one from another. Thus, if AC DE BH G F, &c., be several objects, the situations of which D Fig. 15. G are to be laid down on a map, and they are within the lines, A C D E B H G F, accurately calculated. It is supposed that the stations A and B are chosen such as that all the others can be seen from each of them. Then from the extremity A, measure the angles E A B, D AB, CA B, &c., HA B, GA B, FA B, &c. And from the other extremity B, measure the angles, CBA, DBA, E B A, &c., F B A, G B A, H B A, &c. And as the common base, A B, and the several angles of all the triangles are now known, the sides, A C, A D, A E, &c. may be determined by simple proportion, for as the natural sine of ACB:AB:: sine CAB: CB and so is sine A B C to CA, and so through all the triangles, the three sides being thus found in each triangle, the area is easily found, as shown in section 8th of this treatise. But to insure accuracy the objects C D E, etc., should be all intersected from some third station, O in the base A B, otherwise the figure may appear in the plotting to be right when it is not so, and there will be no means of knowing whether the angles have been correctly taken without going over the work again. 27th. Here follows an example of a triangle con A. R. P. taining a mean area of 1135.2.12.79. The sides of which were traced by a transit instrument, and poles placed at the several points marked thus ; this being done, the respective distances of the sides were ascertained by a mean of measures as follows, viz: B A 14643 links, or 9664.38 feet, A C 17814 links, or 11777.24 feet, B C 16588 links, or 10948. 08 feet. The angles were taken by a theodolite as they are marked in the figure. Now to determine the area of the triangle, A BC: 1st. From the data, A B, and the three angles of 2d, by B C, and the three angles, the area will be 1135.3.029 3d, by C A, and the three angles, the area will be 1135.0.38.6 4th, by data A B, and the two adjacent angles, we have by the known formula, A B Xsine BXsine A, 2 sine (B+A) The area will be 5th, and by B C, and the two ad jacent angles 6th, by a similar formula from A C, and the two adjacent angles, the area will be 7th, by data A B, and the adjacent angle A, and the remote angle C, we have by the known formula, (A B) Xsine AXsine (C+A) 2 sine C. A. R. P. 1135.2.25.7 1135.3.01.9 1135.0.37.99 8th, by a similar formula from hav ing A B, and the angles, B and C; area 1135.2.28.2 9th, by having C B and the angles, C and A; area 1135.3.03.58 10th, by having C B and the angles B and A; area 1135.3.04.38 11th, by a similar formula data C A, and the angles, C and B, gives the area 12th, by a similar formula from da ta CA, and the angles A and B; area 13th, by data A BXB C, and the contained angle, we have A. R. P. 1135.0.39.66 1135.1.00.12 A BXB CXsine B=1135.2.35.06 2 14th, by A CXA B, and the con tained angle 1135.1.32.92 15th, by A CXB C, and their con tained angle C 16th, by data, A BXB C, and the angle, A, we have by a known 1135.2.00.79 area 1135.2.394 =sine C, and A BXB C, sine (A+C) 2 17th, by the application of similar formula to the data, A BXB C, and angle, C; area 1135.2.30.4 |