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34

TRIGONOMETRY.

CASE III.

Fig. 49.

C

Two Sides and their contained Angle given, to find the other Angles and Side. Fig. 49.

180

AL 36 40

B

240 The solution of this Casa depends on the following PROPOSITION.

In every Plane Triangle, as the sum of any two Sides is to their difference, so is the Tangent of half the sum of the two opposite Angles to the Tangent of half the difference between them. Add this half difference to half the sum of the Angles and

you will have the greater Angle, and subtract the half difference from the half sum and you will have the lesser Angle.

In the Triangle ABC, given the Side AB 240, the Side AC
180, and the Angle at A 360 40' to find the other Angles and
Side..
Side AB

240
AB

240 AC

180
AC

180

420

Sum of the two Sides

Difference

60

The given Angle BAC 360 40', subtracted from 1800, leaves 1430 20' the sum of the other two Angles, the half of which is 710 40'. As the sum of two Sides, 420

2.62325 Their difference 60

1.77815 :: Tangent half unknown Ang. 710 40'

10,47969

:

12.25784 2.62325

: Tangent half difference, 230 20%

9.63459:

The half sum of the two unknown Angles,
The half difference between them,

710 40

23 20

Add, gives the greater Angle ACB

95 00

Subtract, gives the lesser Angle ABC

48 20 CASE IV.

Fig. 50.

85

The three sides given to find the Angles. Fig. 50.

75 39

A 105 D The solution of this Case depends on the following PROPO

SITION.

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 to half the sum of the Segments, that is, to half the 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 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 36

TRIGONOMETRY.

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 1.92942 As Hyp. BC, 50

1.69897 : Radius 10.00000 : Radius

10.00000 :: Seg. AD, 75 1.87506 :: Seg. BD, 30

1.47712

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The Angle DCA 610 55' subtracted from 900 leaves the Angle CAD 280 5.

The Angle DCB 360 50 subtracted from 900 leaves the Angle CBD 530 10'.

The Angle DCA 610 55' added to the Angle DCB 360 50 gives the Angle ACB 980 45'.

This Case may also be soived 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

who is ignorant of TRIGONOMETRY, can be an accomplished

one

SURVEYING,

SURVEYING 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, iš 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 Hundredihs. 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 PROBLEMIS.

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 apnex 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 eyen 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 Decind Parts.

38

SURVEYING.

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 84 Acres and 6 Tenths.

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 Superficial Con

tents, is expressed in Square Chains and Links, the whole may be conşidered 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 oft by 40, and again cut off 5 figures, and you have the Kods.

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

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