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and IA, IB, IC, &c. according to the given bearings ; then will the intersections A, B, C, &c. of the corresponding bearings HA and IA, HB and IB, HC and IC, &c. be the angular points of the field.
In each of the triangles IHA, IHB, IHC, &c. we have the side IH, and from the bearings of the sides, we have all the angles, to find the sides IA, IB, IC, &c.
Then in each of the triangles, IAB, IBC, ICD, &c. we have two sides and the included angle; whence the areas may be found by prob. III, sect. 1.
From the sum of the areas of the triangles IAB, IBC, ICD, and IDE, which is equal to the area IABCDEI, subtract the sum of the areas of the triangles IAG, IGF and IFE, which is equal to the area IAGFEI, the remainder will be the area of the field ABCDEFGA.
Note. In working the proportions for finding the sides IA, IB, &c., it will be unnecessary, when the area only is required, to take out the natural numbers corresponding to the logarithms of those sides; because in the proportions for finding the areas it will be sufficient to know the logarithms of the sides, without knowing their real lengths.
To find the double area of the triangle ICD.
To find the double area of the triangle IDE.
To find the double area of the triangle IEF.
To find the double area of the triangle IFG.
To find the double area of the triangle IAG.
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 two