Suppose ABC a triangular piece of ground, which by an old survey we find to be thus: AB 260, AC 160, BC 150 perches; the mearing lines, AC and BC, are destroyed, or plowed down, and the line AB, only remaining. What angles must be set off at A and B, to run new mearings by, exactly where the old ones were?

This is performed by case 4. of oblique angled trigonometry, thus ;

1. As AB: AC+BC:: AC-BC: AD-DB=11. 92; then,

130+5.96=135.96=AD.
130-5.96-124.04=DB.

2. As AD: R:: AC: Sec. A,=31° 47′

3. As BC: S. A:: AC: S. B,=34° 10′.

Let D and C, be two trees in a bog, to which you can have no nearer access than at A and B; there is

49°, and the line AB 113 perches. Required the distances of the trees DC.

180°-the sum of DBA and DAB-ADB=31°. 180°-the sum of CAB and CBA-ACB=22.30.

In the triangle ABD, find DB, thus;

1. As S. ADB : AB :: S. DAB : DB,=216. And in the triangle ABC, find BC thus ;

2. As S. ACB: AB:: S. CAB: BC, 175.6

In the triangle DBC, you have DBC-ABC-ABD =729; likewise the sides BD, BC, as before found, given to find DC.

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

T. of DCB-CDB,=8° 5′

54°+8° 05-62° 05-DCB.
540-8° 05-45° 55-CDB.

4. As S. CDB: BC:: S. DBC: DC, 232.5

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. 4.) and CBE is a right angle (by cor. 5. theo. 7. sect. 1.) and ADC: whence ECB-ACD. The triangles CEB, CAD, are therefore mutually equiangular, and (by theo. 16. sect. 1.) DC : AC : : CB: CE, or DC : AC: CB CE. Q. E. D.

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

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,=516,

Then, 1.254.258=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,=1.736.

Then, by the preceding lemma.

2. As CD: AC:: CB: the diameter,=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.

From the centre E, let fall a perpendicular upon any of the sides, as EF, and it will bisect it 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. As AF: R.:: AE: Sec. FAE, 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 ;

As AE: R.:: AF: S. AEF,=60° 18',

Its complement 29° 47', is FAE, as before.

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