Examples.

9. A ship sails southeasterly from the South Point of the Great Bank of Newfoundland a distance of 2821 miles, when it has made a departure of 910 miles; find the position at which it has arrived.

Ans. Its position is 208 miles north of Cape St. Roque.

Area of triangle.

CHAPTER VI.

SURVEYING.

56. The object of Surveying is to determine the dimensions and areas of portions of the earth's surface. In the application of Plane Trigonometry, the portions of the earth are supposed to be so small that the curvature of the earth is neglected. They are, in this case, nothing more than common fields bounded by lines either straight or curved.

57. Problem. To find the area of a triangular field, when its angles and one of its sides are known.

Solution. Let ABC (fig. 2) be the triangle to be measured, and c the given side. The area of the triangle is equal to half the product of its base by its altitude, or

58. Problem. To find the area of a triangular field, when two of its sides and the included angle are known.

Solution. Let ABC (fig. 2) be the triangle to be measured, b and c the given sides, and A the given angle. Then, by (273),

or, the area of a triangle is equal to half the continued product of two of its sides and the sine of the included angle.

59. Problem. To find the area of a triangular field, when its three side are known.

Solution. Let ABC (fig. 1) be the given triangle. Then,

in which s denotes the half sum of the three sides of the triangle.

Area of triangle.

Hence

and

b c sin. A = 2 √ [s (s — a) (s — b) (s — c) ] ;

area of ABC = √ [s (s — a) (s — b) (s — c)]; (276) or, to find the area of a triangular field, subtract each side separately from the half sum of the sides; and the square root of the continued product of the half sum and the three remainders is the required area.

60. EXAMPLES.

1. Given the three sides of a triangular field, equal to 45.56 ch., 52.98 ch., and 61.22 ch.; to find its area.

Solution. In (fig. 1) let a = 45.56 ch., b 52.98 ch.,

c = 61.22 ch.

=

Area of ABC = 1173. 1 sq. ch. 3.06932

Ans. The area 117 A. 1 R. 9 r.

2. Given the three sides of a triangular field equal to 32.56 ch., 57.84 ch., and 44.44 ch.; to find its area.

Ans. The area 71 A. 3 R. 29 r.

Area of rectilinear field.

3. Given one side of a triangular field equal to 17.95 ch., and the adjacent angles equal to 100° and 70°; to find its

area.

Ans. The area 85 A. 3 R. 17 r.

4. Given two sides of a triangular field equal to 12.34 ch. and 17.97 ch., and the included angle equal to 44° 56′; to find its area.

Ans. The area 7 A. 3 R. 13 r.

61. Problem. To find the area of an irregular field bounded by straight lines.

First Method of Solution. Divide the field into triangles in any manner best suited to the nature of the ground. Measure all those sides and angles which can be measured conveniently, remembering that three parts of each triangle, one of which is a side, must be known to determine it.

But it is desirable to measure more than three parts of each triangle, when it can be done; because the comparison of them with each other will often serve to correct the errors of observation. Thus, if the three angles were measured, and their sum found to differ from 180°, it would show there was an error; and the error, if small, might be divided between the angles; but if the error was large, it would show the observations were inaccurate, and must be taken again.

The area of each triangle is to be calculated by one of the preceding formulas, and the sum of the areas of the triangles is the area of the whole field.

This method of solution is general, and may be applied to surfaces of any extent, provided each triangle