Draw the tangents, radii, and curves, fixing the P. R. C. midway of D. Draw the chords G I, I E, the line B F perpendicular to G I, and the line E H in prolongation of radius CE to an intersection with B H passed through centre B parallel to tangents. That I falls midway of D, follows from the necessary symmetry of the figure; and GIE must be a straight line, because the radii BI, IC, perpendicular to a common tangent at the same point, form a straight line, to which the chords GI, IE, are equally inclined. CH CB = .. GE =cos. A; but CH2 RD, and C B = 2 R. ... cos. A = (2 R— D) ÷ 2 R. = = GFR sin. † A; GE=4 GF. 4 R sin. A, and GI or IE = 2 R sin. A. Observe, that, in the right triangles G KE and BG F, angles at G and B are each equal to the A: hence the triangles 1,600 X 0.1725 = BH may then be found 2 R sin. A 276 feet, and laid off from the P. C. at G to K, the point E being fixed by a right angle from K. = = Or GE may be found 4 R sin. A 3,200 X 0.866 277.1 feet, and laid off from G to E, the point I being fixed 138.5 feet from G, and angle KG E made equal to half of A = 4° 58'. 2. The distances G K and D given, to find R. In triangle GKE, KE = D. = Or, having found GE, we have from the congruity of triangles G KE, BFG, D: GE GE or GF: R. TO CONNECT TWO PARALLEL TANGENTS BY A REVERSED CURVE HAVING UNEQUAL RADII. 1. Given the perpendicular distance, D, between two parallel tangents, and the unequal radii, R and r, of a reversed curve, to find the central angles, A, the chords, and the = = Cos. A CH BC; but CH (R − r) — D, and BC=R+r. The straight reach G K = BH = (R+ r) sin. A. The sum of the chords G E = GK÷cos. GI= 2 R sin. A. IE=2r sin. A = GE — GI. Example. D= 28, R= 955, r = 574. A. Cos. A(R+ r−D) ÷ (R+ r) = 1,501 1,529. 2. The distances GK and D, and one of the unequal radii, R, given, to find the other radius, r, and the central A REVERSED CURVE HAVING UNEQUAL ANGLES. Given the angles A and B, and the length A B of a straight line connecting two diverging tangents, to find theradius of a reversed curve to close the angles. AI=RX tan. A; BI = RX tan. B. .. AB=RX (tan. § A + tan. § B). A REVERSED CURVE BETWEEN FIXED POINTS. Given the angles N and K, and the length of the straight line E F connecting two divergent tangents, to find the radius of a reversed curve from E to F, connecting the tangents. 1. Denote the angle EIC or DIF by I; the angle CEI, complement of N, by n; and the angle DFI, complement of K, by k. Then, in triangle E CI, |