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[blocks in formation]

IA

IB

9.99335

9.80807

1.44716

11.25523

log. 1.26188

To find the double area of the triangle IAB.

As rad.

10.00000

: sin. AIB 13° 45'

9.37600

log. 1.46937

:: IAXIB,

1.54943

: 2IAB 248.2

2.39480

To find the double area of the triangle IBC.

As rad.

10.00000

: sin. BIC 9° 15'

9.20613

SIB

:: IBXIC,

log. 1.54943

IC

1.47463

To find the double area of the triangle ICD.

As rad.

10.00000

9.52350

log. 1.47463 1.48780

:: IC× ID,

sin. CID 19° 30'

SIC

ID

: 2.ICD 306.15

To find the double area of the triangle IDE.

As rad.

10.00000

: sin. DIE 26° 30'

9.64953

log. 1.48780

1.34698

:: ID×IE, IE

{

: 2.IDE 305.007

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ID

2.48593

To find the double area of the triangle IEF.

As rad.

10.00000

: sin. EIF 24° 00′

9.60931

:: IE-IF,{

SIE

log. 1.34698

IF

1.13542

: 2.IEF 123.511

2.48431

T

To find the double area of the triangle IFG.

As rad.

10.00000

9.36818

:: IF×IG, {

: sin. FIG 13° 30 SIF

log. 1.13542

IG

1.26188

: 2.IFG 58.274

2.09171

1.76548

To find the double area of the triangle IAG.

As rad.

10.00000

9.71809

log. 1.46937
1.26188

2.44934

2.IAB

sin. AIG 31° 30'

:: IA×IG, IG
SIA

: 2.1AG 281.412

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2.IBC

2,1CD

2.IDE

2.IABCDEI 2.IAGFEL®

169.9

306.15

305.007

1029.257

463.197

2.ABCDEFGA 566.060

ABCDEFGA

EXAMPLE 2.

2.IFG 58.274

2.IAG 281.412

2.IAGFEI 463.197

283.03 Ch. 28 A. 1 R. 8 P.

The bearings and distances of the sides, if required, might readily be obtained. For, having found the distances IA, IB, we have in the triangle IAB, two sides and an included angle; whence the angle IAB and side AB may be found. The angle IAB applied to the bearing of IA, will give the bearing of AB. In the same manner the bearings and distances of the other sides may be found.

Being required to calculate the area of a field, the owner of which refuses permission to go on it, I choose

all the angles of the field are visible. The bearing and distance of the stations, and the bearings of the angles, from each station, are as follow. What is the area of the field?

The station G bears from the station F, N. 43° W. 20 ch.

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LAYING OUT AND DIVIDING LAND.

PROBLEM 1.

To lay out a given quantity of land in a square form.

Reduce the given quantity to chains or perches and extract the square root, which will be the length of a side, of the same denomination to which the given quantity is reduced.

RULE.

EXAMPLES.

1. Required the side of a square that shall contain 9 A. 3 R. 28 P.

40)28 Per.

4)3.7 R.

9.925 A. -99.25 Ch.

Ch.

99.25(9.96 Ch. the length of a side.

81

189)1825

1701

1986)12400

11916

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