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And in the triangle ABC, find BC, thus;

2. S. ACB ABS. CAB : BC. 22° 30′ 113 36° 30′ 175.6.

In the triangle PBC, you have DBC-ABCABD=72°; likewise the sides BD, BC, as before found, given to find DC.

3. BD+BC: BD-BC :: T. of DCB+CDB:

391.6

40.4

T..of DCB-CDB.

8° 05'

54°+8° 05-62° 05-DCB.
54°-8° 05=45° 55-CDB.

4. S. CDB: BC :: S. DBC: DC.

54°

45° 55' 175.6

72° 232.5

LEMMA.

Plate VI. fig. 10.

If from a point C, of a triangle ABC, inscribed in a circle, there be a perpendicular CD let fall upon the opposite side AB; that perpendicular is to one of the sides, including the angle, as the other side, including the angle, is to the diameter of the circle, i. e. DC: AC: CB CE.

Let the diameter CE be drawn and join EB; it is plain the angle CEB CAB (by cor. 2. theo. 7. sect. 1.) and CBE is a right angle (by cor. 5. theo. 7. sect. 1.) and = ADC: whence ECB

ACD.

ACD. The triangles CEB, CAD, are therefore mutually equiangular, and (by theo. 16. sect. 1.) DC: AC:: CB : CE, or DC : CB :": AC : CE, Q. E. D.

PROB. V.

Plate VI. fig. 5.

Let three gentlemen's seats, A, B, C, be situate in a triangular form: there is given, AB 2. 5 miles, AC 2, 3, and BC 2. It is required to build a church at E, that shall be equi-distant from the seats A, B, C. What distance must it be from each seat, and by what angle may the place of it be found?

Geometrically.

By prob. 15. sect. 1. Find the centre of a circle that will pass through the points A, B, C; and that will be the place of the church; the measure of which, to any of these points, is the answer for the distance: draw a line from any of the three points to the centre, and the angle it makes with either of the sides that contain the angle it was drawn to; that angle laid off by the direction of an instrument, on the ground, and the distance before found, being ranged thereon, will give the place of the church required.

By Calculation.

1. AB: AC+BC:: AC-BC: AD-DB.

2.5

4.3

.3

.516.

1,25+.258 1.508-AD.

By cor. 2. theo. 14. sect. 1. The square root of the difference of the squares of the hypothenuse AC, and given leg AD, will give DC.

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the half of which, viz. 1.325 is the semi-diameter. br distance of the church from each seat, that is, AE, CE, BE.

From the centre E, let fall a perpendicular upon any of the sides, as EF, and it will bisect in E: (by theo. 8. sect. 1.)

Wherefore AF-CF-AC-1.15.

In the right angled triangle AFE, you have AF 1.15, and AE the radius 1.325 given, to find FAE, thus;

3. AF R.:: AE: Sec. FAE. 1.15 90° 1.325 29° 47'.

Wherefore directing an instrument to make an angle of 29° 47', with the line AC; and measuring 1.325 on that line of direction, will give the place of the church, or the centre of a circle that will pass through A, B, and C.

The

The above angle FAE, may be had without a secant, as before, thus ;

AE R.: AF: S. AEF. 1.325 90° 1.15 60° 13'.

Its complement 29° 47', will give FAE, as be fore.

The questions that may be proposed on this head being innumerable, we have chosen to give only a few of the most useful.

OF

SECT. III.

Containing a particular Description of the several Instruments used in Surveying, with their respective Uses. And first,

OF THE

THE

CH AI N.

HE stationary distance, or mearings of ground, are measured either by Mr. Gunter's chain of four poles or perches, which consist of 100 links; (and this is the most natural division) or by one of 50 links, which contains two poles or perches but because the length of a perch differs in many places, therefore the lengths of chains and their respective links will differ also.

The English statute perch is 5 yards, the twopole chain is 11 yards, and the four-pole one is 22 yards: hence the length of a link in a statutechain is 7.92 inches.

There are other perches used in different parts of England, as the perch of woodland-measure, which is 6 yards; that of church land-measure, which is 7 yards (or the same with the plantation-perch) and the forest measure perah, which is 8 yards.

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