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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 judicious tutor. Nor 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 ythe 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 2 Multiply mat. cos 44° = .996.9173 By . . . . . . . . . . . . . . . . . 400

Product gives hor. dist. = 898-7669200

. 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 prosesses 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 4° 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 formulae 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 C distance of the tower from each end of the measured base.

In the annexed figure are given

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conseq. ca 180°– (A+B) = 57°34'. P
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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 B 64° 51’’. . . . 9.9567437
To Ac. .. 643°49. . . . 2-8085443.

Remark. These distances might have been deter

mined without the aid of an instrument to measure

angles. Thus, suppose in the continuations of the respective lines CA, cB, two distances, AD, BE were mea-. sured, each = 120; and suppose, on measuring the distances BD, A E, the former is found to be 660, the latter 672 yards; from these measures the required angles may be determined. For, in the triangle ABE thus formed, we have (Euc. ii. 12) AE* = AB* + BE* + 2EB. BP. Whence, by trans- - - - - - AF* – AB” – BE.” position and division, BP = —ass- But BP is Ee the cosine of the angle ABC, to the radius AB; so that, dividing the preceding expression by AB, we shall have the cosine of that angle to radius 1. A like process will give the cosine of CAB. AE? — AB” – BE.” 6722 – 600? – 1203 ;III- = 240. 600 = 425 = cos 64° 51’ BD” – AB" – AD” 660 – 600? – 120° and, cos BAc = −.I-- 240. 600 = .536 = cos 57° 35'. Hence, the angles ABC, and BAC, being determined, the distances are found as before.

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Thus, cos ABC =

ExAMPLE II.

In order to find the distance between two trees A and B, which could not be directly measured because of a pool which occupied much of the intermediate space, I measured the distance of each of them from a third object c, viz. Ac = 588, Bc = 672, and then at the point c took the angle ACB between the two trees = 55° 40'. Required their distance.

This is an example to case 2 of plane triangles, in which two sides and the included angle are given. The work, therefore, is left to exercise the student: the answer is 593-8

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Wanting to know the distance between two inaccessible objects, which lay in a direct line from the bottom

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of a tower on whose top I stood, I took the angles of depression of the two objects, viz. of the most remote 25%", of the nearest 57°. What is the distance between them, the height of the tower being 120 feet? A The figure being constructed, as in the margin, AB = 120 feet, the altitude of the tower, and AH the horizontal line drawn through its B top; there are given, 11 AD = 25° 30', hence BAB = BAH – 11 AD = 64° 30' . HAc = 57° 0', hence BAC = BAHI – HAc = 33° 0'. Hence the following calculation.

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Conseq. BD – BC = 173656 feet, the distance required, Such is the way in which this problem is usually solved: it may, however, be done more easily and concisely, o means of the natural tangents. For, if A B be regarded as radius, BD and BC will be the tangents of the respective angles BAD, BAC, and CD the difference of those tangents. It is, therefore, equal to the product of the difference of the natural tangents of those angles into the height AB. Thus, nat. tan 6

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ExAMPLE IV.

From the top of a hill I observed two mile-stones on a horizontal road, which ran straight from its bottom, and took their respective angles of depression below the horizontal plane passing through the place of my eye; that of the nearer mile-stone was 14° 3', that of the farther was 3° 56'. Required the height of the hill.

The figure being drawn, it will be found analogous to that in exam. 3, to which, therefore, we shall refer. There are given CD = 1760 yards, the distance between the two mile-stones; ADB = HAD = 3° 56'; AcB = HAc = 14° 3". This admits of three distinct modes of determination, as below.

1st Method.

The angle AcB is equal to the sum of the angles CAD and cDA (Euc. i. 32); therefore CAD = AcB – ADB = 10° 7'. Then, in the triangle AcD, it will be, as sin cAD : CD :: sin cDA : cA. And, in triangle AcB, it will be, as rad. or sin B : Ac:: sin c : AB = 166-85; and so, if it be required, is sin cAB : cB = 666-75.

According to this method the logarithmic process will require eight lines.

- 2d Method. Since CAD is the difference of AcB and ADC, we have,

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sinc sin D cosec (c.— o:
Or, making the terms homogeneous (ch. iv.17),
AB. rad} = cD sin c sin D cosec (c — D).
The logarithmic formula is, therefore, this:

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