350 



Mr. Adams's Solution of a Problem. 



[MAYy 



z m = cos. m A + \/ cos. 2 m A — 1 = cos. m A + */ — 1 



sin. »? A. 



Therefore 



(cos. A + V — 1 sin. A) m = cos. rri A + \/ — 1 sin. wj A. 



Hence the following equations 

 (cos. A + */ — 1 sin. A)" 1 = cos.m A + V — 1 sin. mA 

 (cos. A — */ — lsin. A) m = cos. mA- V-l sin. mA. 



By addition and subtraction, 



_ . (cos. A + sin. A ■%/ — l) m + (cos. A — sin. A -/ — l) m 

 COS. 7W A = ' ' 2 ' 



(cos. A + sin. A a/ — l) m — (cos. A — sin. A V — l) m 



Sin. m A — 



2 v^H 



Article VII. 



Solution of a Problem. By James Adams > Esq. 



(To Dr. Thomson.) 



SIR, Stonekouse, Jan, 15, 1818. 



At page 259, vol. iii. Dr. Hutton's Course of Mathematics, 

 we have the following problem : 



" To determine the thickness of the wall at the top when the 

 face is not perpendicular, but inclines as the front of a fortifica- 

 tion wall usually is." 



The solution to which is there given by a quadratic equation. 



If, in your opinion, the following solution by a simple equation 

 be an improvement, your inserting it in the Annals of Philosophy 

 will much oblige Your humble servant, 



James Adams. 



Let A B M E be a vertical section of a 

 bank of earth, and the triangular part 

 ABE that which is supported by means 

 of a wall, the vertical section of which is 

 A E F G. Draw G D perpendicular to 

 F E, and conceive two weights, W and 

 w, to be suspended from the centres of 

 gravity of the rectangle A D, and the 



Ww 



triangle G D F, and to be proportional to their areas respectively. 

 Put A E = a, D F = -, and EF = i= breadth at the base. 



ThenEF -DF = ED = i - - = c -^-? .-. aED= ^^ 



c • " 2 e 



