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.

C

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 cbв (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

4AB

= ab (= = 4) CB; hence CB is given; and be

AB

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 copsecutive terms in the series of natural numbers (by

const.) From c upon AB demit the perp. CD, set off .Db DB, and draw bc. Then, AB (='5): AC + CB (=6+ 4) :: AC — CB (#6 — 4) : 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 b and radius ba, or bc; and consequently (Euc. iii. 20) angle cỏв = CBA = 2 angle CAB.

2dly. Algebraically.

Denote BC 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

x2+4x

2.x 1

x + 1

AE. But AB :

2

2x + 2

2x + 2 2x + 2

and (by the quest.) this angle2A. But, equa. (1),

By equating these two values of sin B, we have

x + 1

x-1

=

; whence x2+ 2x + 1 = x2 + 3x − 4,

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

15, and 6. The cosines of the angles are, cos AT

And the angles are, a = 41° 24′ 34′′ 34′′ c55° 46′ 16′′ 18′′

B=82° 49′ 9′′ 8′′"

CHAPTER V.

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

11.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 are. 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.

cious tutor.

3. The space would be improperly occupied in giving particular descriptions of these instruments, and the manner of adjusting them for use. These matters will be much better learnt from an examination of the instruments, and a few explanatory remarks from a judiNor shall we here embarrass the computation and diagrams, by showing how to allow for the height of the instrument, for that is a matter which requires only a single hint from the person who teaches the use of the instrument employed. So again, in reference to chains, tapes, and other contrivances for measuring lines, descriptions are suppressed. It ought, however, to be observed, that whenever a base, or distance between two stations, is measured on sloping ground, it must be reduced to the corresponding horizontal line, before it is employed in the general computation, if horizontal angles are taken at its extremities with a theodolite. This is in all cases easily effected: for, if the sloping or hypothenusal line be regarded as radius, the corresponding horizontal line will be the cosine of the inclination of the plane on which the sloping line was measured. It is, therefore, simply requisite to multiply the natural cosine of the angle of inclination, by the length of the line measured, to obtain the true horizontal line.

Suppose that on a plane inclined to the horizon in an angle of 4 degrees, a distance of 400 yards were measured, what would be the corresponding horizontal distance?

Multiply nat. cos 41° =
By...

•9969173

400

Product gives hor. dist. = 398-7669200

Here the difference is not 14 yard, or not a 320th part of the measured line. As this is not a greater deviation from accuracy than will occur in the usual processes for measuring distances with a chain or a tape,

the reduction from the sloping to the horizontal line, may, in common cases, be disregarded, except the angle of inclination exceed 41° or 5°.

4. There are several simple methods of approximating to the heights of both accessible and inaccessible objects, by means of shadows, mirrors, unequal vertical staffs, &c. But as these depend solely upon the principle of similar triangles, and do not require the theorems and formulæ of trigonometry, properly so called, they are not described here. The student may, without being detained by farther observations, proceed to the solution of the following problems.

EXAMPLE I.

In order to ascertain my distance from a tower, which was inaccessible by reason of an intervening river, I measured on a horizontal plane a base line of 600 yards, and at each end took the angle included between the other end and the tower, finding them to be 57° 35′ and 64° 51′ respectively. Required the distance of the tower from each end of the measured base.

In the annexed figure are given

600

AB

CAB

57° 35′

CBA

64° 51′

conseq. c 180° (A+B) = 57° 34'. D

C

B

E

Logs

Hence, sin c....... 57° 34′.... 9·9263507

To AB... 600

2.7781513

So is sin A 57° 35′ .... 9.9264310 To BC... 600.11 2.7782316 and, So is sin в 64° 51′.. 9-9567437

....

To AC... 643-49.... 2·8085443.

Remark. These distances might have been determined without the aid of an instrument to measure angles. Thus, suppose in the continuations of the res