78 Wrights Mathematical Papers. 
Find by the foregoing directions all the including angles of a field, 
taking the remainder, after subtracting an outward angle from 360°, 
for an including angle; add them together; then to this sum add 
360, and divide the result by 90; if the courses have been correctly 
taken, the quotient will be exactly double the number of angles in 
the survey without a remainder. 
Ex. S. 84° W: thence N. 59° E: thence S. 34° W: thence 
N. 59° E: thence S. 27° W. 
BAGS BOE clo Baa 5D aa 25 
59 «884 BOs 27 84 25 
Bee Pp ey ee 835 
25.1128 25 2h 22 57 32 
360 180 123 
— = 360 
335 128 tues 
; 90)900(10 
No. III.— general method of finding the area of an irregular 
Polygon, having the sides and contained angles given. 
Suppose the several sides of the polygon AB, BC, CD, DE, EA, 
to be hypothenuses of right angled triangles, of which the perpendicu- 
lars Bf, Cg, Dh, Ez, are parallel with the first side AB, with which 
the calculation commences. 
Let the contained angles ABC, BCD, CDE, DEA, be exchang- 
ed for the angles fBC, gCD, ADE,:EA. These latter may be 
termed angles of commutation, and are obtained in the way here- 
after described. 
The positions of the bases and perpendic- 
ulars, are next to be discovered and desig- 
nated. ‘The perpendiculars Bf, Cg on one 
side of the bases may be considered affirm- 
ative, denoted by A, and the perpendiculars 
Dh, Ei on the opposite side, negative, deno- 
ted by N, placed before the angle of com- 
mutation ; likewise the bases fC, gD, on one 
side of the perpendiculars, may be consid- 
ered affirmative, denoted by A, and the bases AE, 7A, on the opposite 
side, negative, denoted by N, placed after the angle of commutation. 
When a side happens to be at right angles with the first side, the 
perpendicular vanishes; and when a side is parallel with the first 
side, the base and angle of commutation vanish: yet, for the pur- 
pose of discovering the positions of the subsequent bases and per- 
