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In the Triangle ABC, given the Side AB 240, the

Side AC 180, and the Angle at A 36° 40′ to find the

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The given Angle BAC 36° 40', subtracted from

180°, leaves 143° 20′ the Sum of the other two Angles;

the half of which is 71° 40′.

As the Sum of two Sides, 420

2.62325

: Their Difference, 60

1.77815

:: Tangent half unknown Ang. 71° 40′

10.47969

12.25784

2.62325

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The Side BC may be found by CASE I or II.

CASE IV.

The three Sides given to find the Angles. Fig. 50. The solution of this CASE depends on the following PROPOSITION..

In every Plane Triangle, As the longest side; Is to the Sum of the other two Sides; So is the Difference between those two Sides; To the Difference between the Segments of the longest Side, made by a Perpendicular let fall from the Angle opposite that Side.

Half the Difference between these Segments, added

length of the longest Side, will give the greatest Segment; and this half Difference subtracted from the half Sum will be the lesser Segment. The Triangle beingth us 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
BC

.85

AC

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Sum of the two Sides 135

Difference

As the longest Side AB, 105

: Sum of the other two Sides, 135

:: Difference between those Sides, 35

85

50

35

2.02119

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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.

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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 igle ACB 98° 45′.

This Case may also be solved according to the following PROPOSITION.

In every plane Triangle, As the Product of any two Sides containing a required Angle; Is to the Product of half the Sum of the three Sides, and the Difference between that half Sum and the Side opposite the Angle required; So is the Square of Radius; To the Square of the Co-Sine of half the Angle required.

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 Surveyors 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.

SURVEYIN.

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; as the Survey can be more accurately taken.

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.

Note. The reverse of this Rule, that is, dividing by 4, will reduce 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

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