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perpendicular KH, which make equal to c ;and through H draw HA, making the angleH = 56°"28', and meeting KC in A ; makeAB-KH and AC = AH, then produce HAto meet CE, drawn parallel to KH in E,thenshall B be the vertex, and E another pointin the catenary curve BE. For by com-paring the triangle HAK with with ABE),drawn by the other method, we have HK= AB, the angle H=DAB, each=56 28',the angle K=ABD, and consequently thetriangles are equal in all respects; henceAH—AD=tf, and AK=BD== 2 , the wholeKC is therefore=£-f-#.

CASE II.

When x is given, to find the rest.

Make BC = x, and through the point Cdraw CD, making the angle BCD =

-—-=—-—=61 46 , meeting the

perpendicular BD in D; through D drawDA, making the angle CDA equal to theangle BCD, and meeting CB produced inA. Then through A draw AE perpendi-cular to AD, meeting CE, drawn parallel to