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Let Dand C, be two trees in a bog, to which you can have no nearer access than at A and B; there is given, DAB 100°, CAB 36o. 30', CBA 121.. DBA 49°, and the line AB 113 perches. Required, the distances of the trees DC.

180o--the sum of DBA and DAB=ADB=31o. 180°—the sum of CAB and CB4=ACB=22. 30.

In the triangle ABD, find DB, thus ;

1. S. ADB: AB ::S. DAB · DB.

31° 113::1000 216.

And in the triangle ABC, find BC, thus

2. S. ACB: AB::S: CAB: BC.

22° 30' 113 36° 30' 175.6.

In the triangle DBC, you have DBC=ABCABD=720; likewise the sides BD, BC, as before found, given to find DC.

3. BD+BC: BD-BC:: T. of 1 DCB+CDB : 391.6 40.4

540

T. of DCB-CDB.

8° 05'.

549+8° 05'=620 05' =DCB.
54° -8° 05'=45° 55'=CDB.

4. S. CDB: BC:: S. DBC: DC.
45° 55. 175.6 72° 232.5.

LEMMA.

PL. 6. fig. 12.

If from a point C, of n 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. geom.) and CỔ E is a right angle (by cor. 5. theo. 7. geom.) and=ADC : whence ECB=ACD. The triangles CEB, CAD, are therefore mutually equiangular, and (by theo. 16. geom.) DC : AC:: CB: CE, or DC: CB :: AC:ČE. 2. E. D.

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PROB. V.

Pl. 6. fig. 5.

a

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?

By Construction.

By proh. 15. geom. 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 :: ACBC : AD-DB. 2.5 4.3

.3 ,516.

1.254.258=1.508=AD.

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

That is, 5.29—2,274064=3.015936.

Its square root is 1.736=CD.

=

Then by the preceding lemma,

2. CD: AC ::CB : the diameter.

1.736 2.3 2 2.65.

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

a

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

Wherefore AF=CF= AC=1.15.

In the right angled triangle A FE, 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 above angle FAE, may be had without a secant, as before, thus ;

AE:R::AF: S. AEF.

i
1.325 909 .115 60o. 13'

Its complement 29o. 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.

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MENSURATION OF AREAS, OR THE VARIOUS METHODS
OF CALCULATING THE SUPERFICIAL CONTENT OF

ANY FIELD.

DEFINITION.

THE

HE area or content of any plane surface, in perches, is the number of square perches which that surface contains.

Pl. 7. fig. 1.

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Let ABCD represent a rectangular parallelogram, or oblong: let the side AB, or DC, contain 8 equal parts ; and the side AD, or BC, three of such parts; let the line AB be moved in the direction of AD, till it has come to EF; where AE, or BF (the distance of it from its first situation) may be equal to one of the equal parts. Here it is evident, that the generated oblong ABEF, will contain as many squares as the side AB contains equal parts, which are 8; each square having for its side one of the equal parts, into which AB, or AD, is divided. Again, let AB move on till it comes to GH, so as GE, or HF, may be equal to AE, or BF; then it is plain that the oblong AGHB, will contain twice as many squares as the side AB contains equal parts. After the same manner it will appear, that the oblong ADCB will contain three times as many squares as the side AB contains equal parts ; and in general, that every rectangular parallelogram, whether square or oblong, contains as many squares as the product of the number of equal paris in the base, multiplied into the number of the same equal parts in the height, contains units, each square having for its side one of the equal parts.

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