316 OF THE POSITIONS OF EQUILIBRIUM. 



Therefore, by substituting the respective values of a, b and c, as 

 given in the question, and the value of d as computed above, we shall 

 have the following values of cos.0 and cos.^'. 



Thus, for the absolute numerical value of cos.0, it is 



4(28* + 23.99*) 18 2 _ A 



COS '* = -- 8X28X23.99 ~ ' 95166 ' 



and for the corresponding value of cos. 0', we have 



4(23 2 -4-23.99 2 18 2 

 COS - = 8X23X23.99 =- 92747 ' 



Having ascertained the numerical values of cos. and cos.^>', let the 

 respective quantities be substituted in equation (239), and it becomes 



28X23X435X0.92747 



4 0. 



1000 

 28 2 X23 2 X435 2 



1000 2 

 from which, by computing the several terms, we get 



* 4 45.66* 3 -f 12466* = 78478.36. 



The root of this equation will be most easily discovered by approxi- 

 mation, and for this purpose, we shall adopt the method of trial and 

 error, which Dr. Hutton has so successfully applied to the resolution 

 of every form and order of equations, however complicated may be 

 their arrangement. 



By a few simple trials, indeed it is almost self evident, that the 

 value of x will be found between 15 and 16; consequently, by sub- 

 stitution we obtain 



15 4 45.66X15 8 4- 12466X15 78478.36 = 15113.64 too little. 

 16 4 45.66X16 3 + 12466X16 78478.36 = 509.72 too great. 

 Here it is manifest that the errors are of different affections, the one 

 being in defect and the other in excess ; hence we have 



15113.64 + 509.72 : 16 15 :: 509.72 : 0.032; 

 consequently, for the first approximation, we get 



x = 16 0.032 = 15.97 very nearly. 



Supposing therefore, that x lies between 15.9 and 16 ; by repeating 

 the process, we shall have 



16 4 45.66X 16 3 -f 12466X 16 78478.36 = 509.72 too great. 

 15.94 __ 45.66 X15.9 3 4- 12466 X 15.9 78478.36 zz!05.393too little. 

 Here again, the e/rors are of different affections, the one being in 

 excess and the other in defect; consequently, we have 



