NOTE. This theorem follows immediately from Theorems I. and II., but we prove it algebraically in order to show an elegant application of the addition theorem. Proof. Put a, ß, y, etc., the angles which the straight lines make with one of the lines of projection; π, the angle which the first two lines of projection make with each other. π, γ π, etc., will be the angles which the lines make with the second line of projection. By hypothesis we have a cos a + b cos B+c cos y + etc. = a cos (a π) + b cos (ẞ -- π) + etc. = 0. The last equation, by the addition theorem, reduces to cos л (α cos a + b cos ẞ + c cos y + .) ) = 0. sin (a sin a + b sin ẞ + c sin y + The first term of this equation vanishes by (a). Hence, the whole sum being zero, the second term must also vanish, which requires that we either have which will give = 0 or 180°,-in which case the two lines would be parallel,—or (b) a sin ab sin 6+ c sin y + etc. = 0. Since, by hypothesis, the two lines are not parallel, the equation b must hold true. Now let be the angle which any third line of projection forms with the first line. The angles which the lines a, b, c, etc., form with this third line will then be Therefore the sum of the projections upon this line will be which reduces to cos 0) + b cos (8-0) + c cos (y) + etc., (a cos a + b cos ẞ + c cos y + etc.) + sin (a sin a + b sin y+c sin y + etc.), a sum which vanishes by (a) and (b), whatever be the value of 0, thus proving the theorem. 72. Cor. From §§ 69 and 71 it follows that if all the sides of a polygon but one are given in length and direction,—a, b, c, etc., being the lengths, and a, ß, y, etc., the angles with a fixed base line, and if / be the omitted side and 2, its angle, we shall have which will determine 7 and 2, by Prob. I. Chap. V. Example. A surveyor measures off courses and distances as follows: I. North 80° 28' east, 42.68 metres. II. North 23° 22′ east, 22.79 What distance and direction will means r° east of north. Taking the east and west line OX as the base from which we measure angles, we readily find Ex 66 66 D B X the angles made by each side with the base, as shown in the following table, which also gives the values of a sin a, b sin ß, etc., and a cos a, b cos ß, etc., as computed from the given data: from which we find 7 = 24.575; 2301° 18'.1. Expressed in the language of surveyors, the angle 2 indicates the direction, South 31° 18'.2 East. It will be seen that we have taken as the positive direction of the last line that from the point last reached to the starting-point. This is in accordance with the convention that the positive directions of the several sides of a polygon are so taken that in passing around it the directions shall all be positive or all negative. But we might equally consider the problem: Having proceeded along a series of connected lines, AB, BC, etc., of which the lengths and directions are given, to E, what is our distance and direction from our starting-point A? It is evident that the distance and direction are the length and direction of the line AE, which is simply the negative of the side EA necessary to complete the polygon. If we call & the angle of direction of AE, the equations for determining land & would be 7 sin & a sin a+b sin ẞ+ etc.; l cos ε = a cos a + b cos ß, etc.; and, in the preceding example, we should have (3) 1. A surveyor ran a course S. 12° 13' E., 289.26 metres, and thence N. 82° 49′ E., 92.68 metres. What is his direction and distance from his starting-point? What is the direction and distance of the starting-point from him? 2. Five sides of an irregular hexagon taken in order have the following lengths and directions relative to a fixed line: a. Length, 297.43 metres; direction, 332° 6'.8; What is the length and direction of the remaining side? 73. Areas of polygons. When the sides of a polygon are all given in length and direction, the area may be computed by a process demonstrated in geometry, but which we shall describe here. Let ABCDE be any polygon, and OX the base line from which we measure angles. The area of this polygon is B equal to Area A'ABCC' minus Area A'AEDCC'. Α' E' B D The first area is equal to the sum of the areas of the two trapezoids A'ABB' and B'BCC'. The second is equal to the sum of the areas of the three trapezoids . C'CDD', D'DEE', and E'EAA'. It will be noted that there is one trapezoid for each side of the polygon, of which the non-parallel sides are the side of the polygon and its projection upon the base line. We have for the area of the first trapezoid, noting that the base line OX is perpendicular to the parallel sides, Area A'ABB' = {(AA' + BB') A'B'. Putting, as before, a, b, c, etc., for the length of the sides AB, BC, CD, etc.; a, B, y, etc., for the angles which they form with OX; putting, also, we have p, the length AA', AA' =P; A'B' a cos a. Substituting these values, we have, for the area of the first trapezoid, Area A'ABB' = (2p+ a sin a) a cos a. Passing on to the other trapezoids, we have, for the lengths of the several perpendiculars, BB' = p + a sin a; CC'pa sin a+b sin ß; DD' = p + a sin a +b sin ß+c sin y ; EE' pa sin a+b sin ẞ+c sin y+d sin o Also, for the altitudes of the trapezoids between their parallel (2p+2a sin a + 2b sin ẞ+c sin y) c cos y; (2p+2a sin a + 2b sin ẞ + 2e sin y+d sin d) d cos &; (2p+2a sin a + 2b sin ẞ+ 2c sin y +2d sin de sin ε) e cos ε. (a) It has been shown that the required area of the polygon is found by subtracting the last three of these areas from the first two. But in reaching this conclusion we took no account of algebraic signs, and so virtually considered the areas all positive. Now, as the figure is drawn, cos a and cos B are positive, and the cosines of y, d, and ɛ are negative. Hence if we put the sign+before each of the areas (a), the subtraction will be indicated by the negative character of those products which have cos y, cos d, and cos & as factors, and so the algebraic expression will be correct. If we add up the quantities (a), beginning with the terms in p, we see that the coefficient of Ρ is sion for the area. Since p is defined as the distance below A at |