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length of the longest Side, will give the greater Segment; and this half Difference subtracted from the half Sum will give the lesser Segment. The Triangle being thus divided becomes two Right Angled Triangles, in which the Hypothenuse and one Leg are given to find the Angles.

In the Triangle ABC, given the Side AB 105, the
Side AC 85, and the Side BC 50; to find the Angles.
Side AC
85 AC

85
BC
50 BC

50

Sum of the two Sides 135

Difference

35

As the longest Side AB, 105 : Sum of the other two Sides, 135 : : Difference between those Sides, 35

2.02119
2.13033
1.54407

3.67440
2.02119

: Difference between the Segments, 45

1.65321

Half the Side AB
Half the Difference of the Segments

52.5 22.5

Add, gives the greater Segment AD

75.0

Subtract, gives the lesser Segment BD

30.0

Thus the Triangle is divided into two Right Angled Triangles, ADC and BDC ; in each of which the Hypothenuse and one Leg are given to find the Angles. To find the Angle DCA.

To find the Angle DCB. As Hyp. AC, 85 1.929.12 As Hyp. BC, 50 1.69897 : Radius 10.00000 : Radius

10.00000 :: Seg. AL', 75 1.87506 :: Seg. BD, 30

1.47712

11.87 506
1.92942

11.47712 1.69897

: Sine DCA, 61° 55' . 9.94564.

: Sine DCB, 36° 50'

9.77815

The Angle DCA 61° 55' subtracted from 90° leaves the Angle CAD 28° 5'.

The Angle DCB 36° 50' subtracted from 90% leaves the Angle CBD 53° 10'.

The Angle DCA 61° 55' added to the Angle DCB 36° 50' gives the Angle ACB 98° 45'.

Those who make themselves well acquainted with TRIGONOMETRY will find its application easy to many useful purposes, particularly to the mensuration of Heights and Distances; called ALTIMETRY and LonGIMETRY. These are here omitted because, as this work is designed principally to teach the Art of common FIELD SURVEYING, it was thought improper to swell its size, and consequently increase its price, by inserting any thing not particularly connected with that Art.

It is recommended to those who design to be Sur. veyors to study Trigonometry thoroughly ; for though a common Field may be measured without an acquaintance with that Science, yet many cases will occur in practice where a knowledge of it will be found very beneficial ; particularly in dividing Land, and ascertaining the boundaries of old Surveys. Indeed no one, who is ignorant of TriGONOMETRY, can be an accomplished Surveyor.

SURVEYING.

SURY

URVEYING is the Art of measuring, laying out and dividing Land.

PART I.

MEASURING LAND.

The most common measure for Land is the Acre ; which contains 160 Square Rods, Poles or Perches ; or 4 Square Roods, each containing 40 Square Rods.

The instrument most in use, for measuring the Sides of Fields, is GUNTER's Chain, which is in length 4 Rods or 66 Feet; and is divided into 100 equal parts, called Links, each containing 7 Inches and 92 Hundredths. Consequently, 1 Square Chain contains 16 Square Rods, and 10 Square Chains make 1 Acre.

In small Fields, or where the Land is uneven, as is the case with a great part of the Land in New-England, it is better to use a Chain of only two Rods in length;

SECTION I.

PRELIMINARY PROBLEMS. PROBLEM I. To reduce Two Rod Chains to Four Rod Chains.

RULE. If the number of Two Rod Chains be even take half the number for Four Rod Chains, and annex the Links if any : Thus, 16 Two Rod Chains and 37 Links make 8 Four Rod Chains and 37 Links.

But if the number of Chains be odd, take half the greatest even number for Chains, and for the remaining number add 50 to the Links : Thus, 17 Two Rod Chains and 42 Links make 8 Four Rod Chains and 92 Links.

PROBLEM II. To reduce Two Rod Chains to Rods and Decimal Parts.

Rule. Multiply the Chains by 2 and the Links by 4, which will give Hundredths of a Rod : Thus, 17 Two Rod Chains and 21 Links make 34 Rods and 84 Hundredths ; expressed thus 34.84 Rods.

If the Links exceed 25 add 1 to the number of Rods and multiply the excess by 4 : Thus, 15 Two Rod Chains and 38 Links make 31.52 Rods.

PROBLEM III. To reduce Four Rod Chains to Rods and Decimal Parts.

Rule. Multiply the Chains, or Chains and Links, by 4; the Product will be Rods and Hundredths : Thus, 8 Chains and 64 Links make 34.56 Rods. Notr. The reverse of this Rule, that is, dividing by 4, will re

duce Rods and Decimals to Chains and Links : Thus, 105.12 Rods make 26 Chains and 28 Links. PROBLEM IV. To reduce Square Rods to Acres.

Rule. Divide the Rods by 160, and the Remainder by 40, if it exceeds that number, for Roods or Quarters of an Acre : Thus, 746 Square Rods make 4 Acres, 2 Roods and 26 Rods.

PROBLEM V. To reduce Square Chains to Acres.

RULE. Divide by 10; or, which is the same thing, cut off the Right hand figure : Thus, 1460 Square Chains make 146 Acres; and 846 Square Chains make

PROBLEM VI. To Reduce Square Links to Acres.

RULE. Divide by 100000; or, which is the same thing, cut off the 5 Right hand figures : Thus, 3845120 Square Links make 38 Acres and 45120 Decimals. Note. When the Area of a Field, by which is meant its Super

ficial Contents, is expressed in Square Chains and Links, the whole may be considered as Square Links, and the number of Acres, contained in the field, found as above. Then multiply the figures cut off by 4, and again cut off 5 figures, and you have the Roods ; multiply the figures last cut off by 40, and again cut off 5 figures, and you have the Rods.

EXAMPLE. How many Acres, Roods and Rods are there in 156 Square Chains and 3274 Square Links ?

15)63274 Square Links

4

2)53096

40

21)23840

Answer. 15 Acres 2 Roods and 21 Rods.

PROBLEMs for finding the Area of Right Lined Figures, and also of Circles.

PROBLEM VII. To find the Area of a Square or Parallelogram.

RULE. Multiply the length into the breadth ; the Product will be the Area.

PROBLEM VIII, To find the Area of a Rhombus ør Rhomboides.

Rule. Drop a Perpendicular from one of the Angles to its opposite Side, and multiply that Side into the Perpendicular ; the Product will be the Area.

PROBLEM IX. To find the Area of a Triangle.

Rule 1. Drop a Perpendicular from one of the Angles to its opposite Side, which may be called the Base ; then multiply the Base by half the Perpendicular, or the Perpendicular by half the Base ; the Product will be the Area. Or, multiply the whole Base by the whole Perpendicular, and half the Product will

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