane triangle whose sides are three consecutive terms in the natural series of integer numbers, and whose largest angle is just double the smallest. Required its sides and angles.

That the student may compare the two methods, we shall present both a geometrical and an algebraical solution to this problem.

1st. Geometrically.

Analysis. Suppose the triangle ABC to be that whose sides CB, BA, AC, are respectively as three consecutive terms in the increasing series of integer numbers, and the greater angle ABC equal to twice the less angle BAC. From c upon AB let fall the perpendicular CD, make Db DB, and join BC. Then, A because the angle cbB (= ABC) is equal

6DB

to 2CAB, the points a and c are in the circumference of a circle whose radius is ba, or bc, and centre b. But AB: AC + CB (≈ 2ab) :: AC

CB (= 2) : AD

DB:

4) CB; hence CB is given; and be

cause ABCB + 1 = AC 1 (by hypothesis),, the three sides are given, to construct the triangle.

Construction. Let the right line AB be made equal to 5, on any scale of equal parts; from centres A and B with radii 6 and 4 respectively, describe arcs to-intersect each other in c; draw AC, BC, and ABC is the required triangle.

Demonstration. The sides CB, BA, AC, are three consecutive terms in the series of natural numbers (by

const.) From c upon AB demit the perp. cn, set off Db = DB, and draw bc. Then, AB (= 5): AC + CB (=6+4) :: AC CB (64): AD - DB Ab 4. Therefore Ab = BC = bc: and hence the points A and c are in the circumference of the circle whose centre is ₺ and radius ba, or bc; and consequently (Euc. iii. 20) angle cỏв = CBA = 2 angle CAB.

2dly. Algebraically.

Denote вс by x- 1, BA by x, and AC by x + 1; then, (BE being perpendicular to AC the longest side) we have AC AB + BC :: AB -BC AE

EC, that is,

Half this added to half

2x

1

x + 1

By equating these two values of sin B, we have

x + 4

x + i

; whence x2 + 2x + 1 = x2 + 3x 4,

.

5. The sides, therefore, of the triangle are 4,

The cosines of the angles are, cos A =

3

And the angles are, A = 41° 24′ 34′′ 34′′

c = 55° 46′ 16′′ 18"'

B 82° 49′ 9′′ 8""

CHAPTER V.

Application of Plane Trigonometry to the determination of Heights and Distances.

1. ONE of the most familiar, and at the same time, useful, applications of plane trigonometry, is to the determination of the altitudes and distances of remote objects, the former usually designated by the term altimetry, the latter by that of longimetry. It is not intended to treat them separately here, nor to treat them jointly much in detail; but simply to present and solve a selection of such problems as are most likely to occur in practice, and as are best calculated to suggest the modes of procedure in other cases.

2. The instruments employed to measure angles are quadrants, sextants, and other circular instruments. For military men, perhaps, the best instrument is a pocket sextant, which, if accurately constructed by a skilful artist, will enable a careful observer to take angles to within a minute of the truth. But for general purposes the most proper instrument for the measuring of horizontal and vertical angles, is a theodolite furnished with a compass and level, one or two telescopes, and a vertical arc. Such an instrument, when each circle is 6 or 7 inches diameter, and has a nonius adapted to it, will enable the observer to ascertain angles to half a minute.